Chapter 9 Redox Reactions by TEACHING CARE Online coaching and tuition classes

Chapter 9 Redox Reactions by TEACHING CARE Online coaching and tuition classes

File name : Chapter-9-Redox-Reactions.pdf

 

 

Chemical reactions involve transfer of electrons from one chemical substance to another. These electron – transfer reactions are termed as oxidation-reduction or redox-reactions.

Redox reactions play an important role in our daily life. These reactions are accompanied by energy changes in the form of heat, light, electricity etc. Generation of electricity in batteries and many industrial

 

processes such as production of caustic soda, are common examples of redox reactions.

KMnO4 , extraction of metals like sodium, iron and aluminium

 

 

  • Molecular equations : When the reactants and products involved in a chemical change are written in molecular forms in the chemical equation, it is termed as molecular

 

Examples :   (i)

(ii)

MnO2  + 4 HCl ® MnCl2  + 2H2O + Cl2

2FeCl3 + SnCl2 ® 2FeCl2 + SnCl4

 

In above examples, the reactants and products have been written in molecular forms, thus the equation is termed as molecular equation.

  • Ionic equations : When the reactants and products involved in a chemical change are ionic compounds, these will be present in the form of ions in the The chemical change is written in ionic forms in chemical equation, it is termed as ionic equation.

 

Examples :      (i)

MnO2 + 4 H + + 4 Cl

®  Mn2+ + 2Cl + 2H 2 O + Cl2

 

(ii) 2Fe 3+ + 6Cl + Sn2+ + 2Cl ® 2Fe 2+ + 4Cl + Sn4+ + 4Cl

In above examples, the reactants and products have been written in ionic forms, thus the equation is termed as ionic equation.

  • Spectator ions : In ionic equations, the ions which do not undergo any change and equal in number in both reactants and products are termed spectator ions and are not included in the final balanced equations.

 

Example :

Zn + 2H + + 2Cl

® Zn2+ + H   + 2Cl

(Ionic equation)

 

2

Zn + 2H +

® Zn2+ + H2

(Final ionic equation)

 

In above example, the balanced equation.

Cl

ions are the spectator ions and hence are not included in the final ionic

 

(4)  Rules for writing ionic equations

  • All soluble ionic compounds involved in a chemical change are expressed in ionic symbols and covalent substances are written in molecular form. H2O, NH3 , NO2 , NO, SO2, CO, CO2, etc., are expressed in molecular
  • The ionic compound which is highly insoluble is expressed in molecular
  • The ions which are common and equal in number on both sides, e., spectator ions, are cancelled.
  • Besides the atoms, the ionic charges must also be balanced on both the The rules can be explained by following examples,

Example : Write the ionic equation for the reaction of sodium bicarbonate with sulphuric acid, The molecular equation for the chemical change is,

NaHCO3 + H2SO4   ® Na2SO4 + H2O + CO2

NaHCO3, H2SO4 and Na2SO4 are ionic compounds, so these are written in ionic forms.

Na+ + HCO + 2H + + SO2 ® 2Na + + SO2 + H  O + CO

3                           4                                   4            2                 2

 

 

 

 

Na+

and

 

SO2-

ions are spectator ions; hence these shall not appear in the final equation.

 

4
3

HCO + 2H +  ®     H 2 O + CO2

 

To make equal charges on both sides,

HCO

should have a coefficient 2.

 

3
3

2HCO + 2H + ® H 2O + CO2

In order to balance the hydrogen and carbon on both sides, the molecules of H2O and CO2 should have a coefficient 2 respectively.

 

3
3

2HCO + 2H + = 2H This is the balanced ionic equation.

2O + 2CO2        or

HCO + H + = H

2O + CO2

 

Conversion of ionic equation in molecular form can be explained by following example,

Example : Write the following ionic equation in the molecular form if the reactants are chlorides.

2Fe3++Sn2+ ®   2Fe2++Sn4+

For writing the reactants in molecular forms, the requisite number of chloride ions are added.

 

2Fe 3+  + 6Cl –     +

Sn2+  + 2Cl

or  2FeCl3 + SnCl2

 

Similarly 8 Cl

ions are added on R.H.S. to neutralise the charges.

 

2Fe 2+ + 4Cl     +

Sn4+ + 4Cl

or  2FeCl2 + SnCl4

 

 

Thus, the balanced molecular equation is, 2FeCl3+SnCl2 = 2FeCl2 + SnCl4

  • Oxidation : Oxidation is a process which involves; addition of oxygen, removal of hydrogen, addition of non-metal, removal of metal, Increase in +ve valency, loss of electrons and increase in oxidation
    • Addition of oxygen
  • 2Mg + O2 ® 2MgO (Oxidation of magnesium)
  • S + O2 ®  SO2                                           (Oxidation of sulphur)
  • 2CO + O2 ® 2CO2                                      (Oxidation of carbon monoxide)
  • Na2SO3+H2O2 ®  Na2SO4+H2O                                            (Oxidation of sodium sulphite)
    • Removal of hydrogen
  • H2S + Cl2 ® 2HCl + S (Oxidation of hydrogen sulphide)
  • 4HI + O2 ® 2H2O + 2I2 (Oxidation of hydrogen iodide)

 

  • C2H5OH

¾¾Cu /¾300¾o¾C ®

CH3CHO + H2                    (Oxidation of ethanol)

 

  • 4HCl + MnO2 ® MnCl2 + 2H2O + Cl2 (Oxidation of hydrogen chloride)
    • Addition of an electronegative element or addition of Non-metal
  • Fe + S ® FeS                                                                         (Oxidation of iron)
  • SnCl2 + Cl2 ® SnCl4 (Oxidation of stannous chloride)
  • 2Fe + 3F2 ® 2FeF3 (Oxidation of iron)
    • Removal of an electropositive element or removal of metal
  • 2KI + H2O2 ® 2KOH + I2 (Oxidation of potassium iodide)

 

 

  • 2K2MnO4 + Cl2 ® 2KCl + 2KMnO4 (Oxidation of potassium manganate)
  • 2KI + Cl2 ® 2KCl + I2 (Oxidation of potassium iodide)
    • Increase in +ve valency and Decrease in – ve valency

 

(a)

(b)

Fe 2+ ® Fe 3+ + e

Sn2+ ® Sn4+ + 2e

(+ve valency increases) (+ve valency increases)

 

(c) [Fe (CN)6 ]4 ® [Fe(CN)6 ]3 + e

(–ve valency decreases)

 

MnO2

® MnO + e

(–ve valency decreases)

 

4
4
  • Loss of electrons (also known as de-electronation)

 

 

          
     

 

 

 

 

(a)

(b)

H 0 ® H + + e

H 0 ® H + + e

(Formation of proton)

(De-electronation of hydrogen)

 

2             2

 

(c)

Fe 2+ ® Fe 3+ + e

(De-electronation of

Fe 2+ )

 

(d)

Mg ® Mg 2+ + 2e

(De-electronation of Magnesium)

 

MnO2 ® MnO + e

(De-electronation of

MnO2 )

 

4                     4                                                                                                                                4

 

  • 2Cl ® Cl2 + 2e

(g) 2Fe 0 ® 2Fe 3+ + 6e

  • Increase in oxidation number

(De-electronation of chloride ion) (De-electronation of iron)

 

(a)

Mg 0 ® Mg 2+

(From 0 to +2)

 

6                                    6

(b) [Fe +2 (CN)  ]4-   ® [Fe +3 (CN)  ]3-

(From +2 to +3)

 

(c)

2Cl ® Cl 0

(From –1 to 0)

 

2
  • Reduction : Reduction is just reverse of oxidation. Reduction is a process which involves; removal of oxygen, addition of hydrogen, removal of non-metal, addition of metal, decrease in +ve valency, gain of electrons and decrease in oxidation
  • Removal of oxygen

(a) CuO + C ® Cu + CO                                                                     (Reduction of cupric oxide)

 

(b)

HO +

Steam

C

Coke

® C–O+ –H2

Water gas

(Reduction of water)

 

 

 

 

(c)

Fe3 O4  + 4 H 2  ® 3Fe + 4 H 2 O

(Reduction of

Fe3 O4 )

 

(d) C6 H5 OH + Zn ® C6 H6  + ZnO

  • Addition of hydrogen

(a) Cl2 + H 2 ® 2HCl

(Reduction of phenol)

 

(Reduction of chlorine)

 

(b)

S + H 2 ® H 2 S

(Reduction of sulphur)

 

(c) C2 H4 + H 2  ® C2 H6

(Hydrogenation of ethene)

 

  • Removal of an electronegative element or removal of Non-metal

 

  • 2HgCl2 + SnCl2  ® Hg 2 Cl2  + SnCl4

(Reduction of mercuric chloride)

 

(b)

(c)

2FeCl3  + H 2  ® 2FeCl2  + 2HCl

2FeCl3  + H 2 S ® 2FeCl2  + 2HCl + S

(Reduction of ferric chloride) (Reduction of ferric chloride)

 

  • Addition of an electropositive element or addition of metal

 

(a)

HgCl2 + Hg ® Hg 2Cl2

(Reduction of mercuric chloride)

 

  • CuCl2 + Cu ® Cu2Cl2

(Reduction of cupric chloride)

 

  • Decrease in +ve valency and Increase in –ve valency

 

(a)

(b)

Fe 3+ ® Fe 2+

Sn4+ ® Sn2+

(+ve valency decreases) (+ve velency decreases)

 

(c) [Fe (CN)6 ]3 ® [Fe(CN)6 ]4

(–ve valency increases)

 

(d)

MnO  ® MnO2

(–ve valency increases)

 

4                    4

  • Gain of electrons (also known as electronation)

 

 

          
     

 

 

 

 

 

 

(a)

(b)

Zn 2+ (aq) + 2e ® Zn(S)

Pb 2+ + 2e ® Pb 0 Mn7+ + 5e ® Mn2+ Fe 3+ + e ® Fe 2+

(Electronation of (Electronation of (Electronation of

(Electronation of

Zn2+ ) Pb 2+ ) Mn7+ )

Fe 3+ )

 

  • Sn4+ + 2e ® Sn2+ (Electronation of Sn4+ )

 

 

 

 

  • Cl + e ® Cl

(g) [Fe(CN)6 ]3 + e ® [Fe(CN)6 ]4

  • Decrease in oxidation number

(Formation of chloride ion) (Eectronation of [Fe(CN)6 ]3 )

 

(a)

Mg 2+ ® Mg 0

(From +2 to 0)

 

6                               6

(b) [Fe(CN)  ]3-   ® [Fe(CN)  ]4 –

2

(c) Cl 0 ® 2Cl

(3)  Redox-reactions

(From +3 to +2)

(From 0 to –1)

 

  • An overall reaction in which oxidation and reduction takes place simultaneously is called redox or oxidation-reduction reaction. These reactions involve transfer of electrons from one atom to another. Thus every redox reaction is made up of two half reactions; One half reaction represents the oxidation and the other half reaction represents the

(ii)  The redox reactions are of following types

  • Direct redox reaction : The reactions in which oxidation and reduction takes place in the same vessel are called direct redox
  • Indirect redox reaction : The reactions in which oxidation and reduction takes place in different vessels are called indirect redox Indirect redox reactions are the basis of electro-chemical cells.
  • Intermolecular redox reactions : In which one substance is oxidised while the other is For example,

2 Al + Fe2O3 ® Al2O3  + 2Fe

 

Here, Al is oxidised to

Al 2 O3

while

Fe2 O3 is reduced to Fe.

 

  • Intramolecular redox reactions : In which one element of a compound is oxidised while the other is For example,

2 KClO3 ¾¾D ® 2 KCl + 3 O2

 

Here, Cl +5

in KClO3

is reduced to Cl -1 in KCl while O 2

in KClO3

is oxidised to O0 .

 

2
  • To see whether the given chemical reaction is a redox reaction or not, the molecular reaction is written in the form of ionic reaction and now it is observed whether there is any change in the valency of atoms or If there is a change in valency, the chemical reaction will be a redox reaction otherwise not. For example,
  • BaO2 + H 2 SO4 ® BaSO4  + H 2 O2
  • CuSO4 + 4 NH 3  ® [Cu(NH 3 )4 ]SO4

In above examples there is no change in the valency of any ion or atom, thus these are not redox reactions.

 

  • Some examples of redox reactions are,

(a)

2HgCl2 + SnCl2  ® Hg 2 Cl2  + SnCl4

+ 2e

Reduction

Here mercuric ion is reduced to mercurous ion and stannous ion is oxidised to stannic ion, i.e., mercuric ion acts as an oxidising agent while stannous ion acts as a reducing agent.

 

 

 

 

 

(b)

SnCl2 + 2FeCl3 ® 2FeCl2 + SnCl4

– 2e

Oxidation

Here ferric ion is reduced to ferrous ion by gain of one electron while stannous ion is oxidised to stannic ion by loss of two electrons. The ferric ion acts as an oxidising agent while stannous ion acts as a reducing agent.

 

2Na2 S2 O3 + I 2 ® Na2 S4 O6 + 2NaI

–2e

Oxidation

Here thiosulphate ion is oxidised to tetrathionate ion by loss of electrons while iodine is reduced to iodide ion by gain of electrons. Thiosulphate ion acts as a reducing agent and iodine acts as an oxidising agent.

 

0       2+                                                              0

Zn+ Cu SO4 ® ZnSO4 + Cu

+2e

Reduction

 

–2e

Oxidation

0       0               +          –

 

H2 + X 2 ® 2 H– X

+2e

Reduction

–2e

Oxidation

(Where X = F, Cl, Br, I)

 

0      0            +2 -1

 

 

ZnI 2 ® Zn I 2

+2e

Reduction

+2e

Reduction

 

0           +1                       0       +2

Zn+ 2 Ag CN ® 2 Ag+ Zn(CN)2

–2e

Oxidation

 

0           +1                     +2                           0

Cu+ 2 Ag NO3 ® Cu(NO3 )2 + 2 Ag

–2e

Oxidation

 

  • Definition : The substance (atom, ion or molecule) that gains electrons and is thereby reduced to a low valency state is called an oxidising agent, while the substance that loses electrons and is thereby oxidised to a higher valency state is called a reducing agent.

Or

 

 

 

An oxidising agent is a substance the oxidation number of whose atom or atoms decreases while a reducing agent is a substance the oxidation number of whose atom increases.

(2)  Important oxidising agents

  • Molecules made up of electronegative elements, g. O2, O3 and X2 (halogens).

 

  • Compounds containing an element which is in the highest oxidation state g.

KMnO4 , KCr2 O7 ,

 

Na2 Cr2 O7 ,  CrO3, H 2 SO4 ,  HNO3 , NaNO3 , FeCl3 , HgCl2 , KClO4 , SO3 , CO2 , H 2 O2 etc.

 

  • Oxides of elements g.

MgO, CuO, CrO3 , CO2 , P4 O10 ,

etc.

 

  • Fluorine is the strongest oxidising

(3)  Important reducing agents

  • All metals g. Na, Zn, Fe, Al, etc.
  • A few non-metals g. C, H2, S etc.
  • Hydracids : HCl, HBr, HI, H2S
  • A few compounds containing an-element in the lower oxidation state (ous), g. FeCl2 , FeSO4 ,

 

SnCl2 , Hg 2 Cl2 , Cu2 O

etc.

 

  • Metallic hydrides g. NaH, LiH etc.
  • Organic compounds like HCOOH and (COOH)2 and their salts, aldehydes, alkanes
  • Lithium is the strongest reducing agent in
  • Cesium is the strongest reducing agent in absence of Other reducing agents are

 

 

Na 2 S2 O3

 

 

 

and KI.

 

  • Hypo prefix indicates that central atom of compound has the minimum oxidation state so it will act as a

 

reducing agent. e.g.,

HPO2 (hypophosphorous acid).

 

(4)  Substances which act as oxidising as well as reducing agents

 

Example :

H2O2 , SO2 , H2 SO3 , HNO2 , NaNO2 , Na2 SO3 , O3  etc.

 

(5)  Tips for the identification of oxidising and reducing agents

  • If an element is in its highest possible oxidation state in a compound, the compound can function as an

 

oxidising agent, e.g.

KMnO4 , K2Cr2O7 , HNO3 , H2 SO4 , HClO4

etc.

 

  • If an element is in its lowest possible oxidation state in a compound, the compound can function only as a reducing agent, g. H 2 S, H 2 C2 O4 , FeSO4 , Na2 S2 O3 , SnCl2 etc.
  • If an element is in its intermediate oxidation state in a compound, the compound can function both as an

 

oxidising agent as well as reducing agent, e.g.

H 2 O2 , H 2 SO3 , HNO2 , SO2 etc.

 

  • If a highly electronegative element is in its highest oxidation state in a compound, that compound can

 

function as a powerful oxidising agent, e.g.

KClO4 , KClO3 , KBrO3 , KIO3

etc.

 

  • If an electronegative element is in its lowest possible oxidation state in a compound or in free state, it can

 

function as a powerful reducing agent, e.g.

(6)  Tests for oxidising agents

I , Br , N 3

etc.

 

  • Aqueous solutions of oxidising agents react with,
  • Hydrogen sulphide to give a milky yellow precipitate of

H 2S + Oxidising agent ® S (milky yellow ppt.)

  • Potassium iodide solution and evolve iodine which gives intense blue colour with starch solution

 

 

 

KI + Oxidising agent ®  I 2  ¾¾Star¾ch s¾olut¾i¾on ® Intense blue colour

 

  • Freshly prepared solution of ferrous ammonium sulphate in presence of

H 2 SO4 . Ferric ions (Fe 3+ )

 

can be detected by adding ammonium thiocyanate solution when a deep red colouration is produced.

3

Oxidising agent + Fe 2+  ® Fe 3+  ¾¾CNS¾-  ® Fe(CNS)   (deep red clouration)

  • Insoluble oxidising agents on,
  • Strong heating evolve oxygen which relights a glowing splinter.
  • Warming with concentrated hydrochloric acid evolve chlorine which bleaches the moist litmus

(7)  Tests for reducing agents

(i) Aqueous solutions of reducing agents react with,

  • Acidified potassium permanganate solution and decolourise
  • Few drops of acidified potassium dichromate solution, green colouration is
  • Few drops of ferric chloride The ferrous ions thus formed give a deep blue colouration with potassium ferricyanide (K3 [Fe(CN)6 ]) .

Insoluble reducing agents on,

  • Heating with concentrated nitric acid, evolve brown fumes of nitrogen
  • Heating with powdered cupric salt, form a red deposit of copper which does not dissolve in warm dilute sulphuric

(8)  Equivalent weight of oxidising and reducing agents

  • Equivalent weight of a substance (oxidant or reductant) is equal to molecular weight divided by number of electrons lost or gained by one molecule of the substance in a redox

Equivalent weight of oxidising agent =                   Molecular weight               

No. of electrons gained by one molecule

Equivalent weight of reducing agent =                 Molecular weight            

No. of electrons lost by one molecule

  • In other words, it is equal to the molecular weight of oxidant or reductant divided by the change in oxidation

Equivalent weight of oxidising agent =          Molecular weight     

Changein O.N. per mole

Equivalent weight of reducing agent =          Molecular weight     

Changein O.N. per mole

 

Equivalent weight of few oxidising/reducing agents

 

 

S2O32-+ 2S4O62-+ 2.50.50.5 × 2 = 1Mol. wt./1
 H2O2– 1 H2O– 211 × 2 = 2Mol. wt./2
 H2O2– 1 O2011 × 2 = 2Mol. wt./2
 MnO+ 7 Mn2++ 255 × 1 = 5Mol. wt./5
(Acidic medium)

 MnO

 

+ 7

 

 MnO2

 

+ 4

 

3

 

3 × 1 = 3

 

Mol. wt./3

(Neutral medium)      
 MnO+ 7 MnO2+ 611 × 1 = 1Mol. wt./1

 

  • Definition : Charge on an atom produced by donating or accepting electrons is called oxidation number

or oxidation state. It is the number of effective charges on an atom.

  • Valency and oxidation number : Valency and oxidation number concepts are In some cases (mainly in the case of electrovalent compounds), valency and oxidation number are the same but in other cases they may have different values. Points of difference between the two have been tabulated below

(3)  Oxidation number and Nomenclature

  • When an element forms two monoatomic cations (representing different oxidation states), the two ions are distinguished by using the ending-ous and The suffix – ous is used for the cation with lower oxidation state and the suffix – ic is used for the cation with higher oxidation state.

For example : Cu+ (oxidation number +1) cuprous : Cu2+ (oxidation number +2) cupric

  • Albert Stock proposed a new system known as Stock system. In this system, the oxidation states are indicated by Roman numeral written in parentheses immediately after the name of the For example,

 

Cu2OCopper (I) oxideSnOTin (II) oxide
FeCl2Iron (II) chlorideMn2O7Manganess (VII) oxide
K2Cr2O7Potassium dichromate (VI)Na2CrO4Sodium chromate (VI)
V2O5Vanadium (V) oxideCuOCopper (II) oxide
SnO2Tin (IV) oxideFeCl3Iron (III) chloride

Note : ® Stock system is not used for non-metals.

(4)  Rules for the determination of oxidation number of an atom

The following rules are followed in ascertaining the oxidation number of an atom,

  • If there is a covalent bond between two same atoms then oxidation numbers of these two atoms will be Bonded electrons are symmetrically distributed between two atoms. Bonded atoms do not acquire any charge. So oxidation numbers of these two atoms are zero.

 

 

 

A : A or A A ® A* + A*

For e.g. Oxidation number of Cl in Cl2, O in O2 and N and N2 is zero.

  • If covalent bond is between two different atoms then electrons are counted towards more electronegative Thus oxidation number of more electronegative atom is negative and oxidation number of less electronegative atom is positive. Total number of charges on any element depends on number of bonds.

A B ® A+ + B :

A B ® A+2 + : B : –2

The oxidation number of less electronegative element (A) is + 1 and + 2 respectively.

  • If there is a coordinate bond between two atoms then oxidation number of donor atom will be + 2 and of acceptor atom will be –

A ® B ® A2+ + B-2

  • The oxidation number of all the atoms of different elements in their respective elementary states is taken to

 

be zero. For example, in

N2, Cl2, H2, P4 , S8 , O2, Br2, Na, Fe, Ag etc. the oxidation number of each atom is zero.

 

  • The oxidation number of a monoatomic ion is the same as the charge on For example, oxidation

 

numbers of

Na + , Mg 2+ and

Al 3+

ions are + 1, + 2 and + 3 respectively while those of Cl , S2

and

N 3-

ions are

 

–1, –2 and –3 respectively.

  • The oxidation number of hydrogen is + 1 when combined with non-metals and is –1 when combined with active metals called metal hydrides such as LiH, KH, MgH2, CaH2

 

  • The oxidation number of oxygen is – 2 in most of its compounds, except in peroxides like

H2O2, BaO2

 

etc. where it is –1. Another interesting exception is found in the compound OF2 (oxygen difluoride) where the oxidation number of oxygen is + 2. This is due to the fact that fluorine being the most electronegative element known has always an oxidation number of –1.

  • In compounds formed by union of metals with non-metals, the metal atoms will have positive oxidation numbers and the non-metals will have negative oxidation For example,
  • The oxidation number of alkali metals (Li, Na, K ) is always +1 and those of alkaline earth metals (Be, Mg, Ca etc) is + 2.
  • The oxidation number of halogens (F, Cl, Br, I) is always –1 in metal halides such as KF, AlCl3, MgBr2, CdI2.
  • In compounds formed by the union of different elements, the more electronegative atom will have negative oxidation number whereas the less electronegative atom will have positive oxidation For example,
  • N is given an oxidation number of –3 when it is bonded to less electronegative atom as in NH3 and NI3, but is given an oxidation number of + 3 when it is bonded to more electronegative atoms as in NCl3.
  • Since fluorine is the most electronegative element known so its oxidation number is always –1 in its compounds e. oxides, interhalogen compounds etc.
  • In interhalogen compounds of Cl, Br, and I; the more electronegative of the two halogens gets the oxidation number of –1. For example, in BrCl3, the oxidation number of Cl is –1 while that of Br is +3.
  • For neutral molecule, the sum of the oxidation numbers of all the atoms is equal to For example, in

NH3 the sum of the oxidation numbers of nitrogen atom and 3 hydrogen atoms is equal to zero. For a complex ion,

 

4

the sum of the oxidation numbers of all the atoms is equal to charge on the ion. For example, in of the oxidation numbers of sulphur atom and 4 oxygen atoms must be equal to –2.

SO2-

ion, the sum

 

  • It may be noted that oxidation number is also frequently called as oxidation For example, in H2O,

the oxidation state of hydrogen is +1 and the oxidation state of oxygen is – 2. This means that oxidation number gives the oxidation state of an element in a compound.

 

 

  • In the case of representative elements, the highest oxidation number of an element is the same as its group number while highest negative oxidation number is equal to (8 – Group number) with negative sign with a few The most common oxidation states of the representative elements are shown in the following table,

 

 

GroupOuter shell configurationCommon oxidation numbers (states) except zero in free state
I Ans1+1
II Ans2+2
III Ans2np1+3, +1
IV Ans2np2+4,+3,+2,+1, –1, –2, –3, –4
V Ans2np3+5,+3,+1, –1, –3
VI Ans2np4+6,+4,+2,–2
VII Ans2np5+7,+5,+3, +1, –1
  • Transition metals exhibit a large number of oxidation states due to involvement of (n –1) d electron besides ns
  • Oxidation number of a metal in carbonyl complex is always zero, e.g. Ni has zero oxidation state in [Ni(CO)4 ].
  • Those compounds which have only C, H and O the oxidation number of carbon can be calculated by following formula,

 

Where, nO is the number of oxygen atom, nH is the number of hydrogen atom,

*

nC is the number of carbon atom.

 

For example, (a)    CH 3 OH ; nH   = 4,nC   = 1,nO    = 1

 

Oxidation number of ‘C’ =

 

*

(1 ´ 2 4) = -2 1

 

(b)

HCOOH ;  nH

= 2,

nO   = 2,nc   = 1

 

Oxidation number of carbon =

(2 ´ 2 2) = +2 1

 

  • Procedure for calculation of oxidation number : By applying the above rules, we can calculate the oxidation numbers of elements in the molecules/ions by the following
  • Write down the formula of the given molecule/ion leaving some space between the
  • Write oxidation number on the top of each In case of the atom whose oxidation number has to be calculated write x.
  • Beneath the formula, write down the total oxidation numbers of each element. For this purpose, multiply the oxidation numbers of each atom with the number of atoms of that kind in the molecule/ion. Write the product in a
  • Equate the sum of the oxidation numbers to zero for neutral molecule and equal to charge on the
  • Solve for the value of

 

Example: 1         The oxidation state of Mn in K2 MnO4

is            [CPMT 1982; 83; 84; Delhi PMT 1982; NCERT 1973; AMU 2000]

 

(a) + 2                            (b) + 7                               (c) – 2                          (d) + 6

 

 

 

 

Solution : (d)

KMnO4

x = 8 – 2 = +6

So, oxidation state of Mn is +6.

2 + x – 2 ´ 4 = 0

 

Example: 2         The oxidation number of Cr in K2Cr2O7 is                   [CPMT 1981; 85, 90, 93, 99; KCET 1992; BHU 1988, 98;

AFMC 1991, 99;  EAMCET 1986;  MP PMT 1996, 99, 2002;   MP PET/PMT 1998;  Bihar  CEE 1995;  RPET 2000]

(a) + 6                            (b) – 7                                (c) + 2                         (d) – 2

 

Solution : (a)

K2 Cr 2O7

2 + 2x – 2 ´ 7 = 0 or 2x – 14 + 2 = 0 or So, oxidation state of Cr is +6.

 

2x = 12 or

x = 12 = + 6

2

 

Example: 3         Oxidation number of N in (NH4 )2 SO4

is                                                                                 [CPMT 1996]

 

(a) –1/3                           (b) – 1                                (c) + 1                         (d) – 3

Solution : (d)      (NH4 ) SO4 ⇌ 2 NH + + SO2

2                            4           4

4

NH +

x + 4 = +1 or x = 1 – 4 = -3

Example: 4         The oxidation number of Sb in K2 H2Sb2O7 is

(a) + 2                            (b) + 5                               (c) – 2                          (d) – 5

Solution : (b)     As K2H2Sb2O7 is a neutral species, therefore the sum of the oxidation number of all the atoms present in

K2 H2Sb2O7 = 0

Let oxidation number per Sb atom = x

 

\     2 + 2 + 2x + (-2)7 = 0

[   Oxidation number of K, H and O are +1, +1 and –2 respectively]

 

or    4 + 2x – 14 = 0 or 2x – 10 = 0 or x = + 5

Example: 5         The oxidation number of Al in Ca3 (AlO3 )2 is

(a) + 3                            (b) – 1                                (c) + 2                         (d) 0

Solution : (a)     Ca3 (AlO3 )2

2 × 3 + 2x – 6 × 2 = 0       or    6 + 2x – 12 = 0 or 2x = 6 or x = + 3

Example: 6         The oxidation state of Cl in Ca(OCl) Cl is

(a) – 1                             (b) + 2                               (c) + 3                         (d) 0

Solution : (a) OCl is known as hypochlorite radical having contribution towards = – 1.

Let the oxidation state of underlined chlorine atom = x

\     2 + ( – 1) + x = 0 or x = -1

The oxidation state of underlined Cl = –1.

While the oxidation state of Cl present in OCl radical = + 1.

Example: 7         Oxidation state of oxygen in hydrogen peroxide is

[DPMT 1984; 91; CPMT 1988; MNR 1994; UPSEAT 2001; RPMT 2002]

(a) –1                              (b) + 1                               (c) 0                            (d) – 2

Solution : (a)     In all peroxides oxygen shows –1 oxidation state.

Example: 8         The brown ring complex compound is formulated as [Fe(H 2O)5 NO + ]SO4 . The oxidation state of iron is

[EAMCET 1987; IIT 1987; MP PMT 1994; AIIMS 1997; DCE 2000]

(a) + 1                            (b) + 2                               (c) + 3                         (d) 0

4

Solution : (b)     [ Fe (H 2 O)5 NO]SO4   [Oxidation number of H2O = 0 ; Oxidation number of NO = 0; Oxidation number of SO2- =–2]

 

x + 0 + 0 – 2 = 0 or

x – 2 = 0 or

x = + 2

 

 

Example: 9         The oxidation state of V in Rb4 Na [H V10O28 ] is

(a) + 2                            (b) + 5                               (c) – 2                          (d) – 5

Solution : (b) The oxidation state of Rb = +1 (Since it is alkali metals) Let oxidation state of V = x

\     4 ´ 1 + 1 + 1 + 10x + 28 ´ ( – 2) = 0

 

or    6 + 10x – 56 = 0 or

10x – 50 = 0 or

x = 5

 

\ The oxidation state of V = +5

Example: 10      The oxidation state of Fe in FeS2 is

(a) – 2                             (b) + 2                               (c)  0                            (d) + 1

S

Solution : (b) The structure of FeS2 is Fe

S

 

Therefore

FeS2 contain one per disulphide linkage, which is similar to that of peroxide linkages, having

 

contribution of – 2 units towards the oxidation state Let the oxidation state of Fe = x

\      x – 2 = 0 or x = 2

The oxidation state of Fe in FeS2 is + 2.

Oxidation number of some elements in compounds, ions or chemical species

ElementOxidation NumberCompounds, ions or chemical species
Sulphur (s)– 2H2S, ZnS, NaHS, (SnS3)2–, BaS, CS2
 0S, S4, S8, SCN
 + 1S2, F2, S2Cl2
 + 4SO2, H2SO3, (SO3)2–, SOCl2, NaHSO3, Ca[HSO3]2, [HSO3], SF4
 + 6H2SO4, (SO4)2-, [HSO4], BaSO4, KHSO4, SO3, SF6, H2S2O7, (S2O7)2–
Nitrogen– 3NH3, (NH4)+, AlN, Mg3N2, (N)3–, Ca3N2, CN
(N)  
 – 2N2H4, (N2H5)+
 – 1NH2OH
 –1/3NaN3, N3H
 0N2
 + 1N2O
 + 2NO
 + 3HNO2, (NO2), NaNO2, N2O3, NF3
 + 4NO2
 + 5HNO3, (NO3), KNO3, N2O5
Chlorine– 1HCl, NaCl, CaCl2, AlCl3, ICl, ICl5, SOCl2, CrO2Cl2, KCl, K2PtCl6, HAuCl4, CCl4
(Cl)0Cl, Cl2
 + 1HOCl, NaOCl, (OCl), Cl2O
 + 3KClO2, (ClO2), HClO2
 + 4ClO2

 

 

 

+ 5                   (ClO3), KClO3, NaClO3, HClO3

+ 7                   HClO4, Cl2O7, KClO4, (ClO4)

Hydrogen         – 1                    NaH, CaH2, LiAlH4, LiH

(H)                    + 1                   NH3, PH3, HF

 

Phosphoru s

– 3                    PH3, (PH4)+, Ca3P2

 

(P)                     0                       P4

+ 1                   H3PO2, KH2PO2, BaH4P2O4

+ 3                   PI3, PBr3, PCl3, P2O3, H3PO3

+ 5                   (PO4)3–, H3PO4, Ca3(PO4)2, H4P2O7, P4O10, PCl5, (P2O7)4–, Mg2P2O7, ATP Oxygen            – 2                    H2O, PbO2, (CO3)2–, (PO4)2–, SO2, (C2O4)2–, HOCl, (OH), (O)2-

(O)                     – 1                    Na2O2, BaO2, H2O2, (O2)2–, Peroxides

– 1/2                KO2

0                       O, O2, O3

+ 1                   O2F2

+ 2                   OF2

Carbon             – 4                    CH4

  • – 3 C2H6
    • 2 CH3Cl, C2H4
    • 1 CaC2, C2H2

0                       Diamond, Graphite, C6H12O6, C2H4O2, HCHO, CH2Cl2

+ 2                   CO, CHCl3, HCN

+ 3                   H2C2O4, (C2O4)2–

+ 4                   CO2, H2CO3, (HCO3), CCl4, Na2CO3, Ca2CO3, CS2, CF4, (CO3)–2 Chromium     + 3                   Cr2(SO4)3, CrCl3, Cr2O3, [Cr(H2O)4Cl3]

(Cr)                   + 6                   K2CrO4, (CrO4)2–, K2Cr2O7, (Cr2O7)2–, KCrO3Cl, CrO2Cl2, Na2Cr3O10, CrO3

Manganese    + 2                   MnO, MnSO4, MnCl2, Mn(OH)2 (Mn)                 + 8/3              Mn3O4

+ 3                   Mn(OH)3

+ 4                   MnO2, K2MnO3

+ 6                   K2MnO4, (MnO4)2–

+ 7                   KMnO4, (MnO4), HMnO4

Silicon              – 4                    SiH4, Mg2Si

(Si)                    + 4                   SiO2, K2SiO3, SiCl4

 

Iron

  • 8 Fe3O4

 

(Fe)                   3

+ 2                   FeSO4 (Ferrous ammonium sulphate), K4Fe(CN)6, FeCl2

+ 3                   K3[Fe(CN)6], FeCl3

 

 

 

  • Exceptional cases of evaluation of oxidation numbers : The rules described earlier are usually

helpful in determination of the oxidation number of a specific atom in simple molecules but these rules fail in the following cases. In these cases, the oxidation numbers are evaluated using the concepts of chemical bonding involved.

Type I.   In molecules containing peroxide linkage in addition to element-oxygen bonds. For example,

  • Oxidation number of S in H2SO5 (Permonosulphuric acid or Caro’s acid).

By usual method; H 2 SO5

2 ´ 1 + x + 5 ´ (-2) = 0 or           x = + 8

But this cannot be true as maximum oxidation number for S cannot exceed + 6. Since S has only 6 electrons in its valence shell. This exceptional value is due to the fact that two oxygen atoms in H2SO5 shows peroxide linkage as shown below,

 

O

H     O     S     O

O

Peroxide linkage

 

Therefore the evaluation of o.n. of sulphur here should be made as follows, 2 × (+1)  +  x               + 3 × (–2) + 2 × (–1)

(for H)             (for S)            (for O)                 (for O–O)

or               2 + x – 6 – 2 = 0    or     x = + 6.

  • Oxidation number of S in H2S2O8 (Peroxidisulphuric acid or Marshall’s acid)

By usual method ; H 2 S2O8

1 × 2 + 2x + 8 (–2) = 0

2x = + 16 – 2 = 14  or    x = + 7

Similarly Caro’s acid, Marshall’s acid also has a peroxide linkage so that in which S shows +6 oxidation state.

Peroxide linkage

 

O

H     O               S O

O

O     O     S     O     H O

 

Therefore the evaluation of oxidation state of sulphur should be made as follow, 2 × (+1)  +  2 × (x)         + 6 × (–2) + 2 × (–1) = 0

(for H)                   (for S)                    (for O)                    (for O–O)

or         2 + 2x – 12 – 2 = 0 or x = + 6.

  • Oxidation number of Cr in CrO5 (Blue perchromate)

 

By usual method

CrO5 ;  x – 10 = 0 or  x = + 10

 

This cannot be true as maximum o.n. of Cr cannot be more than + 6. Since Cr has only five electrons in 3d orbitals and one electron in 4s orbital. This exceptional value is due to the fact that four oxygen atoms in CrO5 are in peroxide linkage. The chemical structure of CrO5 is

 

 

 

 

 

 

Peroxide linkage

O                            O

Cr                                  Peroxide linkage

O                            O

O

 

Therefore, the evaluation of o.n. of Cr should be made as follows

x      + 1 × (– 2) + 4 (–1) = 0

(for Cr)              (for O)               (for O–O)

or        x – 2 – 4 = 0 or x = + 6.

Type II. In molecules containing covalent and coordinate bonds, following rules are used for evaluating the oxidation numbers of atoms.

  • For each covalent bond between dissimilar atoms the less electronegative element is assigned the oxidation number of + 1 while the atom of the more electronegative element is assigned the oxidation number of –1.
  • In case of a coordinate-covalent bond between similar or dissimilar atoms but the donor atom is less electronegative than the acceptor atom, an oxidation number of +2 is assigned to the donor atom and an oxidation number of –2 is assigned to the acceptor

Conversely, if the donor atom is more electronegative than the acceptor atom, the contribution of the coordinate bond is neglected.

Example :

®

  • Oxidation number of C in HC º N and HN =  C

The evaluation of oxidation number of C cannot be made directly by usual rules since no standard rule exists for oxidation numbers of N and C.

In such cases, evaluation of oxidation number should be made using indirect concept or by the original concepts of chemical bonding.

®

  • Oxidation number of carbon in H – N = C

The contribution of coordinate bond is neglected since the bond is directed from a more electronegative N

atom (donor) to a less electronegative carbon atom (acceptor).

®

Therefore the oxidation number of N in HN = C remains – 3 as it has three covalent bonds.

1 × (+ 1) + 1 × (– 3)   + x = 0

(for H)                       (for N)                 (for C)

or          1 + x – 3 = 0    or       x = + 2.

  • Oxidation number of carbon in HC º N

In HC º N , N is more electronegative than carbon, each bond gives an oxidation number of –1 to N. There

are three covalent bonds, the oxidation number of N is HC º N is taken as – 3

 

Now

HC º N

\  +1 + x – 3 = 0  Þ x = + 2

 

Type III. In a molecule containing two or more atoms of same or different elements in different oxidation states.

  • Oxidation number of S in Na2S2O3

By usual method Na 2 S2O3

\      2 × (+1) + 2 × x + 3 (–2) = 0 or 2 + 2x – 6 = 0    or   x = 2.

But this is unacceptable as the two sulphur atoms in Na2S2O3 cannot have the same oxidation number because on treatment with dil. H2SO4, one sulphur atom is precipitated while the other is oxidised to SO2.

 

 

Na2 S2 O3 + H 2 SO4  ® Na2 SO4  + SO2  + S + H 2 O

In this case, the oxidation number of sulphur is evaluated from concepts of chemical bonding. The chemical structure of Na2S2O3 is

 

 

Na +

S

O–     S     O

O

Na +

 

Due to the presence of a co-ordinate bond between two sulphur atoms, the acceptor sulphur atom has oxidation number of – 2 whereas the other S atom gets oxidation number of + 2.

2 × (+1)   +  3 × (–2)  +     x   × 1         + 1 × (– 2) = 0

(for Na)                  (for O)                        (for S)                       (for coordinated S)

or          + 2 – 6  +   x – 2 = 0   or x = + 6

Thus two sulphur atoms in Na2S2O3 have oxidation number of – 2 and +6.

  • Oxidation number of chlorine in CaOCl2 (bleaching powder)

In bleaching powder, Ca(OCl)Cl, the two Cl atoms are in different oxidation states i.e., one Cl having oxidation number of –1 and the other as OCl having oxidation number of +1.

  • Oxidation number of N in NH4NO3

By usual method N2H4O3 ; 2x + 4 × (+1) + 3 × (–1) = 0

2x + 4 – 3 = 0 or 2x = + 1    (wrong)

No doubt NH4NO3 has two nitrogen atoms but one N has negative oxidation number (attached to H) and the

4

other has positive oxidation number (attached to O). Hence the evaluation should be made separately for NH +

 

and

NO

3
4

NH +

3

NO

x + 4 × (+1) = +1 or x = – 3

x + 3 (– 2) = –1 or x = + 5.

 

  • Oxidation number of Fe in Fe3O4

In Fe3O4, Fe atoms are in two different oxidation states. Fe3O4 can be considered as an equimolar mixture of FeO (iron (II) oxide) and Fe2O3 (iron (III) oxide). Thus in one molecule of Fe3O4, two Fe atoms are in + 3 oxidation state and one Fe atom is in + 2 oxidation state.

  • Oxidation number of S in sodium tetrathionate (Na2S4O6). Its structure can be represented as

 

 

follows :

O

 

Na + O     S      S

O

O

 

S     S     O Na +

O

 

The two S-atoms which are linked to each other have oxidation number of zero. The oxidation number of other S-atoms can be calculated as follows

Let oxidation number of S = x.

\            2 × x      +    2 × 0 + 6 × ( – 2) = – 2

(for S)             (for S–S)                (for O)

x = + 5.

Though there are a number of methods for balancing oxidation – reduction reactions, two methods are very important. These are,(1) Oxidation number method, (2) Ion – electron method

  • Oxidation number method : The method for balancing redox reactions by oxidation number change method was developed by Johnson. In a balanced redox reaction, total increase in oxidation number must be equal to the total decrease in oxidation This equivalence provides the basis for balancing redox reactions.

 

 

This method is applicable to both molecular and ionic equations. The general procedure involves the following steps,

  • Write the skeleton equation (if not given, frame it) representing the chemical
  • Assign oxidation numbers to the atoms in the equation and find out which atoms are undergoing oxidation and Write separate equations for the atoms undergoing oxidation and reduction.
  • Find the change in oxidation number in each Make the change equal in both the equations by multiplying with suitable integers. Add both the equations.
  • Complete the balancing by First balance those substances which have undergone change in oxidation number and then other atoms except hydrogen and oxygen. Finally balance hydrogen and oxygen by putting H2O molecules wherever needed.

The final balanced equation should be checked to ensure that there are as many atoms of each element on the right as there are on the left.

  • In ionic equations the net charges on both sides of the equation must be exactly the same. Use H+ ion/ions in acidic reactions and OH ion/ions in basic reactions to balance the charge and number of hydrogen and oxygen

The following example illustrate the above rules,

 

Step : I

Cu + HNO3  ® Cu(NO3 )2  + NO2  + H 2 O

(Skeleton equation)

 

Step: II Writing the oxidation number of all the atoms.

 

0       +1 +5 -2

+2    +5 -2

+4 -2

+1   -2

 

Cu H NO 3

® Cu( N O3 )2N O2 + H 2 O

 

Step: III Change in oxidation number has occurred in copper and nitrogen.

0           +2

 

Cu ® Cu(NO3 )2

……(i)

 

+5               +4

 

H N O3 ® N O2

……(ii)

 

Increase in oxidation number of copper       = 2 units per molecule Cu

Decrease in oxidation number of nitrogen = 1 unit per molecule HNO3

Step: IV To make increase and decrease equal, equation (ii) is multiplied by 2.

Cu + 2HNO3 ® Cu(NO3 )2  + 2NO2  + H 2 O

Step: V Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained.

Cu + 4 HNO3 ® Cu(NO3 )2  + 2NO2  + 2H 2 O

This is the balanced equation.

(2)  Ion-electron method (half reaction method)

The method for balancing redox-reactions by ion electron method was developed by Jette and LaMev

in 1927. It involves the following steps

  • Write down the redox reaction in ionic
  • Split the redox reaction into two half reactions, one for oxidation and other for
  • Balance each half reaction for the number of atoms of each For this purpose,
  • Balance the atoms other than H and O for each half reaction using simple
  • Add water molecules to the side deficient in oxygen and H+ to the side deficient in This is done in acidic or neutral solutions.

 

 

  • In alkaline solution, for each excess of oxygen, add one water molecule to the same side and 2OH ions to the other side. If hydrogen is still unbalanced, add one OH ion for each excess hydrogen on the same side and one water molecule to the other
    • Add electrons to the side deficient in electrons as to equalise the charge on both
    • Multiply one or both the half reactions by a suitable number so that number of electrons become equal in both the
3
  • Add the two balanced half reactions and cancel any term common to both sides. The following example illustrate the above rules

 

Step: I

I 2 + OH ® IO + I + H 2 O

(Ionic equation)

 

3

Step: II Splitting into two half reactions, I 2 + OH ® IO + H 2 O ;          I 2 ® I

(Oxidation half reaction) (Reduction half reaction)

 

Step: III Adding OH ions,

I 2 + 12OH ® 2IO + 6H 2 O

 

3

Step: IV Adding electrons to the sides deficient in electrons,

 

3

I 2 + 12OH ® 2IO + 6H

2O + 10e ;

I 2 + 2e   ® 2I

 

Step: V Balancing electrons in both the half reactions.

 

3

I 2 + 12OH ® 2IO + 6H

2O + 10e ; 5[I 2

+ 2e ® 2I ]

 

3

Step: VI Adding both the half reactions.

 

3

6 I 2 + 12OH ® 2IO + 6H

2O + 10I ; Dividing by 2, 3 I 2

+ 6OH ® IO + 5I + 3H 2 O

 

 

 

  • Turpentine and numerous other olefinic compounds, phosphorus and certain metals like Zn and Pb can absorb oxygen from the air in presence of water. The water is oxidised to hydrogen peroxide. This phenomenon of formation of H2O2 by the oxidation of H2O is known as autoxidation. The substance such as turpentine or phosphorus or lead which can activate the oxygen is called activator. The activator is supposed to first combine with oxygen to form an addition compound, which acts as an autoxidator and reacts with water or some other acceptor so as to oxidise the For example;

 

Pb

(activator)

+ O2 ®

PbO2

(autoxidator)

PbO2 +

H 2O

(acceptor)

® PbO + H 2O2

 

  • The turpentine or other unsaturated compounds which act as activators are supposed to take up oxygen molecule at the double bond position to form unstable peroxide called moloxide, which then gives up the oxygen to water molecule or any other

 

RCH = CHR + O2

®   RHC O

CHR O

 

RHC

O

CHR + 2H 2O        ®

O

RCH = CHR + 2H 2O2

 

2KI + H 2O2 ® 2KOH + I 2

The evolution of iodine from KI solution in presence of turpentine can be confirmed with starch solution which turns blue.

 

  • The concept of autoxidation help to explain the phenomenon of induced

Na2SO3 solution is

 

oxidised by air but

Na3 AsO3

solution is not oxidised by air. If a mixture of both is taken, it is observed both are

 

oxidised. This is induced oxidation.

 

 

Na2 SO3 + O2  ® Na2 SO5

Moloxide

Na2 SO5  + Na3 AsO3  ®  Na3 AsO4  + Na2 SO4 Na2 SO3 + Na3 AsO3 + O2 ® Na2 SO4 + Na3 AsO4

One and the same substance may act simultaneously as an oxidising agent and as a reducing agent with the result that a part of it gets oxidised to a higher state and rest of it is reduced to lower state of oxidation. Such a reaction, in which a substance undergoes simultaneous oxidation and reduction is called disproportionation and the substance is said to disproportionate.

Following are the some examples of disproportionation,

increase

decrease

 

-1                        0                                                                   +5                      +7               -1

 

(1)   H 2O2 + H 2 O2  = H 2O O2

 

–1                                  –2

(2)      4 K Cl O3 ® 3K Cl O4 + KCl

 

increase

 

 

 

 

0

(3)

decrease

decrease

 

 

+1         -3

 

 

0

(4)

decrease

 

 

hot

 

-1              +5

 

4 P + 3NaOH + 3H2O ® 3NaH2 PO2 + PH3                 3Cl2 + 6 NaOH ¾ ¾® 5 Na ClNa ClO3 + 3H2O

increase                                                                                                  increase

Many applications are based on redox reactions which are occuring in environment. Some important examples are listed below;

 

  • Many metal oxides are reduced to metals by using suitable reducing For example

Al 2 O3

is reduced

 

to aluminium by cathodic reduction in electrolytic cell.

Fe 2 O3 is reduced to iron in a blast furnace using coke.

 

  • Photosynthesis is used to convert carbon dioxide and water by chlorophyll of green plants in the presence of sunlight to

 

6CO2(g) + 6H 2 O(l)

¾¾Chl¾orop¾h¾yll  ® C6 H12 O6(aq.)  + 6O2

(g)

 

Sunlight

In this case, CO2 is reduced to carbohydrates and water is oxidised to oxygen. The light provides the energy required for the reaction.

  • Oxidation of fuels is an important source of energy which satisfies our daily need of

Fuels + O2  ® CO2 + H 2 O + Energy

In living cells, glucose  (C6 H12 O6 )  is oxidised to  CO2   and  H 2 O   in the presence of oxygen and energy is

 

released,

C6 H12 O6 (aq.) + 6O2 (g) ® 6CO2 (g) + 6H 2 O(l) + Energy

 

  • The electrochemical cells involving reaction between hydrogen and oxygen using hydrogen and oxygen electrodes in fuel cells meet our demand of electrical energy in space
  • Respiration in animals and humans is also an important application of redox

 

***

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