Electrochemistry deals with interdependence of chemical energies and electrical energies. Such interactions are observed at the boundaries between electrolytes and electrodes dipping in them. Electrochemistry holds a central position in the study of chemistry for a number of reasons:- i)   It acts as a bridge between thermodynamics and the rest of the chemistry. ii)  It is of enormous commercial importance because of the costly destruction caused by corrosion and the exciting possibilities of fuel cells which generate electricity directly from fuel. From these reasons, we can understand the importance of this topic. Here, we are going to see about various types of cells, emf calculations and how to find out thermodynamic quantities from electrochemical measurements.

 

1. IIT–JEE Syllabus

Faraday’s laws of electrolysis; Electrochemical cells and cell reactions; Standard electrode potential and electrochemical series, emf of a galvanic cell; Nernst  equation and significance of DG and DG°.

2. Introduction

Electrochemistry deals with the inter-conversion of electrical energy and chemical energy. A flow of electricity through a substance may produce a chemical change (redox reaction) and also a chemical change (redox reaction) may cause a flow of electricity through some external circuit. The former involves the study of electrolysis and conductance while the latter, the measurement of electromotive force.

3. Electrolysis

The phenomenon of electrolysis involves the breaking of electrolytes when electric current is passed through it. The apparatus used to carryout electrolysis is known as electrolytic cell. The main characteristics of electrolytic cell are as follows:

		Cathode	Anode
a)	Sign	(–ve); as it is linked to the negative end of the external battery	(+ve); as it is linked to the positive end of the battery
b)	Direction of electron movement	Into the cell	Out of the cell
c)	Ions attacked within the cell	Cations	Anions
d)	Half cell reaction	Reduction	Oxidation

The electrolysis of molten salts produces substances which are characteristic of the salt. When aqueous salt solutions are electrolysed, water may be also be involved in the electrode reaction rather than the ion derived from the solute only

The reaction involved during electrolysis of water are follows:

2H₂O ¾® O₂  + 4H⁺ + 4e (oxidation)

2H₂O + 2e ¾® H₂ + 2OH^– (reduction)

The following figure shows the layout of an electrolytic cell used for commercial production of Mg metal from molten MgCl₂.

As in a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode, electrons travel through the external wire from anode to cathode, cations move through the electrolyte toward the cathode, and anions move toward the anode. Unlike the process in a galvanic cell, source drives electrons through the wire in a pre-determined direction, forcing oxidation to occur at one electrode and reduction at the other:

Anode reaction : 2Cl^–(l) ¾¾® Cl₂(g) + 2e^–

Cathode reaction:  Mg²⁺ (l) + 2e^–  ¾¾® Mg(l)

A rechargeable battery functions as galvanic cell when it is doing work and as an electrolytic cell when it is being recharged

Preferential Discharge Theory: If more than one type of ion is attracted towards a particular electrode, then the ion discharged is the one which requires the least energy.

The decreasing order of the discharge potential or the increasing order of deposition for some of the cations and anions is as follows.

Cations:     K⁺, Na⁺, Ca²⁺, Mg²⁺, Al³⁺, Zn²⁺, H⁺, Cu²⁺, Ag⁺, Au³⁺

Anions:      SO₄^2–, NO₃^–, OH^–, Cl^–, Br^–, I^–

4. Faraday’s Laws of Electrolysis

The calculations to find out the amount of product formed in electrolysis are based on the following Faraday’s Laws:

First law: The number of moles of product formed by an electric current is stoichiometrically equivalent to the number of moles of electrons supplied.

Once we know the number of moles of product formed, we can calculate the masses of the products, or if they are gases, their volumes.

For examples, Cu is refined electrolytically by using an impure form of copper metal (blister copper) as the anode in an electrolytic cell. The current supply causes the oxidation of blister copper to copper (II) ions.

Cu (blister) ¾¾®   Cu²⁺ + 2e^–

These ions are then reduced at the cathode:

Cu²⁺ (aq) + 2e^– ¾¾®   Cu(s)

From the stoichiometry of these chemical equations, we know that 2 mole  electrons give one mol of Cu. Therefore, if 4.0 mole electrons is supplied, then the amount of copper produced is 2.0 mol. This law can also be stated in other way:

“The mass of a substance deposited or liberated at any electrode is directly proportional to the amount of charge passed” i.e., w a q  (where w is the mass of the substance deposited or liberated and q is the amount of charge passed). This proportionality can be made into an equality by, w = zq where z is the proportionality constant called the electrochemical equivalent. It is the mass of the substance in grams deposited or liberated by passing one coulomb of charge.

The quantity of electricity passed through the electrolytic cell is normally measured as the current and the time for which the current flows. The charge passed through an electrolytic cell is the product of the current and the time for which it is supplied. Electric current is measured in the SI unit (ampere), A, the rate of flow of charge in coulombs per second (1A = 1CS^–1). Therefore, the charge  supplied is,

Charge supplied = Current  (A) x time(s)

Next, we use the fact that faraday’s constant F, is the magnitude of the charge per mole of electrons.

Charge = moles ´ F

So. Moles of e^– =  =

So, by measuring the current and the time for which it flows, we can determine the moles of electrons supplied.

The amount of product in an electrolytic reaction is calculated from the stoichiometry of the half – reaction and  the current and time for which the current flows.

Second law:   This law states that “the mass of a substance deposited or liberated at any electrode on passing a certain amount of charge is directly proportional to its equivalent weight of the substance”.

That is w a E where w is the mass of the substance in grams while E is its chemical equivalent  i.e. weight in gms per equivalent. This law can be explained as follows.

Consider three reactions, such as:

Na⁺ + e^– ® Na

Cu²⁺ + 2e^–‑ ® Cu

Al³⁺ + 3e^– ® Al

Assume that these three reactions are occurring in three separate electrolytic cells connected in series. When x moles of electrons are passed through the three cells, the mass of Na, Cu and Al  deposited are 23x gms, 31.75x gms and 9x gms respectively. We can see that 23, 31.75 and 9 gm/eq are the chemical equivalent weights of the three elements.

\ w = moles of electrons ´ E

The charge possessed by 1 mole of electrons = 1.6 ´ 10^–19 ´ 6.023 x 10²³ » 96500 C

This charge is called as 1 Faraday.

If we pass one Faraday of charge, it means that we are passing one mole of electron and by passing 1 Faraday of charge 1gm equivalent weight of the substance will be deposited or liberated.

\  Þ W =  Þ W =

(n = n-factor)

By combining the first and second law, we get

Note:   It should be made clear that the cathode is the electrode in which reduction reaction(s) occurs while the anode is the electrode where the oxidation reaction(s) occurs. Do not relate the sign (positive or negative) of the electrode with the nature of the electrode.

Illustration 1:          Copper sulphate solution (250 ml) was electrolyzed using a platinum anode and copper cathode. A constant current of 2 mA was passed for 16 minute. It was found that after electrolysis, the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate ion in the solution to begin with.

Solution:              Q = F = 1.989 ´ 10^–5F

\ Equivalent of Cu²⁺ lost during electrolysis = 1.989 ´ 10^–5

or Mole of Cu²⁺ lost during electrolysis =

This value is 50% of the initial concentration of solution

\ Initial mole of CuSO₄ =  = 1.989 ´ 10^–5

\ Initial molarity of [CuSO₄] =

\ [CuSO₄]_initial= 7.95 ´ 10^–5 M

Illustration 2:          A current of 3 ampere was passed for 2 hours through a solution of CuSO₄. 3 gm of Cu²⁺ ions were discharged at cathode calculate the current efficiency.

Solution:              W_Cu =

Þ 3 =  Þ i = 1.266 ampere.

\ Current efficiency =  =  = 42.2%

Illustration 3:          Chromium metal can be plated out from an acidic solution containing CrO₃ according to following equation

                              CrO_3(aq) + 6H⁺ ¾® Cr_(s) + 3H₂O

                              Calculate:

1. i) How many grams of chromium will be plated out by 24000 coulomb?
2. ii) How long will it take to plat out 1.5 gm of Cr by using 12.5 ampere current.

Solution:              Equivalent weight of Cr =

1. i) 96500 C deposit = gm Cr

\ 24000 coulomb º ´ =  2.1554 gm

1.                  ii)   W_cr= Þ5 =

\ t =  sec

Exercise-1:       For how long a current of three amperes has to be passed through a solution of AgNO₃ to coat a metal surface of 80cm² area with 0.005mm
thick layers? Density of silver is 10.5g/cc and atomic weight of
Ag is 108 gm/mol.

Exercise 2:       H₂O₂ can be prepared by successive reactions.

                        2NH₄HSO₄ ® H₂ + (NH₄)₂S₂O₈

                        (NH₄)₂S₂O₈ + 2H₂O ® 2NH₄HSO₄ + H₂O₂

                  The first is an electrolytic reaction and the second a steam distillation. What current needs to be passed in the first reaction to produce enough intermediate to yield 100 gm of pure H₂O₂ per hour? Assume 50% current efficiency.

5. Electrode Potential

In an electrochemical cell, each electrode is placed in contact with the electrolytic solution. Thus at the surface of separation of the electrode and the electrolyte solution, there exists an electrostatic potential which is known  as electrode potential. Let us see in detail how the electrostatic potential arises?

All metallic element and hydrogen have a tendency to pass into solution in the form of positive ions. This property of the metal is known as the ‘solution pressure’, ‘electrolytic solution pressure’ or ‘solution tension’ of the metal and is constant at a given temperature. Due to the migration of the positive ions, the metallic electrode is left negatively charged and thus electrical double layer is set up at the electrode.

If a metallic electrode is dipped in a solution of one of its salts, the position becomes slightly  different. In such a case, the tendency of the ions is to be deposited on the electrode. This backward reaction is attributed to be osmotic pressure of ions in solution. Thus, the standard potential of a metal is equal to the difference between its solution pressure and the osmotic pressure of its ions. Hence there arises three possibilities:

a)   If the solution pressure is greater than the osmotic pressure, the tendency of the metal to lose ions predominates. A potential difference is therefore set up with the metal left with negative charge with respect to the solution. The net result will be that the positive ions will enter the liquid and leave the metal negatively charged with respect to solution. The formation of double layer prevents the further expulsion of ions from the metal, and thus there is rapidly established a state of equilibrium with a definite potential difference, termed as the electrode potential.
b)   If the solution pressure is less than the osmotic pressure of the metal in solution, it means that the ions have greater tendency to leave the solution and get deposited on the metal. So potential difference is established due to the charge separations as shown in the following diagram.
c)   When the solution pressure becomes equal to that of osmotic pressure, no relative charge is developed and hence no potential difference exists. Such a system is sometimes termed as null electrode as shown below:

Thus, the tendency of an electrode to lose or gain electrons when it is in contact with its own ions in solution is called electrode potential. Since the tendency to lose electrons means also the tendency to get oxidised, this tendency is called oxidation potential. Similarly, the tendency to gain electrons means the tendency to get reduced. Hence this tendency is called reduction potential. Needless to  mention that reduction potential is the reverse of oxidation potential.

So far, we have seen how the potential was developed. Now we are going to see how to determine the electrode potential. It is not possible to determine experimentally the potential of a single electrode. It is only the difference of potentials between two electrodes that we can measure by combining them to give a complete cell.

By arbitrarily fixing potential of one electrode as zero (just as the boiling point of water at atmospheric pressure has been arbitrarily fixed as 100 on temperature scale), it is possible to assign numerical values to potentials of the various other electrodes. Accordingly, the potential of a reversible hydrogen electrode in which the gas at one atmospheric pressure is bubbled through a solution of hydrogen ion of unit activity (or to be approximate, unit concentration has been fixed as zero. This electrode is known as Standard Hydrogen Electrode (SHE) and is represented as, Pt, H₂(1atm), H⁺(C = 1)

All other single electrode potentials are referred to as potentials on the hydrogen scale.

If it is required to find the electrode potential of, say zinc electrode dipping in a solution of zinc sulphate ((i.e.,) Zn, Zn²⁺ electrode), all that is needed is to combine it with the standard hydrogen electrode so as to have a complete cell represented as, .

The emf of the cell is determined potentiometrically, is then equal to the potential of the electrode (on the hydrogen scale) since the potential of the standard hydrogen electrode is taken as zero.

Since , E_cell = E_RP(cathode) – E_­RP(Anode) . In this case S.H.E. is undergoing oxidation. So, whatever reading we are getting in the potentiometer directly gives the potential of Zn/Zn²⁺ electrode.

In case, the electron accepting tendency of the metal electrode is more than that of a S.H.E., its standard reduction potential gets a positive sign. On the other hand, if the electron accepting tendency of the metal electrode is less than that of S.H.E, its standard electrode potential gets a negative sign. According to latest convention, all standard electrode potentials are taken as reduction potentials.

Thus, the electrode at which reduction occurs with respect to S.H.E has positive electrode potential and while the electrode at which oxidation occurs with respect to S.H.E has negative electrode potential.

6. Various Types of Half Cells

 

Sl. No.	Type	Example	Half Cell Reaction
1.	Metal – metal ion half cell	Zn | Zn2+ (aq)	Zn(s) ® Zn2+(aq) + 2e
2.	Gas  ion half cell	Pt(H2)|H+(aq), Cl–|Pt(Cl2)	1/2H2 ® H+(aq) + e Cl ® 1/2Cl2 + e
3.	Metal insoluble salt anion half cell	Ag, AgCl | Cl– (aq), Hg, Hg2Cl2 | Cl–(aq) (calomel electrode)	Ag(s) + Cl–(aq) ® AgCl(s) + e 2Hg(l) + 2Cl–(aq) ® Hg2Cl2(s) + 2e
4.	Oxidation / Reduction Half Cell	Pt | Fe2+/ Fe3	Fe2+ ® Fe3++ e
5.	Quinhydrone Half Cell	Pt | Quinhydron | H+(aq)
6.	Metal metal oxide hydroxide half cell	Hg, HgO | OH–(aq)	Hg(l) + 2OH–(aq) ® HgO(s) + H2O(l) + 2e

7. Electrochemical Series

Just now, we have seen that how to find a single electrode potential. In similar manner, the E° value of other electrodes can also be measured. When the elements are arranged in the order of their standard reduction potential , an important series known as electro – chemical or electro motive series is obtained.  is the quantitative measure of readiness of the element to lose electrons. The series is arranged in the order of increasing strength as oxidant and decreasing strength of reducing agent. Those elements having  negative can easily liberate hydrogen from H⁺ ions in aqueous medium and those with positive value can oxidise H₂ to H⁺ ion. Thus Li occupying the top position in the series is the strongest reducing agent while fluorine is the strongest oxidising agent as it occupies the bottom of the series.

Table: Standard Reduction Potentials At 25° C

	E°, V		E°, V
Li+ + e– Li	–3.03	Cu2+ + 2e– Cu	0.34
K+ + e– K	– 2.93	+ e–	0.355
Cs+ + e–‑ Cs	–2.92	+ e–	0.370
Ba2+ + 2e– Ba	– 2.90	Cu+ + e– —® Cu	0.52
Ca2+ + 2e– Ca	– 2.87	I2 + 2e–‑ 2I–	0.535
Na+ + e– Na	– 2.713	Cu2+ + Cl– + e– CuCl	0.566
Ce3+ + 3e– Ce	– 2.48	O2 + 2H+ + 2e– H2O2	0.69
Mg2+ + 2e– Mg	– 2.37	Fe3+ + e– Fe2+	0.771
H2 + 2e– 2H–	– 2.25	+ 2e– 2Hg	0.79
Al3+ + 3e– Al	– 1.66	Ag+ + e– Ag	0.799
Mn2+ + 2e– Mn	– 1.18	Cu2+ + I–  + e– CuI	0.85
2H2O + 2e– H2 + 2OH–	– 0.828	2Hg2+ + 2e–	0.92
Zn2+ + 2e– Zn	– 0.7628	Pd2+ + 2e– Pd	0.92
Cr3+ + 3e– Cr	– 0.74	OCl– + H2O + 2e– Cl– + 2OH–	0.94
Fe2+ + 2e– Fe	– 0.44	+ 3e– Au + 4Cl–	1.00
Cd2+ + 2e– Cd	– 0.403	Br2 + 2e– 2Br–	1.09
Cu2O + H2O + 2e– 2Cu + 2OH–	– 0.34	+ 12H+ + 10e–  I2 + 6H2O	1.20
Tl+ + e– Tl	– 0.336	O2 + 4H+ + 4e–  2H­2O	1.229
PbSO4 + 2e– Pb +	– 0.31	MnO2 + 4H+ + 2e–  Mn2+ + 2H2O	1.23
Co2+ + 2e– Co	– 0.28	Tl3+ + 2e– Tl+	1.26
Ni2+ + 2e– Ni	– 0.25	Cr2O72– + 14H+ + 6e– 2Cr+3 + 7H2O	1.33
CuI + e– Cu + I–	– 0.17	Cl2 + 2e– 2Cl–	1.36
AgI + e– Ag + I–	– 0.151	Au3+ + 3e– Au	1.50
2CuO + H2O + 2e– ¾® Cu2O + 2OH–	– 0.15	MnO4– + 8H+ + 5e–Mn+2 + 4H2O	1.51
Sn2+ + 2e– Sn	– 0.14	Ce4+ + e– Ce3+	1.70
Pb2+ + 2e– Pb	– 0.126	PbO2 + 4H+ + + 2e– PbSO4 + 2H2O	1.70
2H+ + 2e– H2	0.000	MnO4– + 4H+ + 3e– MnO2 + 2H2O	1.70
+e–¾®Ag+	0.017	H2O2 + 2H+ + 2e– 2H2O	1.77
Sn4+ + 2e– Sn2+	0.13	Co3+ + e– ® Co+2	1.82
Cu2+ + e– Cu+	0.15	S2O82– + 2e–  2SO42–	2.0
+ 2e– Pd + 4I–	0.18	F2 + 2e– 2F–	2.87

A close Scrutiny of the numerical values of the standard electrode potentials  in the electro – chemical series, reveals the following interesting points.

1. Lower values of the standard reduction potential indicate greater tendency of the metal to lose electrons (i..e,) higher metallic or electropositive character.
2. A negative value of reduction potential indicates that the metal has a greater tendency  to lose electron (i.e.) to be oxidised into its own ion than the tendency of its ion to be reduced by gaining electron.
3. A positive reduction potential value indicates that the tendency of the metal ion to be reduced by gaining electron, is more than tendency of the metal to be oxidised by losing electron.
4. An element having higher negative potential will replace all others below it with lower negative potential values from their aqueous salt solutions. Thus Zn will replace Cu²⁺ from an  aqueous solution of CuSO₄ (i.e.,) Zn will reduce Cu²⁺ and itself will be oxidised.

= –0.76V

= 0.34V

Hence the reaction is Zn + Cu²⁺ ¾¾® Cu + Zn²⁺ is spontaneous and not the reverse one.

5. An oxidising agent with higher value can oxidise a system with lower  value. Thus the numerical magnitude of  value is a measure of the strength of an oxidant. All oxidising agents are characterised by higher value of .

E.g. since  = 2.87  volt, F₂ is one of the strongest oxidising agents and no common oxidants are known which can oxidise fluoride ion into fluorine gas.

With this idea let us move on to galvanic cells.

8. Galvanic Cells

Galvanic cell is the device in which chemical energy is converted into electrical energy

Main characteristics of galvanic cells are:

	Cathode	Anode
Sign	Positive, as the electrons are consumed at this electrode	Negative, as the electrons are released at this electrode
Direction of electron movement	Into the cell	Out of cell
Half cell reaction	Reduction	Oxidation

A most remarkable feature of oxidation – reduction reactions is that they can be carried out with the reactants separated in space and linked only by an electrical connection. That is to say, chemical energy is converted to electrical energy. Consider figure – 2,
a representation of a galvanic cell which involves the reaction between metallic
zinc and cupric ion:

Fig.2

Zn(s) + Cu²⁺ (aq) ® Cu(s) + Zn²⁺(aq)

The cell consists of two beakers, one of which contains a solution of Cu²⁺ and a copper rod, the other a Zn²⁺ solution and a zinc rod. A connection is made between the two solutions by means of a “salt bridge”, a tube containing a solution of an electrolyte, generally NH₄NO₃ or KCl. Flow of the solution from the salt bridge is prevented either by plugging the ends of the bridge with glass wool, or by using a salt dissolved in a gelatinous material as the bridge electrolyte. When the two metallic rods are connected through an ammeter, a deflection is observed in  ammeter which is an evidence that a chemical reaction is occurring. The zinc rod starts to dissolve, and copper is deposited on the copper rod.  The solution of Zn²⁺ becomes more concentrated, and the solution of Cu²⁺ becomes more dilute. The ammeter indicates that electrons are flowing from the Zinc rod to the copper rod. This activity is continuous as long as the electrical connection and the salt bridge are maintained, and visible amounts of reactants remain.

Now let us analyze what happens in each beaker more carefully. We note that electrons flow from the Zinc rod through the external circuit, and that Zinc ions are produced as the Zinc rod dissolves. We can summarize these observations by writing,
Zn ® Zn²⁺ + 2e^– (at the zinc rod). Also, we observe that electrons flow to the copper rod as cupric ions leave the solution and metallic copper is deposited. We can represent these occurrences by 2e^– + Cu²⁺ (aq) ® Cu (at the copper rod).

In addition, we must examine the purpose of the salt bridge. Since Zinc ions are produced as electrons leave the zinc electrode, we have a process which tends to produce a net positive charge in the left beaker. The purpose of the salt bridge is to prevent any net charge accumulation in either beaker negative ions, diffuse through the salt bridge, and enter the left beaker. At the same time, there can be a diffusion of positive ions from left to right. If this diffusional exchange of ions did not occur, the net charge accumulating in the beakers would immediately stop the electron flow through the external circuit, and the oxidation reduction reaction would stop. Thus, while the salt bridge does not participate chemically in the cell reaction, it is necessary if the cell is to operate.

9. IUPAC Cell Representation

The galvanic cell mentioned above is represented in a short IUPAC cell notation as follows:

It is important to note that: `

1. First of all the anode (electrode of the anode half cell) is written. In the above case, it is Zn.
2. After the anode, the electrolyte of the anode should be written with concentration. In this case it is ZnSO₄ with concentration as C₁.
3. A slash (|) is put in between the Zn rod and the electrolyte. This slash denotes a surface barrier between the two as they exist in different phases.
4. Then we indicate the presence of a salt bridge by a double slash (||).
5. Now, we write the electrolyte of the cathode half-cell which is CuSO₄ with its concentration which is C₂ .
6. Finally we write the cathode electrode of the cathode half – cell .
7. A slash (|) between the electrolyte and the electrode in the cathode half – cell.
8. In case of a gas, the gas to be indicated after the electrode in case of anode and before the electrode in case of cathode. Example: Pt, H₂/H⁺ or H⁺|H₂,
9. The Nernst Equation

Electrode potential depends upon the concentration of the species in solution, so also emf of a galvanic cell is controlled by the concentration of the ions. Nernst equation gives a relation between emf of a galvanic cell and the concentration of ions in solution. The equation is deduced as follows:

Let us consider a reversible reaction,

a A + bB  gC + dD

Equilibrium constant K =

We know from thermodynamics that decrease of free energy (DG) measures the tendency of a chemical reaction to reach the equilibrium state.

If G₁ = total free energy of the reactants and

G₂ = total free energy of the products, the

DG = G₂ – G₁, If  G₁ > G₂, then DG = negative

A negative DG of a reaction implies its spontaneity. DG is a measure of the ability of a system to do useful work. Since we can draw electricity from a galvanic cell, it implies that a spontaneous reaction is going on inside the cell and hence DG of such reaction must be negative. Electrical work is the result of the decrease of free energy of the cell reaction.

We know that, the electrical work obtained from any spontaneous reaction supplying nF coulomb of electricity at a potential E is given by,

Net electrical work = nFE

Again net electrical work = –DG (decrease of free energy)

\DG = – nFE

Applying standard state,

DG° = –nFE°

Again from thermodynamics, DG is related to K by the following equation

DG = DG° + RT ln K

Substituting the value of DG,

-nFE = – nFE° + RT ln K

E = E°  ln K

(or) E = E° –  ln                               … (i)

Putting R = 8.314J/k^–/mole^–1 T = 298K and F = 96500 Coulomb.

We get, E = E° –                     …(ii)

This is the fundamental form of Nernst equation. This can be applied in both oxidation and reduction process as well as in galvanic cell.

1. i) If we represent an oxidation reaction as follows.

Reductant  oxidant + ne

Here product is OX and reactant is Red.

\ E = E° –  log                          …(iii)

Equation (iii) is one from of Nernst equation. When [OX] = [Red], then E = E°. E° is called standard oxidation potential.

1. ii) if we represent a reduction reaction in the following way

oxidant + ne^–  reductant

then by applying Nernst equation,

E = E° –  log

(or) E = E° +  log  ———- (iv)

Equation (iv) is another form of Nernst equation.

Here also, when [OX] = [Red], E = E°. In this case, E° is called standard reduction potential.

iii)   Equation (iii) and (iv) are both infact, two forms of Nernst equation, but must be remembered that E_OX = – E_red , like .

11. Standard Electrode Potential (E°)

In answering problems involving E°, the following few points should be carefully remembered.

We know that, intensive properties are those whose values do not depend upon the amount of material chosen. Colour, physical state, temperature, density etc., are the examples of intensive properties. Like these E° is also an intensive property. Hence in whatever way we write, its value remains same (i.e.,)

Cl₂ + 2e^–2Cl^–       E° = 1.36 volt

Cl₂ + e^–  Cl^–     E° = 1.36 volt

\

But DG is an extensive property like heat content and mass, because they depend on the amount of material.

If the  combination of two half reactions yields a third half reaction, DG of such a reaction is additive, but the potential E° is not additive. E° of such cell indicating such third half reaction can be evaluated from the DG value e.g.,

Illustration 4:          Given that  = 1.28V and  = 1.51 V. What is the E° for Mn²⁺/Mn⁴⁺ couple

Solution:              We know that DG = – nFE

The half cell reactions are,

Mn²⁺ + 2H₂O   MnO₂ + 4H⁺ + 2e            … (i)

Mn²⁺  Mn³⁺ + e                                        …(ii)

On subtracting,

Mn³⁺ + 2H₂O  MnO₂ + 4H⁺ + e               …(iii)

Now, we are to find

For the couple Mn²⁺/ Mn⁴⁺ for equation (i)

DG = – nFE (or)

DG° = –2 (–1.28)F

= 2.56F

For equation (ii) (i.e.,) for Mn²⁺/Mn³⁺  couple,

DG° = –1(–1.51)F = 1.51F

On substracting DG° = 2.56 F – 1.51 F = +1.05 F

Now in the half reaction Mn³⁺ – e^–  Mn⁴⁺ , n = 1 ( n is the number of electrons transferred)

\ DG° = – nFE

= –1 ´ F ´ E° (or)

1.05 F = –1 ´ F ´ E°

(or) E° = – 1.05 Volt

Thus we see that,  is not (–1.28 – 1.51)

= –2.79 V

Illustration 5:          The standard reduction potential of Cu²⁺/Cu and Ag⁺/Ag electrodes are 0.337V and 0.799 V respectively. Construct a galvanic cell using these electrodes so that its E⁰_cell is +ve. For what [Ag⁺] will the emf of cell at 25°C be zero if [Cu²⁺] is 0.1 M

Solution:              For E⁰ to be +ve, the cell diagram should be.

Cu | CuSO₄ (aq) || AgNO₃(aq) | Ag

Cell reaction Cu + 2Ag⁺ ¾® Cu²⁺+ 2Ag

E_cell = E⁰_cell –

Q E_cell = 0 and [Cu²⁺] = 0.01 M

\ o =  – log

Þ 0 = (0.799 – 0.337) –

\ [Ag⁺] = 1.477 ´ 10^–9 mol/lit.

12. Spontaneity of Cell reaction

We know from Thermodynamics, that the sign of change of Gibbs free energy (DG) is an indication of spontaneity of a reaction.

At equilibrium DG = 0, while if DG = negative, the reaction is spontaneous and for non – spontaneous  reaction DG = positive.

Again DG = -nFE and DG° = -nFE° or  also implies the spontaneity of a reaction. If E_cell or  is positive then DG° is negative and the reaction is spontaneous. For non – spontaneous reaction, E_cell or  is negative

The spontaneity of a cell reaction depends upon the sign of the cell potential. Hence the following rules are to be noted with regard to cell reaction, emf and spontaneity of the cell reaction.

1. i) Any cell reaction is the sum of the half – cell reactions. At the negative electrode (anode), reaction is oxidation while at the positive electrode (cathode) the reaction is reduction.

[Note: In electrochemical cell, the negatively charged electrode is anode and in electrolytic cell, anode is positively charged].

1. ii) The total cell emf is the algebraic sum of the single electrode potential, provided each potential is given with proper sign corresponding to the actual reaction in the half cell. The emf of any cell can however be found out by applying any one of the following three formulae.
2. a) E_cell = (Reduction potential of the right hand side electrode) – (Reduction potential of left hand side electrode).
3. b) E_cell = (oxidation potential of LHE) — (oxidation potential of RHE)
4. c) E_cell = (oxidation potential of LHE) + (Reduction potential of RHE)

iii)   If a cell is written as in (i)  and when the emf is positive, electron will flow from left to right in the external circuit and the cell reaction will be spontaneous. We can get current from such cell.

1. iv) If wrong assumption is made with regard to polarity of the electrode, no current will be drawn from such cell as the reaction is non-spontaneous and a negative emf will be produced. In that case, the two half cells forming the galvanic cell should be interchanged (i.e.,) the reaction is reversed and sign of the emf is changed without changing its magnitude. You will come across this in problems like, in some cases you may be asked to find the correct polarity of the cell. In that case, to set the correct polarity of the cell, the emf of the cell should be positive. If not, then Left hand half cell should be made as right and right hand half cell should be made as left.

Feasibility of cell reaction is shown in the following table:

Nature of cell reaction	DG°	Equilibrium constant
Spontaneous	(–)	>1	(+)
At equilibrium	0	1	0
Non – spontaneous	(+)	<1	(–)

13. Application of EMF-Measurement
14. Determination of Equilibrium constant:

From Nernst equation E_cell = E⁰_cell – logQ (Q = Reaction Quotient)

At equilibrium E_cell = 0, Q = K_eq (equilibrium constant)

\ E⁰_cell = logK_eq Þ E⁰_cell = log K_eq

\ log K_eq =

2. Determination of solubility product: Solubility product of sparingly soluble salts can be evaluated from EMF measurement. Suppose we want to determine K_sp of AgX(s). Its saturated solution is placed in Ag electrode. Reference half cell is Ag electrode in Ag⁺ion.

Ag | Ag⁺, saturated AgX || Ag⁺(C)| Ag

Anode        :     Ag ¾® Ag⁺ (C) (in saturated AgX) + e    E⁰_OX = – 0.80V

Cathode    :     Ag⁺ (in given solution C) + e ¾® Ag       E⁰_cell = + 0.80V

––––––––––––––––––––––––––––––––––––––––––––––

Cell reaction    Ag⁺(C) ¾® Ag⁺ (unknown)                     E⁰_cell = 0.00V

\ E_cell = E⁰_cell –

Hence Þ E_cell =                         …(1)

At equilibrium E_cell = 0

From equation (Ag⁺) can be calculated

= [Ag⁺] [Cl^–] = [Ag⁺]²

3. Ionisation constant of weak acid or weak base.

\ We can determine [H⁺] using suitable reference electrode.

Pt(H₂) | H⁺, HA(C₁) || Cu²⁺(C₂) | Cu

Its E_cell is measured

 

 

[H⁺] =  , (Oswald dilution law)

K_a can be calculated

4. Determination of pH of a solution: We can calculate the pH of a unknown solution by using suitable reference half cell.

Pt(H₂) (1 atm) | H⁺ (pH = ?) || Cl^– (1M) | Hg₂Cl₂, Hg (calomel electrode)

Cell reaction:

At anode    H₂(g) ¾® 2H⁺_(aq) + 2e                       E⁰ = 0

At cathode Hg₂Cl_2(S) + 2e ¾® 2Hg_(l) + 2Cl^–        E⁰_red = 0.2676

––––––––––––––––––––––––––––––––––––––––

H₂(g) + Hg₂Cl₂(s) ¾® 2H⁺_(aq)+ 2Hg(l) + 2Cl^–_(aq)      E⁰_cell = 0.2675V

Q = [H⁺]² [Cl^–1]² = [H⁺]² [Cl^–1]² = 1 M

E_cell = E⁰_cell – log[H⁺]²

Þ E_cell = E⁰_cell  + 0.0591 pH

\ pH =

5. Determination of Thermodynamic data:

DG = DH + T  (Gibbs – Helmhotlz Equation)

Þ – nFE_cell = DH – nFT (DG = – nFE)

DH = – nFE_cell + nFTÞ DH = – nF

= (Temperature coefficient of the emf of the cell)

Also DG = DH – TDS

\ DS = – = –

\ DS =

Illustration 6:         The emf of cells Zn | ZnSO₄ || CuSO₄ } Cu at 25°C is 0.03 V per degree. Calculate the heat of reaction for the change taking place inside the cell.

Solution:              According to Gibbs Helmhotz equation, heat of reaction (DH) can be given as

DH = nF

T = 273 + 25 = 298 K, n = 2, F = 96500 C, E = +0.03V

= 1.4 ´ 10^–4V per degree

DH = 2 ´ 96500 (298 ´ ( – 1.4 ´ 10^–4) – 0.03)

= – 13842 Joule = – 13.842 kJ

14. Concentration Cell

A concentration cell is one which is made from two half cells of same material differing in concentration.

1. Concentration cell in which electrode is reversible with respect to cation.

Zn | Zn²⁺ (C­₁) || Zn²⁺(C₂) | Zn

Cell reaction

At anode:               Zn ¾® Zn²⁺(C₁) + 2e            E⁰ = 0.76V

At cathode:            Zn²⁺(C₂) ¾® Zn + 2e            E⁰ = – 0.76V

–––––––––––––––––––––––––––––––––––––

Net cell reaction    Zn²⁺(C₂) ¾® Zn²⁺(C₁)           E⁰_Cell = 0.00V

E_cell = E⁰_cell  log

\ E_cell =  log

In general E_cell =

For spontaneous cell reaction C_2(RHS) > C_1(LHS)

                Similarly for Cell,  | HCl(C₁) || HCl(C₂) |

E_cell = 0.0591

2. Concentration cell in which electrode is reversible with respect to anion.

Pt(Cl_­2) | Cl^–(C₁) || Cl^–(C₂) |  Pt(Cl₂) 1 atm

E⁰_cell = log

For spontaneous cell reaction C₁ > C₂

3. Concentration cell having electrodes of different concentration dipped into same electrolyte.

Pt(H₂) | HCl (1 M) | Pt(H₂)

At P₁                         at P₂

Cell reaction

At anode    H₂ (P₁) ¾® H⁺ (1 M)  + e

H⁺(1 M) + e ¾® H₂(P₂)                 E⁰ = 0.00

–––––––––––––––––––––––––

H₂(P₁) ¾® H₂(P₂)                             E⁰ = 0.00

E_cell= E⁰_cell –

\ E_cell = 0.0591 log

For spontaneous cell reaction P₁ > P₂

For concentration cells E° = 0.

15. Condition for Disproportionation

If intermediate oxidation state is stronger oxidising agent then next higher oxidation state, than that eintermediate oxidation state will undergo disproportion.

Illustration 7:          The reduction potential for Cu in acid solution is

                             

                              Calculate X. Does Cu⁺ disproportionate in solution?

Solution:              Cu²⁺ + e ¾® Cu²⁺                              E⁰₁ = 0.15V              DG₁⁰                 …(1)

Cu⁺ + e  ¾® Cu                          E₂⁰ = 0.5V                DG₂⁰                 …(2)

Cu²⁺ + 2e ¾® Cu                        E₃⁰ = ?                     DG₃⁰                 …(3)

DG₃⁰ = DG₁⁰ + DG₂⁰ Þ 2 ´ E₃⁰  ´ F = – 1 ´ 0.15 ´ F + (1 ´ 0.5 ´ F)

Þ E₃⁰ =  = 0.325 volt

\ X = 0.3225 V

For disproportionation

Cu⁺ ¾® Cu⁺ + e                          E⁰_OP = – 0.15 V

Cu⁺ + e ¾® Cu                           E⁰_RP = 0.50 V

––––––––––––––––

2Cu⁺ ¾®Cu²⁺ + Cu                     E⁰ = 0.15 + 0.5

\ E⁰ +  = – 0.15 + 0.5

= 0.35V

16. Solution to Exercise

Exercise 1:     Assuming that the coating has to be done on only side of the surface, volume of Ag required for coating =  =  0.04cc

Mass of silver = 10.5 ´ 0.04 = 0.42gm

Moles of silver =  = 0.00389

Moles of electrons = 0.00389 (since Ag⁺ + e^– ® Ag)

Charge passed = 0.00389 ´ 96500 = 375.385 coulombs

\ Time =  = 125.1212 sec.

Exercise2:      100g of pure H₂O₂ will be  moles of H₂O₂

                        Þ moles of (NH₄)₂ S₂O₈ that would have reacted =

                        Þ moles of (NH₄)₂ S₂O₈ that needs to be formed by the first reaction

=

Þ moles of NH₄HSO₄ that needs to react =

When 1 mole of NH₄HSO₄ converts to (NH₄)₂S₂O₈, 2 moles of electrons would be liberated. This is because in NH₄HSO₄ all the oxygen atoms have an oxidation state of –2. In (NH₄)₂S₂O₈, two oxygen atoms are linked to each other (peroxide linkage) because of which their oxidation state becomes –1. Therefore two oxygen atoms have got oxidized from –2 state to –1 state. So when 2 moles of NH₄HSO₄ reacts, 2 moles of electrons are liberated.

\moles of electrons given by  of NH₄HSO₄  = moles

Þ charge = coulombs

Þ current with 50% efficiency =  = 315.36 A

 

 

 

 

 

 

 

 

17. Solved Problems

 

17.1     Subjective

Problem 1:       Calculate the quantity of electricity that would be required to reduce 12.3 gm of nitrobenzene to aniline, if current efficiency is 50%. If potential drop across the cell is 3.0 volt how much energy will be consumed?

Solution:         + 6H⁺ + 6e ¾®  + 2H₂O

Change in oxidation no per mole = 3 – (–3) = 6

\Eq. wt. of nitrobenzene =

Þ 12.3 =  (\ Current efficiency = 50%)

i_ot (Q) = 115800 coulomb

Consumption of energy = Q ´ N =  = 347.4 kJ

Problem 2:       An aqueous solution of NaCl on electrolysis gives H_2(g), Cl_2(g) and NaOH according to reaction

                        2Cl^– + 2H₂O ¾® 2OH^– + H_2(g) + Cl_2(g)

                        A direct current of 25 ampere with current efficiency of 62% passed through 20 litre of NaCl solution (20% by weight)

1. a) Write down cell reaction
2. b) How long will take to produce 1 kg of Cl₂?
3. c) what will be the molarity of solution with respect to OH^–? Assume no loss in volume due to evaporation.

Solution:        a)   At anode          2Cl^– ¾® Cl₂ + 2e

At cathode       2H₂O + 2e ¾® H₂ + 2OH^–

1. b) Þ 10³ =

\ t = 48.71 hour                                              = 35.5

i = 25 ´

28. c) of OH^– formed = Eq. of Cl₂ =  Eq. of Cl₂ =  = 28.17

\ Mol. of OH^– formed = 28.17

\ [OH^–] = =  = 1.408 mol/lit

Problem 3:       A silver electrode is immersed in saturated Ag₂SO₄(aq). The potential difference between the silver and standard hydrogen electrode is found to be 0.711 V. Determine K_sp(Ag­₂SO­₄) Given  = 0.799V

Solution:        The given cell

Pt(H₂) | H⁺(1M) || Ag₂SO₄(aq) | Ag

Cell reaction

At anode                H₂ ¾® 2H⁺ + 2e                    E⁰ = 0.00V

At cathode             2Ag⁺ + 2e ¾® 2Ag(s)             = 0.799V

––––––––––––––––––––––––––––––––––––––

H₂ + 2Ag⁺ ¾® 2H⁺+ 2Ag_(s)   E⁰ = 0.799V

E_cell = E⁰_cell –

Þ 0.711 = 0.799 – 0.059´ log                   [H⁺] = 1 M

\ [Ag⁺] = 3.2 ´ 10^–2

K_sp(Ag₂SO₄) = (Ag⁺)²(SO₄^2–) = (3.2 ´ 10^–2)²  = 1.6 ´ 10^–5

Problem 4:       What is the standard electrode potential for the electrode MnO₄^–/MnO₂ in solution. Given that

                                                = 1.23 V

Solution:        MnO₄^– + 8H⁺ + 5e ¾® Mn²⁺ + 4H₂O                 E₁⁰ = 1.51 V    …(1)

DG₁⁰ = – 5 ´ 1.51 ´ F = – 7.55F

MnO₂ + 4H⁺ + 2e ¾® Mn²⁺ + 2H₂O                              E₂⁰ = 1.23 V    …(2)

DG₂⁰ = –2 ´ 1.23 ´ F = – 2.46 F

Substracting equation (2) from equation (1)

MnO₄^– + 4H⁺ + 3e ¾® MnO₂ + 2H₂O, E₃⁰ = ?

DG₃⁰ = – n₃E₃⁰F = DG₁^0‑ – DG₂⁰

\ – 3E₃^0‑F = – 7.55F + 2.46F

\E₃⁰ =  = 1.70 volt

Problem 5:       How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8 hr. at a current of 8.46 ampere? What is the area of a tray if the thickness of silver plating is 0.00254 cm. Density of silver is 10.5 gm cm^–3.

Solution:       

Q = It = 8.46 amp ´ 8 ´ 3600 sec = 8 ´ 8.46 ´ 8 ´ 3600 C

Mass of silver =  gram

= 272.178 gm

Volume =  = 25.92 cc

Surface area =  =  = 1.02 ´ 10⁴ cm²

Problem 6:       During the discharge of lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139 gm cm^–3. Sulphuric acid of density 1.294 is 39% by weight and that of density 1.139 gm m^–1 is 20% by weight. The battery holds 3.5 L of acid. Calculate the number of ampere hour, assuming volume remains constant.

                              Pb(s) + SO₄^2– ¾® PbSO₄   Discharging

                              PbO₂(s) + 4H⁺ + SO₄^2– + 2e^– ¾® PbSO₄ + 2H₂O      Discharging

Solution:        Amount of H₂SO₄ in solution before discharging

=  gm = 1766.3 gm

Amount of H₂SO₄ in solution after discharging

= gm = 797.3 gm

Amount of H₂SO₄ consumed = (1766.3 – 797.3) gm = 969 gm

=  mol = 9.887 mol

\ Amount of electricity = 9.887 faraday = 9.887 ´ 96500 c

Number of ampere hour =  ampere hr

= 265.02 ampere hr.

Problem 7:       The following galvanic cell was

                       

                        was operated as an electrolytic cell using Cu as anode and Zn as cathode. A current of 0.48A was passed for 10 hour and then the cell was allowed to function as galvanic cell. What would be the emf of the cell at 25°C. Assume that the only electrode reactions occurring were those involving Cu/Cu²⁺ and Zn/Zn²⁺. Given  = –0.76V .

Solution:        During electrolysis some Zn²⁺ will discharge and some Cu²⁺ will pass in solution.

1. of Zn²⁺discharged = Eq. of Cu²⁺ formed =

mole of Cu²⁺ formed = Mole of Zn²⁺ deposited = 0.09

milli mole of Cu²⁺ formed = milli mole of Zn²⁺ deposited = 90

\ milli mole of Zn²⁺ left = 100 ´ 1 – 90 = 10

milli mole of Cu²⁺ final = 100 ´ 1 + 90 = 190

\ E_cell = E⁰_cell –  = 0.34 – (– 0.76) –

\ E_cell= 1.37V

Problem 8:       When metallic copper is shaken with a solution of copper salt, the reaction Cu + Cu⁺⁺  2Cu⁺ proceeds. When equilibrium is established at 298 K, [Cu²⁺] / Cu⁺]² = 1.66 ´ 10⁶. If the standard potential of the Cu²⁺ / Cu half cell is 0.337 V, what is standard potential of Cu⁺/Cu cell?

Solution:        The reaction is

Cu + Cu²⁺  2Cu⁺

and the cell is

Cu(s) | Cu⁺ || Cu⁺⁺ | Cu(s)

and its standard emf is

 

K =

At 25°C,

log (6.024 ´ 10^–7) = – 0.3678 V

Now, we have

2Cu⁺  Cu + Cu⁺⁺       E₁⁰ = 0.3678 V

Cu⁺⁺ + 2e^–  Cu(s)  E₂⁰ = 0.337 V

–––––––––––––––––––––––––––––––

Add:           2Cu⁺+ 2e^–  2Cu   E₃⁰ = ?

E₃⁰ =  = = 0.5209V

\ Cu⁺+ e^–  Cu, E₃⁰ = 0.5209 V

Problem 9:       By how much is the oxidising power of the /Mn⁺² couple decreased if the H⁺ concentration is decreased from 1 M to 10^–4 M at 25°C. Assume other species have no change in concentration.

Solution:                (in acidic medium) is oxidising agent

         + 8H⁺ + 5e^–  ¾® Mn⁺² + 4H₂O (l)

Q E = E⁰ –  logK

= E⁰ –

Q E – E ⁰ = –

This /MnO₂ couple will move to position of less oxidising power by 0.38 volt from its standard value.

Problem 10:     EMF of the following cell is 0.67 V at 298 K

            Pt, H₂ | H⁺ (pH = X) || 1 N KCl | Hg₂Cl₂(s), Hg

            Calculate pH,  (Calomal electrode) = 0.28 volt

Solution:        Half Cell    reaction

L.H.S.        H_2(g) ¾®                            E^o = 0.00V

R.H.S.       Hg₂Cl₂ + 2e^– ¾® 2Hg(l) + 2Cl^–(aq),  = 0.28 V

––––––––––––––––––––––––––––––––––––––––––––––––

H_2(g) + Hg₂Cl₂  ¾® 2H⁺ + 2Hg(l) + 2Cl^–(aq)        = 0.28 volt

\ K = [H⁺]² [Cl^–]²  = [H⁺]² (1)²  = [H⁺]²

From Nernst equation

E_cell = E⁰_cell –

0.67 = 0.28 + 0.0591 pH

E_cell =

pH = = 6.6

Problem 11      :     Calculate the equilibrium constant for the reaction:

                        Fe²⁺ + Ce⁴⁺® Ce³⁺ + Fe³⁺

                        Given that, Fe³⁺ + e^– ® Fe²⁺; E⁰ = 0.771 V

                        Ce⁴⁺ + e^– ® Ce³⁺ ; E⁰ = 1.61 V

Solution:        The equilibrium reaction is Fe²⁺ + Ce⁴⁺  Ce³⁺ + Fe³⁺

Since the reaction is at equilibrium, E_cell = 0

=

= 0.0591 log K

– 0.771 + 1.61 = 0.0591 log K

log K =  = 14.1963

\K = 1.57 ´ 10¹⁴

Problem 12:     Calculate the emf of the cell:

                        Pt H₂ |   CH₃ COOH  ||    NH₄OH | H₂, Pt

                        1 atm.      0.1 M            0.01 M      1 atm.

                        K_a for CH₃COOH = 1.8 x 10^-5  and K_b for NH₄OH = 1.8 x 10^-5

Solution:        The reactions are :            at anode : ½ H₂ ®  + e^–

at cathode :  + e^– ® ½ H₂

———————————————————————                                                                            overall reaction           ®

E = E⁰ – 0.059 log

At anode:

From   CH₃COOH  CH₃COO^– +

\= Ca = C

=

= 1.34 ´ 10^–3 M

At cathode:

From   NH₄OH  + OH^–

                                [OH^–] = C a =

\ =  =   =

=  2.359 ´ 10^-11M

\E = 0  –  0.0591 log   = – 0.4582 V

Problem 13:     Electrolysis of a solution of MnSO₄ in aqueous sulphuric acid is a method for the preparation of MnO₂ as per the reaction,

Mn²⁺(aq) + 2H₂O ¾¾® MnO₂ + 2H⁺ + H₂(g)

Passing current of 27 amp for 24 hours gives one Kg of MnO₂. What is the value of current efficiency. Write the reaction taking place at the cathode and at the anode.

Solution:        W = Þ 1000 =  \ I = 25.6 amp

Current efficiency =  = 94.8%

Reactions :            Anode: Mn²⁺ ¾¾® Mn⁴⁺ + 2e

Cathode : 2H⁺ + 2e ¾¾® H₂

Problem 14:     The standard reduction potential for the half cell

                        NO₃^– + 2H⁺ + e ¾® NO_2(g) + H₂O is 0.78 V

1. i) Calculate the reduction potential in 8 M H⁺
2. ii) What will be the reduction potential of the neutral solution. Assume  all the species to be at unit concentration.

Solution:        i)    In 8M H⁺ solution, of all other species has conc. of unity.

E_RP = E⁰_RP –  = 0.78 – 0.0591  = – 0.08862 V

1. ii) In case of neutral solution, concentration of (H⁺) = 10^–7M and conc,. of all the other species are unity, then

E_RP = E⁰_RP –  = 0.78 –  = – 0.0479 V

Problem 15:     The standard reduction potential Cu²⁺/Cu is 0.34 V. Calculate the reduction potential at pH = 14 for the above couple. is 1.0 ´ 10^–19

Solution:         = [Cu²⁺] [OH^–]²

[H⁺] = 10^–14                 \[OH^–] = 10⁰ = 1

[Cu²⁺] =  = 1.0 ´ 10^–19

Now E_RP for the couple Cu²⁺/Cu is

E_RD = E⁰_RP  –

= 0.34 –  = 0.221

17.2     Objective

Problem 1:       In the electrolysis of CuCl₂ solution using Cu electrodes, the weight of Cu anode increased by 2 gram at cathode. In the anode.

(A)  0.2 mole of Cu²⁺ will go into solution.

(B)  560 ml O_­2 liberate

(C)  No loss in weight

(D)  2 gram of copper goes into solution as Cu²⁺

Solution:        During electrolysis of CuSO₄ solution using Cu electrodes, the cell reaction is

Anode: Cu(s) ¾® Cu⁺⁺ + 2e^–

Cathode: Cu⁺⁺ _(aq) + 2e^– ¾® Cu_(s)

The loss in weight of anode is equal to gain in weight of cathode.

                        \ (D)

Problem-2:       During electrolysis of CuSO₄ using Pt-electrodes, the pH of solution

(A)  increases                                 (B) decreases

(C)  remains unchanged                 (D) cannot be predicted

Solution:        In presence of inert electrode, the cell reaction is

Anode                   2H₂O ¾® 4H⁺ + 4e^– + O₂

Cathode:              Cu⁺⁺_(aq) + 2e^– ¾® Cu(s)] ´ 2

–––––––––––––––––––––––––––––––––––––

Net cell reaction  2Cu⁺⁺ aq. + 2H₂O ¾® 4H⁺ + 2Cu(s) + O₂­

Due to increase in [H⁺], pH decreases.

                        \ (B)       

Problem 3:       A gas X at one atm is bubbled through a solution containing a mixture of 1 MY^– and 1 M Z^– at 25°C. If the reduction potential of Z > Y > X, then

(A)  Y will oxidize X and not Z        (B)  Y will oxidize Z not X

(C)  Y will oxidize both X and Z      (D)  Y will reduce both X and Z

Solution:        The tendency to gain electron is in the order

Z > Y > X

\ Y + e ¾® Y^–

X  ¾® X⁺ + e

                        \ (A)

Problem 4:       The standard reduction electrode potential values of elements A, B, C are + 0.68, –2.50, and – 0.50 V respectively. The order of their reducing power is

(A)  A >B > C                                  (B)  A > C > B

(C)  C > B > A                                 (D)  B > C > A

Solution:        More is the reduction potential, more is the power to get itself reduced or lesser is reducing power or greater is oxidizing power.

\ (D)

Problem 5:       The value of equilibrium constant for feasible cell reaction is

(A)  < 1                                           (B)  zero

(C)  = 1                                           (D)  > 1

Solution:        E⁰ = logK, if E⁰ is +ve, thus K > 1

\ (D)

Problem 6:       How much will potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralized to pH = 7

(A) increase by 0.0591 V                 (B) decrease by 0.0591 V

(C)  increase by 0.413 V                  (D) decrease by 0.413 V

Solution:        H⁺+ e^–  H₂(g),

                        [E = E⁰ –  log Q]

= 0.0  –  log =

= – 0.0591 ´ 7 ´ log10 = – 0.413 V

\ (D)

Problem 7:       Given that

= – 0.44 V, and

= 0.77V

What is = ?

(A)  – 0.11V                                     (B) 0.11V

(C)  – 0.04 V                                    (D) 0.04 V

Solution:       

= –0.04V

                        \ (C)                                           

Problem 8:       Following are some standard reduction potential values for the given half cell:

1. i) A⁺⁺ + 2e^–  A                       E⁰ = 1.27 V
2. ii) B⁺ + e^–  B                          E⁰ = – 0.7 V

iii)  C⁺⁺ + 2e^–  C                       E⁰ = – 0.54 V

1. iv) D⁺ + e^–  D                          E ⁰ = 1.05 V

The combination of which two half cells will give galvanic cell having maximum possible emf.

(A)  (i) and (ii)                                 (B) (I) and (iv)

(C)  (ii) and (iii)                               (D) (iii) and (iv)

 

Solution:        Since all the values are standard reduction potential and so the two half cells having maximum and minimum reduction potential values will give a cell of maximum possible emf.

 

\ (A)

Problem 9:       By how much would the oxidising power of the  couple change if the H⁺ ions concentration is decreased 100 times?

                        (A) increases by 189 mV                 (B) decreases by 189 mV

                        (C) will increase by 19 mV              (D) will decrease by 19 mV

Solution:         + 5e^– + 8H⁺ ¾® Mn²⁺ + 4H₂O

According to Nernst equation,

E_red =  –  log        Let [H⁺]_initial = X

E_red(initial) =      [H⁺]_final =

E_red(final) =

E_red(final) – E_red(initial) =  log 10¹⁶ = – 0.1891 V

This E_red decreases by 0.189 V. The tendency of the half cell to get reduced is its oxidising power. Hence the oxidising power decreases by 0.189V

                        \ (B)

Problem 10:     A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at cathode and anode are respectively.

                        (A) H₂,O₂                                         (B) O₂,H₂

                        (C) O₂,Na                                        (D) O₂,SO₂

Solution:        Na⁺ ions are not reduced at cathode and  ions are not oxidized at anode

Cathode: 2H₂(l) ¾® O₂ (g) + 4H⁺ (aq) + 4e^–

Anode: 2H₂O(l) + 2e^– ¾® H₂(g) + 2OH^– (aq)

                        \ (A)

Problem 11:     A solution containing one mole  per litre of each Cu(NO₃)₂; AgNO₃; Hg₂(NO₃)₂ ; Mg(NO₃)₂ is being electrolysed using inert electrodes. The values of standard electrode potentials (reduction potentials in volts are Ag/Ag⁺ = 0.80 V, 2Hg/Hg₂⁺⁺ = 0.79V, Cu/Cu⁺⁺ = + 0.24 V, Mg/Mg⁺⁺ –2.37 V. With increasing voltage, the sequence of deposition of metals on the cathode will be

                        (A) Ag, Hg, Cu                                            (B) Cu, Hg, Ag

                        (C) Ag, Hg, Cu, Mg                                     (D) Mg, Cu,Hg, Ag

 

Solution:        Greater the value of standard reduction potential, greater will be it’s tendency to undergo reduction. So the sequence of deposition of metals on cathode will be Ag, Hg, Cu. Here, magnesium will not be deposited because it’s standard reduction potential is negative. So it has strong tendency to undergo oxidation. Therefore, on electrolysis of Mg(NO₃)₂ solution, H₂ gas will be evolved at cathode.

                        \(A)

Problem 12:     One coulomb of charge passes through solution of AgNO₃ and CuSO₄ connected in series and the conc. of two solution being in the ratio 1:2. The ratio of weight of Ag and Cu deposited  on Pt electrode is

                        (A) 107.9 : 63.54                              (B) 54 : 31.77

                        (C) 107.9 : 31.77                              (D) 54 : 63.54

Solution:        Faraday’s IInd Law  = constant

So,                            (Ag⁺ + e^– ® Ag, E_Ag =  )

\             (Cu⁺² + 2e^– ® Cu, E_Cu =  )

=

\ (C)

Problem 13:     Electrolysis of dil H₂SO₄ liberates gases at anode and cathode

                        (A) O₂ & SO₂ respectively                (B) SO₂ & O₂ respectively

                        (C) O₂ & H₂ respectively                  (D) H₂ & O₂ respectively

Solution:        At anode:   2H₂O ® O₂ + 4H⁺ + 4e^–

At cathode:      2H⁺ + 2e^– ® H₂

                        \(C)        

Problem 14:     For the electrochemical cell M  | M⁺ || X ^– | X    = 0.44V,
= 0.33V from this data one can deduce that

                        (A)  M + X ¾® M⁺ + X^– is spontaneous reaction

                        (B)  M⁺+ X^– ¾® M + X is spontaneous reaction

                        (C)  E_cell  = 0.77V

                        (D)  E_cell = – 0.77V

Solution:        E⁰_RP for M  > DE_RP for X

\ M⁺ + X^– ¾® M + X (spontaneous), E⁰_cell = – 0.33 + 0.44 = 0.11V

                        \ (B)       

Problem 15:     How many coulomb of electricity will be consumed when 100 mA current passes through a solution of AgNO₃ for half an hour during electrolysis

                        (A) 108                                           (B) 180

                        (C) 1800                                          (D) 18000

Solution:        Charge passed during electrolysis      = i x t

= (100 ´10^–3) ´ (´60´60)

= 180 C

                        \ (B)

 

 

18. Assignments (Subjective Problems)

Level –  I

1. In refining of silver by electrolytic method, what will be the weight of 100 gm of Ag anode if 5A current is passed for 2 hour. The purity of silver anode is 95% by weight.
2. In the electrolysis of KI, I₂ is formed at the anode.

2I^–  ¾® I₂ + 2e

After passage of current of 0.5 amperes for 9650 seconds I₂ formed required 40 ml of
0.1 M Na₂S₂O₃.5H₂O solution in the reaction.

I₂ + 2S₂O₃^2– ¾® S₄O₆^2– + I^–

What is current efficiency?

3. A current of 1.70Å is passed through 300 ml of 0.160 M solution of ZnSO₄ for 230- sec with current efficiency of 90%. Find the molarity of Zn²⁺ after the deposition of Zn. Assume the volume of solution remains constant during electrolysis.
4. A current of 40 microamperes is passed through a solution of AgNO₃ for 32 minutes using Pt electrodes. An uniform single atom thick layer of Ag is deposited covering 43% cathode surface. What is the total surface area of cathode, if each Ag atom covers 5.4 ´ 10^-16 cm²?
5. Determine the standard potential of a cell in which the reaction
   Co³⁺(aq) + 3Cl^–(aq) + 3Ag(s) ¾¾® 3AgCl(s) + Co(s) from the standard potentials of the couples Ag/AgCl, Cl (+0.22V), Co³⁺/Co²⁺ (+1.81V) and Co²⁺/Co(–0.28V).
6. The emf of the following cell was 0.2699 V at 293 K and 0.2669V at 303 K

Pt | H₂ | HCl (aq) || Hg₂Cl₂ | Hg

Calculate (i) temperature coefficient

(ii) DG, DH and DS at 293 K

7. Consider the disproportionation 2Cu⁺ Cu²⁺ + Cu (s). At equilibrium the value of  is 1.8 ´ 10⁶. If the standard potential for Cu²⁺ | Cu⁺ is +0.15V, calculate ^o
8. Show that = –
9. For the reaction

4 Al(s) + 3O₂(g) + 6H₂O + 4OH^– ¾® 4 [Al(OH)₄]^–

= 2.73 V

if [OH^–] = –157 kJ / mol

= – 237.2 kJ / mol

Determine [Al(OH)₄]^–, (free energy of formation of [Al(OH)₄]^–).

10. A current of 2 amperes is passed for 11 hours through 500 mL of 2M solution of Ni(NO₃)₂ using platinum electrodes. What will be the molarity of the solution at the end of the electrolysis ?
11. A 100 – watt, 110 – volt incandescent lamp is connected in series with an electrolytic cell containing CdSO₄ solution. What weight of cadmium will be deposited by the current flowing for 10 hours?
12. In an industrial electrolytic cell with current efficiency of 75% it is desired to produce 48 Kg of Mg metal per hour. Calculate the current required.
13. What volume of hydrogen and oxygen would be obtained at 27°C and 740 mm of Hg by passing a current of 25 amperes through acidulated water for 24 hours?
14. 10 g fairly concentrated solution of CuSO₄ is electrolysed by passing 10 ampere current for 1.5 minutes. Calculate.
15. a) the weight of resulting solution
16. b) the number of equivalents of acid or alkali produced in solution  (Cu = 63.5)
17. Cadmium amalgam is prepared by electrolysis of a solution of CdCl₂ using mercury cathode. Find how long should a current of 5 ampere is passed in order to prepare 12% Cd – Hg amalgam on a cathode of 2 g mercury (Cd = 112.40)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Level –  II

1. The Edison storage cell is represented as

Fe(s) | FeO(s) | KOH_(aq) | Ni₂O₃(s) | Ni(s)

The half cell reactions are

Ni₂O_3(s) + H₂O_(l) + 2e ¾® 2NiO_(s) + 2OH^–_(aq)     E⁰ = 0.40 V

FeO_(s) + H₂O_(l) + 2e ¾® Fe_(s) ¾® Fe_(s) + 2OH^–            E⁰= –0.87V

1. i) What is the cell reaction?
2. ii) What is emf of the cell? How does it depend on the concentration of KOH?

iii)   What is the maximum amount of electrical energy that can be obtained from one mol of Ni₂O₃.

2. Calculate the cell potential of a Daniel cell having 1.0 M Zn²⁺and originally having 1.0 M Cu²⁺ after sufficient ammonia has been added to cathode compartment to make NH₃ concentration 2.0M. Given and  are 0.76 and – 0.34V respectively. Also equilibrium constant for the [Cu(NH₃)₄]²⁺ formation is 1 ´ 10¹².
3. Calculate the equilibrium constant for the reaction 2Fe²⁺+ 3I^– 2Fe²⁺ + I₃^–.  The standard reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for Fe³⁺/Fe²⁺ and I₃^–/I^–.
4. In an electrolysis experiment, current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold and the second contains CuSO₄ solution. 9.85 g of gold was deposited in the first cell. If
   the oxidation number of gold is +3. Find the amount of Cu deposited on
   cathode in second cell. Also calculate the current strength in ampere (Au = 197 and Cu = 63.5)
5. In a fuel cell H₂ and O₂ react to produce electricity,. In the process H₂ gas is oxidised at the anode and O₂ at cathode. If 67.2 litre of H₂ at STP reacts in 15 minutes. What is the average current produced? If the entire current is used for electro deposition of Cu from Cu⁺², how many grams of Cu are deposited.
6. Consider the following half reaction

PbO₂(s) + 4H⁺  + (aq) + 2e^– ¾® PbSO₄(s) + 2H₂O,  E° =  1.70 V.

PbSO₄(s) + 2e^– ¾® Pb(s) + (aq),  E° = –0.31 V.

Calculate E_cell of the spontaneous reaction at 298K, given   = 2.0M, pH = 1 and other species are at  unit concentration.

7. A current of 0.193 amp is passed through 100 ml of 0.2M NaCl for an hour. Calculate pH of solution after electrolysis if current efficiency is 90%. Assume no volume change.
8. The voltage of the following cell is 0.987 V

Pt (H₂) (1 atm) | HOCN(1.4´ 10^–3 M) || Ag⁺(0.8M) | Ag(s)

Calculate K_a of HOCN,   = 0.80 V

9. Given the cell

Cd | Cd(OH)₂(s) | NaOH (0.01 mol kg^-1) | H₂(1 atm) | Pt

With E_cell = 0.0 V at 298 K. If = – 0.40 V, Calculate K_sp for Ca(OH)₂

10. The overall formation constant of following reaction is 1 ´ 10¹⁹.

Co⁺²  + 6CN^–‑  [Co(CN)₆]^–4

Calculate formation of following complex

Co⁺³ + 6CN^–  [Co(CN)₆]^–3

Given that  [Co(CN)₆]^–4 ¾® [Co(CN)₆]^–3 + e^–, E⁰ = + 0.83 V

Co⁺³ +e^– ¾® Co⁺², E⁰ = 1.82 V

11. For the galvanic cell

Ag|AgCl(s), KCl (0.2M) || KBr (0.001M), AgBr(s) |Ag

calculate the EMF generated and assign correct polarity to each electrode for a spontaneous process after taking an account of cell reaction at 25°C.

Given K_SP (AgCl) = 2.8 ´ 10^–10 , K_SP (AgBr) = 3.3 ´ 10^–13.

12. Zinc granules are added in excess to 750 ml of 1.5 M Ni (NO₃)₂ solution at 30°C until the equilibrium is reached. If EMF of cell Zn(s) | Zn⁺²(1M) || H⁺(1M) | H₂(g) (1atm),Pt is 0.76 volt and that of cell  Ni | Ni⁺² (1M) || H⁺(pH = 0) | H₂(g) (1 atm), Pt is 0.24 volt, then  calculate [Ni⁺²] at equilibrium. Also find the amount of Zn consumed.
    (Zn = 65)
13. The density of copper is 6.94 g mL^–1. Find the number of coulomb needed to plate an area of 10 ´ 10 cm² to a thickness of 10^–2 cm using CuSO₄ solution as an electrolyte. Atomic weight of Cu is 63.5.
14. 19 g fused SnCl₂ was electrolysed using inert electrodes when 0.119 g
    was deposited at cathode. If nothing was given out during electrolysis, calculate
    the ratio of weight of SnCl₂ and SnCl₄ in fused state after electrolysis
    (At weight Sn = 119).
15. How many moles of iron metal will be produced by the passage of 4 ampere of current through 1 litre of 0.1M Fe³⁺ solution for 1 hour? Assume that only iron (II) is reduced to iron.

 

 

 

 

 

 

 

 

 

 

Level –  III

1. Two litre solution of a buffer mixture containing 1.0 M NaH₂PO₂ and 1.0 M Na₂HPO₄ is placed in two compartments (one litre in each) of an electrolytic cell. The platinum electrodes are inserted in each compartment and 1.25 ampere current is passed for 212 minute. Assuming electrolysis of water only at each compartment. What will be the pH in each compartment after passage of above charge? pK_a for H₂PO₄^–= 2.15.
2. Two electrochemical cells are assembled in which these reactions occur.

V³⁺ + Ag⁺ + H₂O ¾® 2V³⁺ + H₂O                      E⁰_cell = 0.66V

V³⁺ + Ag⁺ + H₂O ¾® VO²⁺ + 2H⁺ + Ag_(s)           E⁰_cell= 0.439 V

If  = 0.80V, calculate E⁰ for half cell reaction

V³⁺+ e ¾® V²⁺

3. Determine the potential of the Daniel cell, initially containing 1.00 L each of 1.00 M copper (II) ion and 1.00 M Zinc, after passage of 0.100 mc charge

= –0.076V  = 0.34 V

4. Calculate the solubility product constant for AgI from the following values of standard electrode potentials

= – 0.15 Volt at 25°C,

5. Knowing that K_sp for AgCl is 1.0 ´ 10^–10. Calculate E for Ag/AgCl electrode immersed in 1.00 M KCl at 25°C. = 0.799 V
6. The emf of the following cell is – 0.46V

Pt(H₂) |  (0.4M), (6.4 ´ 10^–3) || Zn²⁺(0.3M)| Zn

If  = –0.76 V. Calculate pK_a of  (i.e.,) for the equilibrium
H⁺ +

7. An acidic solution of copper (II) sulphate containing 0.4 g Cu²⁺ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 mL and the current of 1.2 amperes. Calculate the volume of gases evolved at STP during the entire electrolysis.
8. Per di-sulphuric acid (H­₂­S₂O₈) can be prepared by electrolytic oxidation of H₂SO₄ as 2H₂SO₄ ® H₂S₂O₈ + 2H⁺ + 2e^–. O₂ and H₂ are byproducts. In such an electrolysis 0.87 g of H₂ and 3.36 g of O₂ were generated at STP. Calculate the total quantity of current passed through the solution to carry out electrolysis. Also report the weight of H₂S₂O₈ formed.
9. Find  from the following data  = 1.51 V and  = 1.23 V
10. A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag₂C₂O₄ at 25°C. The measured potential difference between the rod and the SHE is 0.589V, the rod being positive. Calculate the solubility product of silver oxalate. Given  = + 0.8V
11. When AgCl is dissolved in a large excess of NH₃, practically all silver can be assumed to exist in the form of a single species . Compute the values of m and n using the following two cells.

Ag | 0.379 ´ 10^-3M AgCl,  1M NH₃ || 37.9 ´10^-3M AgCl, 1M NH₃ | Ag

E_cell = 0.1185V at 298K

Ag | 3.4 ´10^-3M AgCl,  1M NH₃ || 3.4 ´ 10^-2M AgCl, 0.1M NH₃ | Ag

E_cell = 0.1263V at 298K

12. An excess of liquid mercury was added to a 10^-3M acidic solution of Fe⁺³. It was found that only 4.6% of the iron remained as Fe⁺³ at equilibrium at 25°C. Calculate . Assume that only following reaction takes place.

Given is 0.771 V. 2Hg + 2Fe⁺³   + 2Fe⁺²

13. From the following information, calculate the overall stability constant Ks of at 25°C

Ag⁺ + e^– ® Ag                  E° = 0.799V

^– ® Ag + 2S₂O₃^-2;              E° = 0.017 volt

14. The EMF of cell

Ag | AgCl, 0.05M  KCl || 0.05M AgNO₃ | Ag  is 0.788 Volt.

Find the solubility product of AgCl.

15. Calculate the EMF of the following cell

Pt (H₂, 1atm) | C₂H₅COOH (0.15M) ||  (0.01 M) NH₄OH | H₂, Pt (1atm)

K_a for C₂H₅COOH = 1.4 ´ 10^–5; K_b for NH₄OH = 1.8 ´ 10^–5

 

 

 

 

 

 

 

 

 

 

 

 

19. Assignments (Objective Problems)

Level –  I

1.         Which  are true for a standard hydrogen electrode

(A) The hydrogen ion concentration is 1 M.

(B) Temperature is 25°C.

(C) Pressure of hydrogen is 1 atmosphere.

(D) It contains a metallic conductor which does not adsorb hydrogen.

2. For the cell Tl½ Tl⁺ (.001M) ½½ Cu²⁺ (.1M) ½ Cu. E_Cell at 25°C is. 83V which can be increased

(A) by increasing [Cu²⁺]                            (B) by increasing [TI⁺]

(C) by decreasing [Cu²⁺]                          (D) by decreasing [TI⁺]

3.         Which represents disproportionation

(A) 2Cu⁺ ¾® Cu⁺² + Cu                          (B) 3I₂ ¾® 5I ^– + I ⁺⁵

(C) Cu⁺² + Zn ¾® Zn⁺² +  Cu                 (D) None

4. The standard potentials at 25 °C for the following half reactions are given against then

Zn²⁺ + 2e^– ® Zn, E⁰ = –0.762 V

Mg²⁺ + 2e^– ® Mg, E⁰ = 2.37 V

When zinc dust is added to a solution of magnesium chloride

(A) No reaction will take place                  (B) Zinc chloride is formed

(C) Zinc dissolve in solution                      (D) Magnesium is precipitated

5.         Emf of the cell

Ni | Ni²⁺ (0.1 M) || Au³⁺ (1.0 M) | Au will be

= 0.25,                                        = 1.5 V

(A) 1.75 V                                                  (B) + 1.7795 V

(C) + 0.7795 V                                          (D) – 1.7795 V

6. The quantity of electricity needed to electrolyse completely 1 M solution of
   CuSO_­4, Bi₂(SO₄)₃, AlCl₃ and AgNO₃ each will be

(A) 2F, 6F, 3F, and 1F respectively

(B) 6F, 2F, 3F and 1F respectively

(C) 2F, 6F, 1F and 3F respectively

(D) 6F, 2F, 1F and 3F respectively

7. If the pressure of H₂ gas is increased from 1 atm to 100 atm. Keeping H⁺ concentration constant at 1 M, the voltage of Hydrogen half cell at
   25° C will be

(A) 0.059 V                                                (B) 0.59 V

(C) 0.0295 V                                             (D) 0.118 V

8. If aqueous solutions of AgNO₃ is electrolysed using inert electrode the gas evolved at anode is

(A) NO₂                                                     (B) O₂

(B) H₂                                                        (D) N₂O

9. The standard cell potential for the electrochemical cell

Ag/Ag⁺ || AgI/Ag;     = –0.151 V, = 0.799 V

(A) + 0.950 V                                            (B) – 0.950 V

(C) – 0.648 V                                            (D) + 0.648 V

10. If equal quantities of electricity are passed through three voltmeter containing FeSO₄, Fe₂(SO₄)₃ and Fe(NO₃)₃ then

(A) The amount of iron deposited in FeSO₄  and Fe₂(SO₄)₃ is equal

(B) The amount of iron deposited in Fe(NO₃)₃ is 2/3 of the amount of iron deposited in FeSO₄

(C) The amount of iron deposited is Fe₂(SO₄)₃ and Fe(NO₃)₃ is equal

(d) Same gas will evolve in all three cases at anode.

11. Two platinum electrodes are dipped in a solution of HCl, and H₂ & Cl₂ gases are made to bubble through them at constant pressure. One half cell is
    Pt, H₂(g)|H₃O⁺ (aq) with reaction ½ H₂ + H₂O ® H₃O⁺ + e^–.

The electrode at which this half cell reaction would occur is

(A) Cathode                                              (B) Anode

(C) May be anode or cathode                  (D) Neither anode nor cathode

12. A Chemist found that the standard electrode potential , of the zinc electrode is –0.763Volt. Which cell will give this value of EMF.

• Pt, H₂ (1 atm) | H⁺ (10M) || Zn⁺² (10M) | Zn
• Pt, H₂ (1 atm) | H⁺ (1M) || Zn⁺² (1M) | Zn
• Zn | Zn⁺² (1M) || H⁺ (1M) | H₂ (1 atm), Pt
• Zn | Zn⁺² (10M) || H⁺ (10M) | H₂ (1 atm), Pt

13. At the reduction-electrode, this reaction takes place

A^+z  + Ze^–   A

Which is/are correct relation(s) for electrode potential?

(A) E =                          (B) E =

(C) E =                          (D)       E =

14. Which of the following is/are function(s) of salt bridge.

(A) It completes the electric circuit with electrons flowing from one electrode to other through external wires and a flow of ions between the two compartments through salt bridge.

(B) It prevents the accumulation of the ions

(C) Both (A) and (B) are correct

(D) None of these

15. Given: Fe⁺² + 2e^–  Fe                       E⁰ = – 0.44V

Co⁺²  + 2e^–  Co                                  E⁰ = – 0.28V

Ca⁺²  + 2e^–  Ca                                  E⁰ = – 2.87V

Cu⁺²  + 2e^–  Cu                                  E⁰ = + 0.337V

Which is the most electropositive metal?

(A) Fe                                                        (B)       Co

(C) Ca                                                       (D)       Cu

16. Which of the following graphs will correctly correlate E_cell as a function of concentrations for the cell (for different M and M^¢)

Zn(s) + Cu⁺²(M) ® Zn⁺²(M^¢) + Cu(s),  = 1.10 V

X-axis: , Y-axis: E_cell

(A)	(B)
(C)	(D)

17. The amount of ion discharged during electrolysis is directly proportional to :

(A) Resistance                                          (B) Time

(C) Current                                                (D) Chemical equivalent of the ion

18. The reaction Cu⁺² (aq) + 2Cl^–(aq) ® Cu(s) + Cl₂(g) has = -1.02V. This reaction

• can be made to produce electricity in a voltaic cell
• can be made to occur in an electrolytic cell
• occurs whenever Cu⁺² and Cl^– are brought together in aqueous solution.
• can occur in acidic solution but not in basic solution.

19. Cu⁺ + e^– ® Cu, E⁰ = x₁ V, Cu⁺² + 2e^– ® Cu, E⁰ = x₂ V

For Cu⁺² + e^– ® Cu⁺, E⁰ will be

(A) x₁  – 2x₂                                                               (B) x₁  + 2x₂

(C) x₁  – x₂                                                                 (D) 2x₂  – x₁

20. Apollo-II was the spacecraft used by American astronauts during the moon explorations in 1969. Which of the following do you think would be best source of energy for the space craft?

(A) Dry cell                                                (B) Lead-acid (storage) battery

(C) Fuel cell                                              (D) Alkaline battery

 

 

 

 

 

 

 

Level –  II

 

1. Given E° for Cu²⁺ ¾® Cu⁺ is +0.15 V  and  Cu⁺ ¾® Cu is  + 0.50 V

Calculate E° for Cu²⁺ ¾® Cu

(A) + 0.325 V                                            (B) + 0.125 V

(C) + 0.250 V                                            (D) + 0.160 V

2. A solution is one molar in each of NaCl, CdCl_­2, ZnCl₂ and PbCl₂. To this, tin metal is added. Which of the following is true? Given that,

= – 0.126V;  = –0.136V

= – 0.40V;  = –0.763V;  = –2.71V

(A) Sn can reduce Na⁺ to Na                   (B) Sn can reduce  Zn²⁺  to Zn

(C) Sn can reduce Cd²⁺ to Cd                  (D) Sn can reduce Pb²⁺  to Pb

3. A certain current liberates 0.5g of hydrogen in 2 hour. How many grams of copper can be liberated by the same current flowing for the same time through a copper sulphate solution?

(A) 12.7 grams                                          (B) 15.9 g

(C) 31.8g                                                   (D) 63.5 g

4. The same quantity of electricity is passed through H₂SO₄ and HCl solutions of same concentration. The amount of hydrogen liberated from H₂SO₄ as compared to that from HCl is

(A) the same                                             (B) twice as such

(C) one half as such                                 (D) dependent on concentration.

5. The reaction: H₂(g) + AgCl(s) ¾® H⁺(aq) + Cl^–(aq) + Ag(s) occurs in the galvanic cell,

(A) Ag/AgCl(s) / KCl (solution/ AgNO₃(soln.)/Ag

(B) Pt/H₂(g) /HCl (soln)/AgNO₃(soln.)/Ag

(C) Pt/H₂(g)/HCl(soln)/AgCl(s)/Ag

(D) Pt/H₂(g)/KCl(soln)/AgCl(s)/Ag

 

6. Given that E°(Zn²⁺/Zn) = – 0.763V

and E° (Cd²⁺/Cd) = – 0.403V, the emf of the following cell:

Zn/Zn²⁺ (a = 0.04) ||Cd²⁺ (a = 0.2) / Cd is given by,

(A) E = –0.36 + [0.059/2] [log(0.004/0.2)]

(B) E = + 0.36 + [0.059/2] [log (0.004/0.2)]

(C) E = – 0.36 + [0.059/2] [log(0.2/0.004)]

(D) E = + 0.36 + [0.059/2][log (0.2/0.004)]

 

7. The standard reduction potential for Cu²⁺/Cu is +0.34 V. The reduction potential at pH = 14 for the above couple is, [K_sp [Cu(OH)₂] = 1.0 ´ 10^–19].

(A) – 0.22 V                                               (B) + 0.22V

(C) – 0.34 V                                              (D) + 0.34 V

8. The emf of the following three galvanic cells:
9. Zn/Zn²⁺ (1M) || Cu²⁺(1M)/Cu
10. Zn/Zn²⁺ (0.1M) || Cu²⁺(M) / Cu
11. Zn/Zn²⁺ (1M) || Cu²⁺(0.1M)/Cu

are represented by E₁,E₂,E₃. Which of the following statement is true?

(A) E₁ > E₂ > E₃                                        (B) E₃ > E₂ > E₁

(C) E₃  > E₁ > E₂                                       (D) E₂ > E₁  > E₃

9. If the half cell reaction A + e^– ¾® A^– has a large negative reduction potential, it follows that

(A) A is readily reduced                            (B) A is readily oxidised

(C) A^– is readily reduced                          (D) A is readily oxidised

10. Four alkali metals P,Q,R and S are having respectively, standard electrode potentials as –3.05, –1.66, –0.40 and 0.80 V. Which one will be the most reducing?

(A) P                                                          (B) Q

(C) R                                                         (D) S

11. A solution of CuSO₄ is electrolysed for ” 7 minutes” with a current of 0.6A. The amount of electricity passed is equal to

(A) 4.2 C                                                   (B) 2.6 > 10^–3 F

(C) 126C                                                   (D) 36C

12. An electrolytic cell is constructed for preparing hydrogen. For an average current of 1 ampere in the circuit, the time required to produce 450 mL of hydrogen at NTP is approximately.

(A) 30 mts                                                 (B) 1 hour

(C) 2 hours                                                (D) 5 hours

13. Which statement is true in regard to a spontaneous redox reaction?

(A)  is always negative                       (B)  is always positive

(C)  is always positive                        (d)  is always positive

14. Normal aluminium electrode coupled with normal hydrogen elecrode gives an emf of 1.66V. So the standard electrode potential of aluminium is,

(A) –1.66V                                                 (B) + 1.66V

(C) –0.83V                                                (d) + 0.83 V

15. Which one of the following will increase the voltage of the cell?

Sn(s) + Ag⁺(aq) ¾® Sn²⁺(aq) + 2Ag(s)

(A) increase in concentration of Sn²⁺       (B) Increase in size of silver rod

(C) Increase in the concentration of Ag⁺  (d) none of these

16. How much will the reduction potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralized to pH = 7?

(A) Increase by 0.059V                                   (B) Decrease by 0.059V

(C) Increase by 0.41V                                     (D) Decrease by 0.41V

17. Passage of one ampere current through 0.1M Ni(NO₃)₂ solution using Ni electrodes brings in the concentration of solution to ……… in 60 seconds.

(A) 0.1M                                                          (B) 0.05M

(C) 0.2M                                                          (D) 0.025M

18. The charge required for the oxidation of one mole Mn₃O₄ intoin presence of alkaline medium is

(A) 5 x 96500 C                                              (B) 96500 C

(C) 10 x 96500 C                                            (D) 2 x 96500 C

19. The cost at 5 paise/KWH of operating an electric motor for 8 hours which takes 15 amp at 110V is

(A) Rs.66                                                         (B) 66 Paise

(C) 37 Paise                                                    (D) Rs.6.60

20. In the electrolysis of aqueous solution of NaOH, 2.8 litre of oxygen was liberated at the anode at NTP. How much hydrogen was liberated at the cathode?

(A) 5.6 litre                                                      (B) 56 ml

(C) 560 ml                                                       (D) 0.056 litre

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

20. Answers to Subjective Assignments

Level – I

1. 57.59 gm 2.         8%
2. 0.154M 4.         602.8 cm²
3. 0.20V 6.         –52.09 kJ, –69.055 kJ, –57.9 JK^–1
4. 0.5197V 9.         1.3 ´ 10³ KJ/mole
5. 1.179 11.        19.95 g                                                                                               
6. 1.43 ´ 10⁵ coloumb 13.        141.46 L
7. (a) 9.63 g, (b) 9.326 ´ 10^–3 15.        93.75 sec.

Level – II

1. 2.45 ´ 10² kJ/mol 2.         0.71V
2. 6.26 ´ 10⁷ 4.         0.804 amp.
3. 190.5 g 6.         1.91 V
4. 9.812 8.         43.65 gm                    
5. 3.36 ´ 10^–15 M² 10.        8.23 ´ 10⁶³
6. –0.037

[Note: to get E_cell positive, polarity of the half cells should be reversed)

12. 7 ´ 10^–18 M, 73.125 gm 13.        27171.96 coloumb
13. 71.340 15.          0.0245

Level – III

1. LHS 2.005, RH 2.295 2.          = – 0.255V
2. 1.085 V 4.         8.4 ´ 10^–17
3. 0.208 V 6.         7.13
4. 158.14 ml 8.         22.27 min                                 
5. 1.69 V 10.        9.7 ´ 10^–12 M³
6. m = 1, n = 2 12.        0.791 V
7. 1.705 ´ 10¹³ 14.        5 ´ 10^–14
8. –0.46

 

 

 

 

21. Answers to Objective Assignments

Level – I

1. A,B,C                                                        2.         A,D
2. A,B                                                            4.         A
3. B                                                               6.         A
4. A                                                               8.         B
5. B                                                               10.       B,C,D
6. B                                                               12.       C
7. B                                                               14.       C
8. C                                                               16.       B
9. B,C,D                                                        18.       B
10. D                                                               20.       C

Level – II

1. A                                                               2.         D
2. B                                                               4.         B
3. C                                                               6.         B
4. A                                                               8.         D
5. B                                                               10.       A
6. B                                                               12.       B
7. B                                                               14.       a
8. C                                                               16.       D
9. A                                                               18.       C
10. B                                                               20.       A

 

 

Electrochemistry

 

    Hints for Subjective Problems

 

Level – I

 

2. Meq. of Na₂S₂O₃ required after passage of current

= Meq. of I₂ produced = Moles of electron (nF) actually required

 

6. DG = – nFE_cell, DH = nF

DS =

 

9. DG⁰= DG⁰_P – DG⁰_R

DG⁰= – nfE⁰

 

Level – II

 

2. Cu²⁺ + 4NH₃ [Cu(NH₃)₄]²⁺

x          2                    1

K_f =  = 1 ´ 10¹²

As K_f is very very large hence almost all Cu²⁺ ion will be converted into complex form very very small amount of X will be left.

 

5. Cell reaction in fuel cell

O₂ + 2H₂O + 4e ¾® 4OH^–   (At cathode)

2H₂ + 4OH^– ¾® 4H₂O + 4e (At anode)

–––––––––––––––––––––––

2H₂ + O₂ ¾® 2H₂O              (Net cell reaction)

 

6. Cell reaction: Pb_(s)+ SO₄^2–_(aq.) ¾® PbSO_4(s)+ 2e                           E⁰ = 0.31V

PbO₂ + 4H⁺ + SO₄^2–_(aq) + 2e ¾® PbSO_4(s) + 2H₂O      E⁰ = 1.70 V

––––––––––––––––––––––––––––––––––––––––

Net cell reaction    PbO₂ + 4H⁺+ SO₄^2– ¾® PbSO_4(s) + 2H₂O

Q =

 

10. Co²⁺ + 6CN^– [Co(CN)₆]^4–                 K_i = K_f = 1 ´10¹⁹         …(1)

Co³⁺+ 6CN^–  [Co(CN)₆]^3–                  K₂                                …(2)

Also                 [Co(CN)₆]^4– ¾® [Co(CN)₆]³⁺+ e       E⁰ = 0.83 V

Co³⁺+ e ¾® Co²⁺                               E⁰ = 1.82V

–––––––––––––––––––––––––––––––

Net reaction    Co³⁺ + [Co(CN)₆]^4– ¾® Co²⁺ + [Co(CN)₆]^3–                              …{3)

Level – III

 

6. Cell reaction

L.H.S.                    H₂ ¾® 2H⁺+ 2e               E₀ = 0

R.H.S.                   Zn²⁺ ¾® Zn(s)                 E⁰ = – 0.76

––––––––––––––––––––

Net cell reaction    Zn²⁺ + H₂ ¾® 2H⁺ + Zn

E_cell = E⁰_cell      –

Calculate H⁺

HSO₃^–  H⁺ + SO₃^2–

0.4 M           0       0

(0.4 M – x)        x         x

x = [H⁺] = [SO₃^2–]

 

9. MnO₄^– ¾® MnO₂ + 3e, E₁⁰ = ?                              …(1)

MnO₄^–¾® Mn²⁺ + 5e                  E₂⁰ = 1.51 V                …(2)

DG₁⁰ = – 3 ´ F ´ 1.51             DG₂⁰ = – 5 ´1.23 ´F

MnO₂ + 2e ^–¾® Mn²⁺           E₃⁰ = 1.23V                 …(3)

(2) – (3) gives equation (1)

 

11. All the Ag⁺_(aq) from AgCl in each of the half cells is complexed to form
    Ag_m [NH₃]­_n^m+

For given cell

L.H.S.        mAg_(s) + nNH₃ ^–¾® Ag_m(NH₃)_n^m+ + me       E_0ox= EV

LHS           LHS

R.H.S.       Ag_m(NH₃)_n^m+ + me  ¾®mAg_(s) + nNH₃        E⁰_red = –EV

RHS                                   R.H.S.

–––––––––––––––––––––––––––––––––––

Ag_m(NH₃)_n^m+_(RHS) + nNH_3(LHS) ¾® Ag_m(NH₃)_n^m+ + nNH_3(RHS)    E⁰ = 0

From given cell derive two equations and get the value of m and n.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

    Subjective Problems

 

Level – I

 

1. Only pure Ag from anode is oxidised which in turn is deposited at cathode. By Faraday’s law

W_Ag = Zit =  5 ´ 2 ´ 60 ´ 60 = 40.29 gm

But anode is 95% pure with respect to Ag

\ Actual loss = 40.29 ´  = 42.41 gm

Wt. of Ag anode after electrolysis = 100 – 42.41 = 57.59 gm

 

2. 40 ml of 0.1 M hypo solution = mol of hypo = 4 ´ 10^–3 mol of hypo

\ Mole of I₂ produced =  = 2 ´ 10^–3

W = Zit Þ W =

\ i = 0.04

\ Current efficiency =  = 8%

 

3. (i) = amp

\ Eq. of Zn²⁺  lost =  = 3.646 ´ 10^–3

                \ Meq. of Zn²⁺ lost = 3.646

Initial Meq. of Zn²⁺ = 300 ´ 0.160 ´ 2 = 96

\ Meq. Of Zn²⁺ left in solution = 96 – 3.646 = 92.354

\ [ZnSO₄] =  = 0.154 M

 

4. Total charge passed = (40 ´10^–6 ´ 32 ´ 60) C =  0.0768 C

Ag⁺ + e^– ® Ags

So, weight of Ag deposited at cathode surface    = gm

= 8.6 ´ 10^-5 gm

                Now, total no. of Ag atoms deposited        =

= 4.8 ´ 10¹⁷

So, surface area of cathode which is covered by Ag atoms

= (4.8 ´ 10¹⁷ ´ 5.4 ´ 10^–16) = 259.2 cm²

\Total surface area         = cm²

                                                                        = 602.8 cm²

5. We need to obtain E° for the couple,
6. Co³⁺ (aq) + 3e^– ¾¾® Co(s)

From the values of E° for the couples

1. Co³⁺(aq) + e^– ¾¾® Co²⁺ (aq)           = 1.81V
2. Co²⁺(aq) + 2e^– ¾¾® Co(S)              = –0.28V

We see that 3 = 1 + 2, therefore

E₃ =  = 0.42 V

RHE reaction: Co³⁺ (aq) + 3e^– ¾¾® Co(s)

LHE reaction:

3Ag(s) + 3Cl^–(aq) ¾¾® AgCl(s) + 3e^–

RHE – LHE  : Co³⁺(aq) + 3Cl^–(aq) + 3Ag(s) ¾¾® 3AgCl(s) + Co(s)

E° =  –

            = (0.42 – 0.22)V = 0.20V

6. i) Temperature coefficient =

=  = -3 ´ 10^–4 VK^–1

 

1. ii) at 293 K

DG = –nFE_cell [Q E_cell at 293 K = 0.2699 volt]

=– 2 ´ 96,500 ´ 0.2699 J = – 52.09 kJ

DH_293K = nF

= 2 ´ 96,500 [293 ´ -3 ´ 10^–4 – 0.2699] = – 69.055 KJ

DS_{at 293 K} =

= 2´ 96,500 ´ – 3 ´ 10^–4 = – 57.9 JK^–1

 

7. Disproportionation reaction

Cu⁺® Cu²⁺ +e                     E°= -0.15V

 

E_cell=E°_cell–

At eqm, E_Cell= 0

\ 0 = (x-0.15) –

\x =0.15+0.0591 log(1.8´10⁶)

\ x = 0.519 V

8. Hint:  +
9. Given D G°=-nFE°

= – 12 ´ 2.73´96500 J

= – 3.163´10³ KJ

[4 Al° ® .4 Al³⁺+12e

3O₂^° + 12e ® 6O^2-

D G°=4.6 ´G_f°_[OH]–

Þ 3.163´10³ =4´6´(-237.2)-4´(-157)

\ =1303 KJ.Mol^-1

 

10. Quantity of electricity passed = 2 ´ 11 ´ 60 ´ 60

= 79200 coulombs  =  = 0.8207 F

Amount of Ni(NO₃)₂ undergoing electrolytic dissociation  = 0.8207 equiv.

= 820.7 milli equiv.

Amount of dissolved Ni(NO₃)₂ before electrolysis = 500 ´ 2 =  1000 millimole

= 2000 milli equiv.

(Q Total units of cationic charge or anionic charge  = 2 \ E = )

Amount of dissolved Ni(NO₃)₂  after electrolysis             = 2000 – 820.7

= 1179.3 milli equiv

Normality  =  =  = 2.3586

\ Molarity of solution after electrolysis  = Normality /2  = 1.1793

 

11. Current in ampere =  =

Quantity of electricity passed  = ampere ´ second  =  ´10 ´ 60 ´ 60

= 32727.272 Coulombs  =  = 0.339 F

Amount Cd deposited = 0.339 equiv. = 0.339 ´ 56.2  = 19.05 g

(Atomic mass of Cd = 112.4 and Valency  = 2. So Eq. wt. = 112.4 /2  = 56.2)

 

12. Atomic mass of Mg is 24 and its valency is 2 so its Eq. wt. will be 24/2 i.e. 12.

Number of equiv. of Mg to be produced per hour  =   = 4000

Quantity of electricity required per hour  = 4000 F

= 4000 ´ 96500 = 3.86 ´ 10⁸ coloumbs

Ampere required per hour  =  = = 107222.22

Taking into account 75% current efficiency,

The current required per hour  =

= 1.42962 ´ 10⁵  coulombs

13. Quantity of electricity passed = 25 ´ 24 ´ 60 ´ 60

= 2160000 coloumbs =  = 22.383 F

Half reactions:

H⁺  +  ¾¾®  (at cathode)

4OH^–    ¾¾® 2H₂O + +

Volume of H₂(g) evolved at STP = 11.2 ´ 22.383  = 250.689 L

Volume of O₂(g) evolved at STP  =  ´ 22.383  = 125.344 L

Volume of H₂(g) evolved at 27°C and 740 mm =              = 282.92 L

Volume of O₂(g) evolved at 27°C and 740 mm Hg  = 141.463 L

14. a) Electrolytic half reactions:

2H₂O   ¾¾® 4H⁺      +          O₂        +          4e^–

or                                                                      (at anode)

4OH^–   ¾¾® 2H₂O   +          O₂        +          4e^–

Cu²⁺    +          2e^–       ¾¾®             Cu(s)               (at cathode)

Note that  is non – dischargeable ion.

Quantity of electricity passed  = ampere ´ second

= 10 ´ 1.5 ´ 60 = 900 Coloumbs

=  = 9.326 ´ 10^–3 F

Weight of Cu deposited at cathode  = 9.326 ´ 10^–3 equiv.

=  g

(Q Eq. wt. of Cu = At mass  / valency = 63.5/2)

Weight of O₂(g) evolved at anode  = 9.326 ´ 10^–3 equiv

= 9.326 ´ 10^–3  ´ 8 g   = 0.0746 g

\ Weight of resulting solution  = 10 –(0.26 + 0.0746)            = 9.6294 g

9. b) Number of equiv. of acid or H⁺ ion produced  = 326 ´ 10^–3
10. 12% Cd – Hg amalgam means 100 g of amalgam contains 88 g Hg and 12 g Cd. So, 2 g Hg must be associated with g Cd i.e, 0.275 g Cd so as to have an amalgam of desired composition.

 

w = z.i.t          = i.t. =  i.t.

t=   =  = 93.75 Seconds

Level – II

 

1. i) Given cell is Fe_(s) | FeO_(s) | KOH_(aq) | Ni_(s)

                  Cell                       Reaction

L.H.S. half cell      Fe_(S) + 2OH^–_(aq) ¾® FeO_(s) + H₂O_(l) + 2e     E⁰_OX = 0.87V

R.H.S. half cell      Ni₂O₃ + H₂O_l) + 2e ¾® 2NiO_(s) + 2OH^–_(aq)   E⁰_red = 0.40V

–––––––––––––––––––––––––––––––––––––––––––––––

Net                                                                                                Fe_(s) + Ni₂O_3(s) ¾® FeO_(s) + 2NIO_(s) E_cell = 1.27V

1. ii) E⁰_cell = E⁰_ox + E⁰_red = 1.27V

Since net reaction does not contain OH^– hence emf of cell does not depend on concentration of KOH

iii)   Maximum amount of electrical energy

DG⁰ = – nFE⁰ = – 2 ´ 96500 ´ 1.27

= 2.45 ´ 10² kJ/mol

2. Cu²⁺+ 4NH₃ [Cu(NH₃)₄]²⁺, K_f = 1 ´ 10¹² =

x            2                1

\ x = 6 ´ 10^–14M

Due to high value of K_f almost all Cu²⁺ ions are converted to Cu(NH₃)₄²⁺ ion

E_cell = E⁰_cell –  =

= 0.76 + 0.34 –  = 0.71V

 

3. For the change 2Fe³⁺ + 3I^– 2Fe²⁺ + I₃^–

E = E⁰ –  At equilibrium E = 0, Q = K_C

O = E⁰ –  =

Þ = 0.77 – 0.54 = 0.23V

\logK_C =  \ K_C =  6.26 ´ 10⁷

or  = ,          (Au³⁺ + 3e ¾¾® Au

(Cu²⁺ + 2e ¾¾® Cu)

or W =

= 4.7629 g

Again

W =

\C =              =

= 0.804 ampere

 

5. Half reactions:

O₂ + 2H₂O + 4e^–  ¾¾® 4OH^– (at cathode)      Net cell reaction

2H₂ + 4OH^– ¾¾® 4H₂O + 4e^–  (at anode)        2H₂ + O₂ ¾® 2H₂O

Number of moles of H₂ reacting  =  = 3

Equivalent of H₂ reacting  = 3 ´ 2  = 6

Now,   =

6  =

\ i =  = 643.333 ampere

Equiv. of H₂ = Equiv. of Cu formed

\ Wt. of Cu deposited  =   = 190.5 g

6. Net cell reaction

Pb(s)+ SO₄^2-_(aq)® Pb SO₄+2e                             E°=0.31 V

Pb(s)+ 4H⁺+SO₄^2-_(aq) +2e®PbSO_4(s)+2H₂O       E°=1.70 v

E_cell = E°_cell^–

[SO₄^2-]=2.0M        [H⁺]=10^-1

E_cell for spontaneous reaction can be calculated.

 

7. Total charge passed        = 0.193 ´ 60 ´ 60 C

= 694.8 C

But, charge is used up

due to 90% current efficiency = 694.8 ´  = 625.32 C

At anode    :           2Cl^– ¾® Cl₂ + 2e^–

At cathode :           2Na⁺ + 2H₂O + 2e^–  ¾® 2NaOH + H₂

Number of moles of NaOH produced during entire electrolysis =

= 6.48 ´ 10^–3

So, molarity of NaOH       = = 6.48 ´ 10^–5 M

\ [OH^–] = 6.48 ´ 10^–5 M

pOH    = – log [OH^–] = – log (6.48 ´ 10^–5) = 4.188

\ pH   = 14 – 4.188

                        = 9.812

8. Hint: E_cell =

[H⁺] = Ca =  =

9. Cell Reaction:

LHS:          Cd ® Cd²⁺+2e                        E°=0.40v

RHS:

E_cell = E°_cell –

At eqm E_cell=0,      K=

\ 0=0.40 –

Þ 0=0.40 –

Thus,  Can be calculated

 

10. Co⁺² + 6CN^– [Co(CN)₆]^–4

 

and Co⁺³ + 6CN^–  [Co(CN)₆]^–3

 

\

[Co(CN)₆]^–4 ¾® [Co(CN)₆]^–3 + e^–          E° = + 0.83 V

Co⁺³ + e^–  ¾® Co⁺²                                E° = 1.82 V

Net cell reaction:   Co⁺³ + [Co(CN)₆]^–4  ¾® Co⁺² + [Co(CN)₆]^–3

= (0.83V + 1.82V)

at equilibrium E_Cell = 0

\

0.99 =

 

as K₁ = 1 ´ 10¹⁹

K_­2 = 8.23 ´ 10⁶³

\ [Ag⁺] _LHS=

[Ag⁺] [Br]= Ksp _{(Ag Br)}=3.3´10^-3

                \ [Ag⁺]_RHS =

Cell Reaction :

LHS       Ag ® Ag⁺+e                   E°_Ag/Ag+= xv

 

E=

E=0-

To make E_cell positive, LHS, should be cathode, (+ve)-half Cell

 

12. Cell reaction is       Zn(s)    +    Ni⁺²    Zn⁺² + Ni(s)

t = 0                                               1.5             0          0

at equilibrium                                 (1.5–x)       x         –

At equilibrium E_cell is 0

\

=  log

0.76 – 0.24 =  log

= log       ( Where 1.5–x = a \ x = 1.5–a)

log

= 4.6 ´ 10^–18

but 1.5 >>> a

\ a = 1.5 ´ 4.6 ´ 10^–18 M  6.9 ´ 10^–18 M » 7 ´ 10^–18 M

And, weight of Zn consumed = 1.5 ´ 0.75 ´ 65 gm  = 73.125 gm

 

13. Volume of Cu deposited on plate = 10 ´ 10 ´ 10^–2  = 1cm³

Wt. of Cu deposited on plate  = volume ´ density  = 1 ´ 8.94  = 8.94 g

W =  =

\Q =  = 27171.96 Coulombs

14. Electrolytic dissociation of SnCl₂:

Sn⁺²     +    2e^–       ¾®     Sn

2Cl^–     ¾® Cl₂ + 2e^–

The Cl₂(g) formed at anode reacts with SnCl₂ to give SnCl₄ since nothing is given out the cell

SnCl₂ + Cl₂  ¾® SnCl₄

No. of equiv. of SnCl₂ decomposed electrolytically.

= No. of equiv. of Cl₂  formed  = No. of equiv. of Sn formed  =  = 2 ´ 10^–3

Again, No of equiv. of SnCl₂ converted into SnCl₄.

= No. of equiv of SnCl₄ formed  = No.of equiv of Cl₂ formed  = 2 ´ 10^–3

So,

Total loss in the number of  equiv. of SnCl₂  = 2 ´ 10^–3  + 2 ´ 10^–3 = 4 ´ 10^–3

No. of equiv. of SnCl₂ left in solution  = No. of equiv. of SnCl₂ initially taken  – No. of equiv  of SnCl₂ lost.

=  – 4 ´ 10^–3 = 0.2 – 0.004            = 0.196

=  = 71.340

 

15. No. of Faradays passed =  = 0.149

Fe³⁺     +    e^–         ¾¾® Fe²⁺

After complete conversion of Fe³⁺ to Fe²⁺, the number of Faradays left

= 0.149 – 0.1  = 0.049

Fe²⁺     +    2e^–                   ¾¾® Fe

Number of equiv. of Fe formed  = 0.049

Number of moles of Fe formed  =  = 0.0245

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Level – III

 

1. Cell reaction: At cathode: 2H⁺+ 2e ¾® H₂

At anode: 2OH^– ¾® H₂O + 2e + O₂

Equivalent of H⁺ and OH^– will be discharged at anode and cathode respectively.

= moles of (H⁺) or OH^– =  = 1.65 ´ 10^–1 M

For buffer mixture at anode (H⁺) will increase by 1.65 ´ 10^–1 M

HPO₄^2– + H⁺  H₂PO₄^–               pH = pK_a + log  = 2.15 + log  = 2.005

1            0.165        1

0.835       0             1.165

For buffer mixture at cathode (OH^–) will increase by 1.65 ´ 10^–1 M

H₂PO₄ + OH^–  HPO₄^2– + H₂O

1           0.165             1

0.835   ––                 1.165

pH = 2.15 + log  = 2.295

 

2. Given V²⁺ + VO²⁺ + 2H⁺ ¾® 2V³⁺ + H₂O         E⁰_cell  = 0.616V

V³⁺ + Ag⁺ + H₂O ¾® VO²⁺ + 2H⁺ + Ag_(s)              E⁰_cell = 0.439V

Also     Ag_(s) ¾® Ag⁺ + e                                      = – 0.80V

Adding these equation

V²⁺ = V³⁺ + e                      E⁰ = 0.255V

\V³⁺ + e = V²⁺                   E⁰ = – 0.255V

 

3. Daniel cell is

Zn/Zn²⁺ (1M) || Cu²⁺ (1M) | Cu,  = 1.10V

Therefore, the cell reaction is,

Zn        +    Cu²⁺      Cu +    Zn²⁺

1M                               1M             Initial concentration

–x                                +x              Change concentration

(1–x)M                              (1+x)M      Final concentration

equilibrium concentration

Where x is the mole of Cu²⁺ discharged

E_cell  =  –  log  =   log

First, we determine the value of x, due to passage of 0.100 MC (Mega coloumb) of charge,

Cu²⁺ + 2e^– ¾¾® Cu

0.100MC = 0.100 ´ 10⁶C.

W(Cu) = Qz          = 0.100 ´ 10⁶ ´

Mol of Cu²⁺ discharged  =  =  = 0.518 mol

[Cu²⁺]_eq =  1 – x  = 0.482 M

[Zn²⁺] = 1 + x = 1.518 M

E_cell = 1.10 –  log  = 1.10 – 0.015 = 1.085 V

 

4. K_sp(AgI = [Ag⁺] [I^–]

Ag_(s) ¾® Ag⁺ + e              Anode

AgI + e ¾® Ag_(s) + I^–         Cathode

––––––––––––––––––––––––––––

AgI_(s) ¾® Ag⁺  + I^–

E_cell = E⁰_cell  –

At equilibrium E_cell = 0, E⁰_cell = – 0.80  – 0.15 = – 95V

log[Ag⁺] [I^–] =  \ K_sp(Ag) = 8.4 ´ 10^–17

 

5. In this case, Ag(l) ion collects as solid AgCl on the electrode itself. However, there is some Ag⁺ in equilibrium with AgCl(s) in solution.

Knowing that K_sp for AgCl is 1.0 ´ 10^–10. Calculate E for Ag/AgCl electrode immersed in 1.00M KCl at 25°C.

= 0.799 V

\ K_sp = [Ag⁺] [Cl^–]

[Ag⁺] =  =

Ag⁺ + e^– ¾¾® Ag

E = E° –  log  = 0.799 – 0.591 = 0.208 V

 

6. Half cell Reaction                                  E°

LHS           H₂ ¾¾® 2H⁺ + 2e^–

RHS          Zn²⁺ + 2e^– ¾¾® Zn

——————————

Net reaction          Zn²⁺ + H₂ ¾¾® Zn + 2H⁺    E_cell = –0.76 V

K =

E_cell =   log

[H⁺] = 4.6 ´ 10^–6 M

H⁺   +

0.4M          4.6 ´ 10^–6 M    6.4 ´ 10^–3 M

K_a =  =

= 7.36 ´ 10^–8

            \ pK_a = 7.13

 

7. Electrolytic decomposition reaction

Cu²⁺    +    2e^–       ¾¾® Cu(s)   (at cathode)

H₂O           ¾¾® 2H⁺      +          O₂ + 2e^– (at anode)

After the complete discharge of Cu^{2 +}, ion the half reaction at anode would continue to occur the same but the cathode half reaction will now occur as

2H₂O + 2e ¾¾® H₂ + 2OH^–      (at cathode)

Amount of Cu deposited  =  = 6.299 ´ 10^–3

Amount of Oxygen liberated  =  ´ 6.299 ´ 10^–3  = 3.145 ´ 10^–3 mol

Quantity of electricity passed after discharge of Cu²⁺ = 1.2 ´ 7 ´ 60

= 504 coulombs

Amount of electrons carrying this much of electricity

=  = 5 ´ 22 ´ 10^–3 mole

Amount of O₂(g) liberated  = 1/4  ´ 5.22 ´ 10^–3  = 1.305 ´ 10^–3 mole

Amount of H₂(g) liberated  =  ´ 5.22 ´ 10^–3  = 2.61 ´ 10^–3 mole

Total amount of gases evolved during the entire course

= (3.145 +1.305 + 2.61)´10^–3

= 0.00706 mole

Volume of the gases evolved at STP  = 0.00706 ´ 22400           = 158.14 ml

 

8. Eq of H₂ produced = Eq of O₂ produced = Eq of H₂S₂O₈ formed= Quantity current passed

Quantity of current passed= Eq of H₂=

=0.87´96500C=83955C

=´Eq wt of H₂S₂O₈

 

9. Given

I    + 8H⁺ + 5e^–  ¾¾® Mn⁺² + 4H₂O, DG°  = – 5 ´ 1.51 F

II  MnO₂ + 4H⁺  + 2e^–  ¾¾® M⁺² + 2H₂O, DG° = –2´1.23 F

Eq. I  – Eq.II

+ 4H⁺ + 3e^– ¾¾® MnO₂ + 2H₂O,  DG°  = –5.09 F

\  = + 5.09 F

or  =  = 1.69 V

11. All Ag⁺ (aq) from AgCl in each of the cell is complexed to form [Ag_m(NH₃)_n]^m+

                      Cell      Reaction:

mAg + nNH₃ ® Ag _m(NH₃)_n^m+ + me                          E° _ox=EV

E° _ox=EV

K=

 

E_cell= 0.1185 V

E_cell = E°_cell –  \ m=1

similarly insecond cell, using equation (1)

K=

E_cell =0.1263 v,     m=1

0.1263=0-

n=2

12. 2 Hg + 2Fe³⁺      H_g2²⁺           + 2Fe³⁺

1´10^-3                0                   0

 

E_cell = E°_cell – ;  At Eqm. E_cell=0

\0 = E° Hg/Hg²⁺₂ + 0.77-

 

13. Hint: The equilibrium constant for the formation of complex ion is called stability constant

Ag⁺ +    Ag (S₂O₃)^–3

K_S =

14. For the cell

AgçAgCl 0.05M KCl çç0.05 m AgNO₃çAg

E_cell= E°_cell–

Þ 0.788 v =0-

(Ag⁺) _LHS   can be calculated

Ksp_AgCl= [Ag⁺]_LHS´[Cl^–]

 

15. Hint: [H⁺] _L.H.S. =

[H⁺]_R.H.S. =

 

 

 

 

    Objective Problems

 

Level – I

 

1. \ (A), (B) & (C)

 

 

2. 2Tl + Cu⁺² ® 2Tl⁺ + Cu

 

\ (A) & (d)

 

 

3. Since, Cu⁺ is a oxidised in Cu²⁺ and reduced in Cu

         I₂ is oxidised  to I⁵⁺ and reduced to I^–                      

         \(A) & (B)

 

4. Zn + MgCl₂ ® ZnCl₂ + Mg

\

= + 0.762 – 2.37 =  – 1.608 V

Here, is negative so no reaction will take place.

\  (A)

 

5. Cell reaction : 3Ni + 2Au⁺³ ¾® 3Ni⁺² +  2 Au

 

= (0.25 + 1.5) –

= 1.75 –

= 1.75 + 0.295

= + 1.7795 V

\ (B)

 

6. CuSO₄ + 2e^– ¾® Cu +

         Bi₂(SO₄)₃ + 6e^– ¾® 2Bi +

AlCl₃ + 3e^– ¾® Al + 3Cl^–

AgNO₃ + e^– ¾® Ag +

Since, in 1M CuSO₄ solution, 1M Bi₂(SO₄)₃  solution, 1M AlCl₃ solution and 1M AgNO₃ solution, 2 mole electron, 6 mole electron, 3 mole electron are needed to deposit Cu, Bi, Al and Ag at the cathode respectively.

But one mole electron = 1F electricity

That’s why number of Faraday required to deposit 1M of each CuSO₄, Bi₂(SO₄)₃, AlCl₃  and AgNO₃ solution are 2F, 6F, 3F and F respectively

         \ (A)                                              

 

7. H₂¾® 2H⁺ + 2e^–

(1atm)    (1M)

         \  –

= 0.00 –  = 0.00 V

Now, pressure is increased from 1atm to 100 atm.

\  –

= 0 –            ([H⁺] is constant )                                                                    =  = 0.0591 V

         \(A)     

                                            

 

8. At anode 2H₂O   ¾® O₂ + 4H⁺ + 4e^–

         At cathode    Ag⁺ + e^– ¾® Ag

         \(B)

  

9. Ag|Ag⁺|| Agl|Ag

          = – 0.799 – 0.151= – 0.950 V

         \(B)     

 

10. At Cathode: FeSO₄ + 2e^– ¾® Fe +

                        Fe₂(SO₄)₃ + 6e^– ¾® 2Fe +

Fe(NO₃)₃ + 3e^– ¾® Fe +

In the reaction, one mole of each FeSO₄, Fe₂(SO₄)₃ and Fe(NO₃)₃ required 2,6 and 3 mole electron respectively

But 1 mole electron º 1F electricity

When the same quantity of charge (i.e., 1F) is passed in each cell, then the number of moles of iron which will be deposited are 1/2 , 1/3 and 1/3 respectively for each cell reactions.

Now,  at anode                              2H₂O ¾® O₂ + 4H⁺ + 4e^–

So, same gas will be evolved in each cell.

            \ (B), (C) & (D)  

 

 

 

            Level – II

1. Cu²⁺ + 1e^– ¾® Cu⁺                     E₁ = 0.15                     …(i)

Cu⁺ + 1e^– ¾® Cu                                    E₂ = 0.50                     …(ii)

Cu²⁺ + 2e^– ¾® Cu                      E₃ = ?                          …(iii)

Clearly (iii) = (i) + (ii)

=  +

2 ´ F ´ E₃  = 1 ´ F ´ E₁ + 1 ´ F ´ E₂

E₃ =  = 0.325V

            \(A)  

4. For H₂SO₄  2H⁺ +

2 ´ 96500 C liberates 1 moles of H₂

For HCl ¾® H⁺ + Cl^–

96500 c liberates  mole of H₂

and therefore 2 ´ 96500 C liberates 1 mole of Hydrogen.

            \(B)  

5. Anode : H₂(g) ¾® H⁺(ag)  + e^–

Cathode: AgCl(s) + e^– ¾® Ag(s) + Cl^–(aq)

———————————————————

H₂(g) + AgCl(s) ¾® H⁺(ag) + Ag(s) + Cl^–(aq)

            \(C)  

6. Sine E°Cd²⁺/Cd > E°Zn²⁺/Zn , therefore Zn electrode acts as anode and Cd electrode as cathode.

Anode: Zn(s) ¾® Zn²⁺ (aq) + 2e^–

Cathode : Cd²⁺ + 2e^– ¾® Cd (s)

—————————————————

Zn(s) + Cd²⁺(aq) ¾® Zn²⁺(aq) + Cd(s)

Q = [Zn²⁺] / [Cd²⁺]  = 0.004/0.2

= –0.403V – (–0.763V) = 0.36V

E_cell= 0.36 – V log

            \(B)  

7. pH = 14 ; \ pOH = O, (or) [OH^–] = 10^–0 = 1

Therefore [Cu²⁺] =  = 1 ´ 10^–19

= –  log  = 0.34 V – ´ 19 = –0.22V

         \(A)     

8. E_cell =  –  log

E₁=  –  log  =

E₂ = –  log

E₃ =  –  log

\ E₂ > E₁ > E₃

            \(D)  

9. Negative electrode potential (reduction potential) indicates lesser tendency for the reduction. Hence a is readily oxidized.

            \(B)  

10. The more negative the electrode potential, the lesser the tendency of the metal to undergo reduction and therefore metal would act as stronger reducing agent.

            \(A)  

11. The amount of electricity passed  = I t (in sec.) = 0.6A ´ 7 ´ 60 Sec.

= 252C ´  = 2.6 ´ 10^–3F

            \(B)  

12. 22,400 ml H₂ at STP = 1 mol H₂

\ 450 mL H₂ at STP =  ´ 1 mole H₂ = 0.02 mol

Electrode reaction:

2H⁺ +   2e^– ¾® H₂

2F       1 mol

So, the liberation of 0.02 mole H₂ requires = 2 ´ 96,500 ´ 0.02C = 3,860 C

Time required to produce 0.02 mole H₂ = ==1.07h» 1 hour

            \(B)  

13. DG° =  –nFE° cell

if  is positive, then DG° will be –ve showing that cell reaction is spontaneous.

            \(B)  

14. Al is above hydrogen in the electrochemical series, therefore Al³⁺ has lesser reduction tendency as compared with H⁺. Hence, hydrogen electrode acts as anode when coupled with aluminium electrode.

=

\ 1.66V = 0.0V –

= – 1.66V

         \(A)     

15. E_cell =  log . Increase in the concentration of Ag⁺ decreases the value of log . This results in the increased value of E_cell.

         \(C)