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Electro Chemistry & Electrochemistry Solutions

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°.

  1. 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.

  1. 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:      SO42–, NO3, OH, Cl, Br, I

  1. 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:              WCu =

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

                              CrO3(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)   Wcr= 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.

  1. 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 , Ecell = ERP(cathode) – E­RP(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.

  1. 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
  1. 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+ + 5eMn+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
+e¾®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.

  1. 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.

  1. 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.

  1. 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 EOX = – Ered , like .

  1. 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 + 2e2Cl       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 E0cell 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

Ecell = E0cell

Q Ecell = 0 and [Cu2+] = 0.01 M

\ o =  – log

Þ 0 = (0.799 – 0.337) –

\ [Ag+] = 1.477 ´ 10–9 mol/lit.

  1. 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 Ecell or  is positive then DG° is negative and the reaction is spontaneous. For non – spontaneous reaction, Ecell 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) Ecell = (Reduction potential of the right hand side electrode) – (Reduction potential of left hand side electrode).
  3. b) Ecell = (oxidation potential of LHE) — (oxidation potential of RHE)
  4. c) Ecell = (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 (–)
  1. Application of EMF-Measurement
  2. Determination of Equilibrium constant:

From Nernst equation Ecell = E0cell – logQ (Q = Reaction Quotient)

At equilibrium Ecell = 0, Q = Keq (equilibrium constant)

\ E0cell = logKeq Þ E0cell = log Keq

\ log Keq =

  1. Determination of solubility product: Solubility product of sparingly soluble salts can be evaluated from EMF measurement. Suppose we want to determine Ksp 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    E0OX = – 0.80V

Cathode    :     Ag+ (in given solution C) + e ¾® Ag       E0cell = + 0.80V

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

Cell reaction    Ag+(C) ¾® Ag+ (unknown)                     E0cell = 0.00V

\ Ecell = E0cell

Hence Þ Ecell =                         …(1)

At equilibrium Ecell = 0

From equation (Ag+) can be calculated

= [Ag+] [Cl] = [Ag+]2

  1. 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 Ecell is measured

 

 

[H+] =  , (Oswald dilution law)

Ka can be calculated

  1. 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 Hg2Cl2(S) + 2e ¾® 2Hg(l) + 2Cl        E0red = 0.2676

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

H2(g) + Hg2Cl2(s) ¾® 2H+(aq)+ 2Hg(l) + 2Cl(aq)      E0cell = 0.2675V

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

Ecell = E0cell – log[H+]2

Þ Ecell = E0cell  + 0.0591 pH

\ pH =

  1. Determination of Thermodynamic data:

DG = DH + T  (Gibbs – Helmhotlz Equation)

Þ – nFEcell = DH – nFT (DG = – nFE)

DH = – nFEcell + 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

  1. 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)           E0Cell = 0.00V

Ecell = E0cell  log

\ Ecell =  log

In general Ecell =

For spontaneous cell reaction C2(RHS) > C1(LHS)

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

Ecell = 0.0591

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

Pt(Cl­2) | Cl(C1) || Cl(C2) |  Pt(Cl2) 1 atm

E0cell = log

For spontaneous cell reaction C1 > C2

  1. 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

Ecell= E0cell

\ Ecell = 0.0591 log

For spontaneous cell reaction P1 > P2

For concentration cells E° = 0.

  1. 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                          E0OP = – 0.15 V

Cu+ + e ¾® Cu                           E0RP = 0.50 V

––––––––––––––––

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

\ E0 +  = – 0.15 + 0.5

= 0.35V

  1. 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

 

 

 

 

 

 

 

 

  1. 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%)

iot (Q) = 115800 coulomb

Consumption of energy = Q ´ N =  = 347.4 kJ

Problem 2:       An aqueous solution of NaCl on electrolysis gives H2(g), Cl2(g) and NaOH according to reaction

                        2Cl + 2H2O ¾® 2OH + H2(g) + Cl2(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 ´

  1. 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 Ksp(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

Ecell = E0cell

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

\ [Ag+] = 3.2 ´ 10–2

Ksp(Ag2SO4) = (Ag+)2(SO42–) = (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 = DG10‑ – DG20

\ – 3E30‑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) + SO42– ¾® PbSO4   Discharging

                              PbO2(s) + 4H+ + SO42– + 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

\ Ecell = E0cell –  = 0.34 – (– 0.76) –

\ Ecell= 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.        H2(g) ¾®                            Eo = 0.00V

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

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

H2(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

Ecell = E0cell

0.67 = 0.28 + 0.0591 pH

Ecell =

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, Ecell = 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.

                        Ka for CH3COOH = 1.8 x 10-5  and Kb 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® NO2(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.

ERP = E0RP –  = 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

ERP = E0RP –  = 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 ERP for the couple Cu2+/Cu is

ERD = E0RP  –

= 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 O­2 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,

Ered =  –  log        Let [H+]initial = X

Ered(initial) =      [H+]final =

Ered(final) =

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

This Ered 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, EAg =  )

\             (Cu+2 + 2e ® Cu, ECu =  )

=

\ (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)  Ecell  = 0.77V

                        (D)  Ecell = – 0.77V

Solution:        E0RP for M  > DERP for X

\ M+ + X ¾® M + X (spontaneous), E0cell = – 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)

 

 

  1. 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 + 2S2O32– ¾® S4O62– + I

What is current efficiency?

  1. 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.
  2. 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?
  3. 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).
  4. 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

  1. 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
  2. Show that = –
  3. 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]).

  1. 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 ?
  2. 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?
  3. 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.
  4. 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?
  5. 10 g fairly concentrated solution of CuSO4 is electrolysed by passing 10 ampere current for 1.5 minutes. Calculate.
  6. a) the weight of resulting solution
  7. b) the number of equivalents of acid or alkali produced in solution  (Cu = 63.5)
  8. 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

Ni2O3(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.

  1. 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.
  2. 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.
  3. 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)
  4. 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.
  5. 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 Ecell of the spontaneous reaction at 298K, given   = 2.0M, pH = 1 and other species are at  unit concentration.

  1. 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.
  2. 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 Ka of HOCN,   = 0.80 V

  1. Given the cell

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

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

  1. 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

  1. 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 KSP (AgCl) = 2.8 ´ 10–10 , KSP (AgBr) = 3.3 ´ 10–13.

  1. 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)
  2. 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.
  3. 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).
  4. 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? pKa for H2PO4= 2.15.
  2. Two electrochemical cells are assembled in which these reactions occur.

V3+ + Ag+ + H2O ¾® 2V3+ + H2O                      E0cell = 0.66V

V3+ + Ag+ + H2O ¾® VO2+ + 2H+ + Ag(s)           E0cell= 0.439 V

If  = 0.80V, calculate E0 for half cell reaction

V3++ e ¾® V2+

  1. 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

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

= – 0.15 Volt at 25°C,

  1. Knowing that Ksp 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
  2. 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 pKa of  (i.e.,) for the equilibrium
H+ +

  1. 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.
  2. 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.
  3. Find  from the following data  = 1.51 V and  = 1.23 V
  4. 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
  5. 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

Ecell = 0.1185V at 298K

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

Ecell = 0.1263V at 298K

  1. 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

  1. 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

  1. The EMF of cell

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

Find the solubility product of AgCl.

  1. Calculate the EMF of the following cell

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

Ka for C2H5COOH = 1.4 ´ 10–5; Kb for NH4OH = 1.8 ´ 10–5

 

 

 

 

 

 

 

 

 

 

 

 

  1. 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.

  1. For the cell Tl½ Tl+ (.001M) ½½ Cu2+ (.1M) ½ Cu. ECell 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

  1. 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

  1. The quantity of electricity needed to electrolyse completely 1 M solution of
    CuSO­4, 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

  1. 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

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

(A) NO2                                                     (B) O2

(B) H2                                                        (D) N2O

  1. 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

  1. 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.

  1. 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

  1. 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
  1. 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 =

  1. 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

  1. 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

  1. Which of the following graphs will correctly correlate Ecell 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: Ecell

(A)

 

(B)

 

(C) (D)
  1. The amount of ion discharged during electrolysis is directly proportional to :

(A) Resistance                                          (B) Time

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

  1. 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.
  1. 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

  1. 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

  1. A solution is one molar in each of NaCl, CdCl­2, 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

  1. 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

  1. 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.

  1. 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

 

  1. 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)]

 

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

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

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

  1. The emf of the following three galvanic cells:
  2. Zn/Zn2+ (1M) || Cu2+(1M)/Cu
  3. Zn/Zn2+ (0.1M) || Cu2+(M) / Cu
  4. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. 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 Ecell positive, polarity of the half cells should be reversed)

  1. 7 ´ 10–18 M, 73.125 gm 13.        27171.96 coloumb
  2. 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

 

 

 

 

  1. 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

 

  1. Meq. of Na2S2O3 required after passage of current

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

 

  1. DG = – nFEcell, DH = nF

DS =

 

  1. DG0= DG0P – DG0R

DG0= – nfE0

 

Level – II

 

  1. Cu2+ + 4NH3 [Cu(NH3)4]2+

x          2                    1

Kf =  = 1 ´ 1012

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

 

  1. 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)

 

  1. Cell reaction: Pb(s)+ SO42–(aq.) ¾® PbSO4(s)+ 2e                           E0 = 0.31V

PbO2 + 4H+ + SO42–(aq) + 2e ¾® PbSO4(s) + 2H2O      E0 = 1.70 V

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

Net cell reaction    PbO2 + 4H++ SO42– ¾® PbSO4(s) + 2H2O

Q =

 

  1. Co2+ + 6CN [Co(CN)6]4–                 Ki = Kf = 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

 

  1. 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

Ecell = E0cell      –

Calculate H+

HSO3  H+ + SO32–

0.4 M           0       0

(0.4 M – x)        x         x

x = [H+] = [SO32–]

 

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

MnO4¾® 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)

 

  1. All the Ag+(aq) from AgCl in each of the half cells is complexed to form
    Agm [NH3nm+

For given cell

L.H.S.        mAg(s) + nNH3 ¾® Agm(NH3)nm+ + me       E0ox= EV

LHS           LHS

R.H.S.       Agm(NH3)nm+ + me  ¾®mAg(s) + nNH3        E0red = –EV

RHS                                   R.H.S.

–––––––––––––––––––––––––––––––––––

Agm(NH3)nm+(RHS) + nNH3(LHS) ¾® Agm(NH3)nm+ + nNH3(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

WAg = 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

 

  1. 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%

 

  1. (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

 

  1. 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

  1. We need to obtain E° for the couple,
  2. 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

  1. i) Temperature coefficient =

=  = -3 ´ 10–4 VK–1

 

  1. ii) at 293 K

DG = –nFEcell [Q Ecell at 293 K = 0.2699 volt]

=– 2 ´ 96,500 ´ 0.2699 J = – 52.09 kJ

DH293K = nF

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

DSat 293 K =

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

 

  1. Disproportionation reaction

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

 

Ecell=E°cell

At eqm, ECell= 0

\ 0 = (x-0.15) –

\x =0.15+0.0591 log(1.8´106)

\ x = 0.519 V

  1. Hint:  +
  2. Given D G°=-nFE°

= – 12 ´ 2.73´96500 J

= – 3.163´103 KJ

[4 Al° ® .4 Al3++12e

3O2° + 12e ® 6O2-

D G°=4.6 ´Gf°[OH]

Þ 3.163´103 =4´6´(-237.2)-4´(-157)

\ =1303 KJ.Mol-1

 

  1. 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

 

  1. 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)

 

  1. 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

  1. 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

  1. 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

  1. b) Number of equiv. of acid or H+ ion produced  = 326 ´ 10–3
  2. 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     E0OX = 0.87V

R.H.S. half cell      Ni2O3 + H2Ol) + 2e ¾® 2NiO(s) + 2OH(aq)   E0red = 0.40V

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

Net                                                                                                Fe(s) + Ni2O3(s) ¾® FeO(s) + 2NIO(s) Ecell = 1.27V

  1. ii) E0cell = E0ox + E0red = 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

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

x            2                1

\ x = 6 ´ 10–14M

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

Ecell = E0cell –  =

= 0.76 + 0.34 –  = 0.71V

 

  1. For the change 2Fe3+ + 3I 2Fe2+ + I3

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

O = E0 –  =

Þ = 0.77 – 0.54 = 0.23V

\logKC =  \ KC =  6.26 ´ 107

or  = ,          (Au3+ + 3e ¾¾® Au

(Cu2+ + 2e ¾¾® Cu)

or W =

= 4.7629 g

Again

W =

\C =              =

= 0.804 ampere

 

  1. 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

  1. Net cell reaction

Pb(s)+ SO42-(aq)® Pb SO4+2e                             E°=0.31 V

Pb(s)+ 4H++SO42-(aq) +2e®PbSO4(s)+2H2O       E°=1.70 v

Ecell = E°cell

[SO42-]=2.0M        [H+]=10-1

Ecell for spontaneous reaction can be calculated.

 

  1. 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

  1. Hint: Ecell =

[H+] = Ca =  =

  1. Cell Reaction:

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

RHS:

Ecell = E°cell

At eqm Ecell=0,      K=

\ 0=0.40 –

Þ 0=0.40 –

Thus,  Can be calculated

 

  1. 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 ECell = 0

\

0.99 =

 

as K1 = 1 ´ 1019

K­2 = 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 Ecell positive, LHS, should be cathode, (+ve)-half Cell

 

  1. 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 Ecell 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

 

  1. 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

  1. 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

 

  1. 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

HPO42– + H+  H2PO4               pH = pKa + 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  HPO42– + H2O

1           0.165             1

0.835   ––                 1.165

pH = 2.15 + log  = 2.295

 

  1. Given V2+ + VO2+ + 2H+ ¾® 2V3+ + H2O         E0cell  = 0.616V

V3+ + Ag+ + H2O ¾® VO2+ + 2H+ + Ag(s)              E0cell = 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

 

  1. 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

Ecell  =  –  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

Ecell = 1.10 –  log  = 1.10 – 0.015 = 1.085 V

 

  1. Ksp(AgI = [Ag+] [I]

Ag(s) ¾® Ag+ + e              Anode

AgI + e ¾® Ag(s) + I–         Cathode

––––––––––––––––––––––––––––

AgI(s) ¾® Ag + I

Ecell = E0cell  –

At equilibrium Ecell = 0, E0cell = – 0.80  – 0.15 = – 95V

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

 

  1. 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 Ksp for AgCl is 1.0 ´ 10–10. Calculate E for Ag/AgCl electrode immersed in 1.00M KCl at 25°C.

= 0.799 V

\ Ksp = [Ag+] [Cl]

[Ag+] =  =

Ag+ + e ¾¾® Ag

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

 

  1. Half cell Reaction                                  E°

LHS           H2 ¾¾® 2H+ + 2e

RHS          Zn2+ + 2e ¾¾® Zn

——————————

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

K =

Ecell =   log

[H+] = 4.6 ´ 10–6 M

H+   +

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

Ka =  =

= 7.36 ´ 10–8

            \ pKa = 7.13

 

  1. Electrolytic decomposition reaction

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

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

After the complete discharge of Cu2 +, 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

 

  1. 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

 

  1. 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

  1. All Ag+ (aq) from AgCl in each of the cell is complexed to form [Agm(NH3)n]m+

                      Cell      Reaction:

mAg + nNH3 ® Ag m(NH3)nm+ + me                          E° ox=EV

ox=EV

K=

 

Ecell= 0.1185 V

Ecell = E°cell –  \ m=1

similarly insecond cell, using equation (1)

K=

Ecell =0.1263 v,     m=1

0.1263=0-

n=2

  1. 2 Hg + 2Fe3+      Hg22+           + 2Fe3+

1´10-3                0                   0

 

Ecell = E°cell – ;  At Eqm. Ecell=0

\0 = E° Hg/Hg2+2 + 0.77-

 

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

Ag+ +    Ag (S2O3)–3

KS =

  1. For the cell

AgçAgCl 0.05M KCl çç0.05 m AgNO3çAg

Ecell= E°cell

Þ 0.788 v =0-

(Ag+) LHS   can be calculated

KspAgCl= [Ag+]LHS´[Cl]

 

  1. Hint: [H+] L.H.S. =

[H+]R.H.S. =

 

 

 

 

    Objective Problems

 

Level – I

 

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

 

 

  1. 2Tl + Cu+2 ® 2Tl+ + Cu

 

\ (A) & (d)

 

 

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

         I2 is oxidised  to I5+ and reduced to I                      

         \(A) & (B)

 

  1. Zn + MgCl2 ® ZnCl2 + Mg

\

= + 0.762 – 2.37 =  – 1.608 V

Here, is negative so no reaction will take place.

\  (A)

 

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

 

= (0.25 + 1.5) –

= 1.75 –

= 1.75 + 0.295

= + 1.7795 V

\ (B)

 

  1. 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)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)                                              

 

  1. 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)     

                                            

 

  1. At anode 2H2O   ¾® O2 + 4H+ + 4e

         At cathode    Ag+ + e ¾® Ag

         \(B)

  

  1. Ag|Ag+|| Agl|Ag

          = – 0.799 – 0.151= – 0.950 V

         \(B)     

 

  1. 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)  

  1. 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)  

  1. 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)  

  1. 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

Ecell= 0.36 – V log

            \(B)  

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

Therefore [Cu2+] =  = 1 ´ 10–19

= –  log  = 0.34 V – ´ 19 = –0.22V

         \(A)     

  1. Ecell =  –  log

E1=  –  log  =

E2 = –  log

E3 =  –  log

\ E2 > E1 > E3

            \(D)  

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

            \(B)  

  1. 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)  

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

= 252C ´  = 2.6 ´ 10–3F

            \(B)  

  1. 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)  

  1. DG° =  –nFE° cell

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

            \(B)  

  1. 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)     

  1. Ecell =  log . Increase in the concentration of Ag+ decreases the value of log . This results in the increased value of Ecell.

         \(C)     

 

 

 

 

 

 

 

 

 

 

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