Class 11 - Chemistry - Redox Reactions

Question 8.1

Assign oxidation number to the underlined elements in each of the following species:

  • NaH2PO4 (b) NaHSO4 (c) H2P2O7 (d) K2MnO4 (e) CaO2 (f) NaBH4 (g) H2S2O7(h) KAl(SO4)2.12H2O


  • NaH2PO4

Na   +1

H     +1

P       x

 O     -2

Sum of oxidation number of various atoms is zero.

1(+1) + 2 (+1) + (x) + 4 (-2) = 0

x – 5 = 0

x = +5


  • NaHSO4

1(+1) + 1 (+1) + x + 4 (-2) = 0

x = +6

  • H4P2O7

4(+1) + 2(x) + 7(-2) = 0

x = +5

  • K2MnO4

2(+1) + 1(x) + 4(-2) = 0

 x = +6

  • CaO2

2 + 2x = 0

x = -1

  • NaBH4

1(+1) + x + 4(-1) = 0

x =+3

  • H2S2O7

2(+1) + 2(x) + 7 (-2) = 0

x = +6

  • KAl(SO4)2.12H2O

+1 + 3 + 2x + 8(-2) + 12 x 0 = 0

x = +6


Question 8.2

What are the oxidation number of the underlined elements in each of the following and how do you rationalize your results?

  • KI3 (b) H2S4O6 (c) Fe3O4  (d) CH3CH2OH  (e) CH3COOH


  • KI3

Oxidation state of K is +1. Average oxidation number of iodine = -1/3.

In the structure, of KI3 coordinate bond is formed between I2 molecule and I ion.

Oxidation number of two iodide atoms forming I2 molecule is zero while iodide forming coordinate bond is -1.

Oxidation numbers of three iodine atoms in KI3 are 0, 0 and -1.

  • Structure of H2S4O6


Oxidation number of each of the S atoms linked with each other in the centre is zero.

Oxidation number of Fe = x


3x + 4 (-2) = 0

x = +8/3

Oxidation no. cannot be fractional.

Using stoichiometry, Fe3O4 is


Thus, Oxidation number of one of three atoms of Fe is + 2 and other two atoms of Fe has oxidation number as +3.

(d)Oxidation number of C in CH3CH2OH


2x + 6 (+1) + 1 (-2) = 0

x = -2

(e)By conventional method, CH3COOH 


2x + 4 – 4 = 0

x  =0

Therefore, average oxidation no. of C is 0.

Both the carbon atoms are present in different environments so they cannot have same oxidation no.

Therefore, carbon has oxidation no. of +2 and -2 in CH3COOH.

Question 8.3

Justify that the following reactions are redox reactions:

  • CuO (s) + H2 (g) --> Cu (s) + H2O (g)
  • Fe2O3 (s) + 3CO(g) --> 2Fe(s) + 3CO2(g)
  • 4BCl3 (g) + 3LiAlH4 (s) --> 2B2H6 (g) + 2LiCl (s) + 3AlCl3(s)
  • 2K (s) + F2 (g) --> 2K+F- (s)
  • 4NH3 (g) + 5O2 (g) --> 4NO (g) + 6H2O (g)


Oxidation number in B decreases from +3 in BCl3 to -3 in B2H6 while that of H increases from -1 in LiAlH4 to +1 in B2H6.

BCl3 is reduced while LiAlH4 is oxidised.

(d) 2K (s) + F2 (g) --> 2K+F- (s)

In the above reaction,

Oxidation no. of K is 0.

Oxidation no. of F is 0.

Oxidation no. of K and F in KF is +1 and -1 respectively.

Each K atom loses one electron to form K+ while F2 has gained two electrons to form two F- ions.

K is oxidized while F2 is reduced. Therefore the reaction is a redox reaction.

(e) 4NH3 (g) + 5O2 (g) --> 4NO (g) + 6H2O (g)

In the above reaction,

Oxidation no. of N and H in NH3 is -3 and +1 respectively.

Oxidation no. of O2 is 0.

Oxidation no. of N and O in NO is +2 and -2 respectively.

Oxidation no. of H and O in H2O is +1 and -2 respectively.

The oxidation no. of N increased from -3 in NH3 to +2 in NO.

The oxidation no. of O2 decreased from 0 in O2 to -2 in NO and H2O. That is O2 is reduced.

Therefore, the reaction is a redox reaction.


Question 8.4

Flourine reacts with ice and results in the change:

H2O (s) + F2 (g) --> HF(g) + HOF (g)

Justify that this reaction is a redox reaction.


Oxidation number of elements are


Now, the oxidation number of F decreases from 0 to F2 to -1 in HF and increases from 0 to F2 to +1 in HOF.

F2 is both reduced as well as oxidized. This is a redox reaction and more specifically, it is a disproportionate reaction.


 Question 8.5

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72- and NO3-.

Suggest structure of these compounds. Count for the fallacy.


(i)Oxidation number of S in H2SO5

Oxidation number of S in H2SO5 is

2(+1) + x + 5(-2) = 0

x = +8

This is not possible because maximum oxidation number of sulphur cannot be

more than six since, it has only 6 electrons in the valence shell.

This fallacy is overcome if we calculate oxidation number of sulphur by chemical bonding method.

Let us draw the structure of H2SO5


Therefore, the oxidation no. of S is +6.

(ii) Cr in Cr2O72-

2x + (-2 x 7) = -2

2x -14 = -2

2x = -2 + 14

x = + 6

There is no fallacy about the oxidation no. of Cr in Cr2O72-


Let x be the oxidation no. of N.

Oxidation no. of O= -2


1(x) + 3(-2) = -1

x – 6 = -1

x = +5

There is no fallacy about the oxidation no. of N in NO3-

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

(for O-)      (for = O) for --> O

In case of nitrogen, oxidation number in NO3- can be calculated by conventional or by chemical bonding method.


Question 8.6

Write the formulas for the following compounds:

  • Mercury (II) chloride
  • Nickel (II) sulphate
  • Tin (IV) oxide
  • Thallium (I) sulphate
  • Iron (III) sulphate
  • Chromium (III) oxide


  • HgCl2
  • NiSO4
  • SnO2
  • Tl2SO4
  • Fe2(SO4)3
  • Cr2O3


Question 8.7

Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.




















































Question 8.8

While sulphur dioxide and hydrogen peroxide can act as oxidizing as well as reducing agents in their reactions,

ozone and nitric acid act only as oxidants. Why?


  • In So2, oxidation number of sulphur is +4. S can have minimum oxidation number -2 and maximum +6.
  • S in SO2 can either decrease or increase its oxidation number and hence, it can act both as oxidizing agent as well as reducing agent.

(ii) In H2O2, Oxidation number of oxygen is -1. O can have minimum oxidation number -2 and maximum of zero.

So, O in H2O2 can either decrease its oxidation number from -1 to -2. Or it can increase an oxidizing as well as reducing agent.

(iii) In ozone, oxidation number of oxygen is zero. It can only decrease its oxidation number from zero to -1 or -2.

It cannot increase to +2. Hence, ozone acts only as an oxidant.

(iv) In nitric acid, oxidation number of N is +5 which is maximum.

It can only decrease its oxidation number and it acts as an oxidant only.


Question 8.9

Consider the reactions:

  • 6CO2 (g) + H2O (l) --> C6H12O6 (aq) + 6O2(g)
  • O3 (g) + H2O2 (l) --> H2O (l) + 2O2 (g)

Why it is more appropriate to write these reactions as:

  • 6CO2 (g) + 12H2O (l) --> C6H12O6 (aq) + 6H2O(l) + 6O2(g)
  • O3 (g) + H2O2(l) --> H2O (l) + O2(g) + O2(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.


Reactions are given below:


It is more appropriate to write the equation for photosynthesis as the third one because it emphasizes on 12 water molecules.

As 12 water molecules, are used per molecule of carbohydrate formed and 6 water molecules are produced in this process.


The reason for writing oxygen two times means that oxygen is being obtained from each of the two reactants.


The path can be found with the help of H2​O218​ or O318

Question 8.10

The compounds AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?


In AgF2, oxidation state of Ag is +2 which is unstable. That is why, it accepts an electron to form the more stable +1 oxidation state.

Ag2+ + e- ---> Ag+

AgF2 will act as a strong oxidizing agent.


 Question 8.11

Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound

of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state

is formed if the oxidisng agent is in excess. Justify this statement giving these illustrations.


  • Carbon is a reducing agent while oxygen is an oxidizing agent. If excess carbon is burnt
  • in a limited supply of oxygen, carbon monoxide is formed in which oxidation state to carbon is +2.
  • If excess of oxygen is used, initially formed Carbon monoxide gets oxidized to Carbon dioxide in which oxidation state of carbon is + 4.


(ii) Sodium is a reducing agent while oxygen is an oxidizing agent. 

When excess of sodium is used, sodium oxide is formed in which oxidation state of O is -2.

If there is excess of oxygen, Na2O2 is formed in which oxidation state of O is -1 which is higher than -2.


(iii) P4 is a reducing agent while Cl2 is an oxidizing agent. When there is excess of P4 is used,

PCl3 is formed in which oxidation state of P is +3. If there is excess of Cl2 , initially PCl3 is formed further PCl5 is

formed in which oxidation state of P is +5.


Question 8.12

How do you count for the following observations?

  • Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidanats,
  • yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant.
  • Why? Write a balanced redox equation for the reaction.
  • When concentrated sulphuric acid is added to an inorganic mixture containing chloride,
  • we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?


  • Toluene can be oxidsed to benoic acid in acidic, basic and neutral media. Redox equations are given below:

In acidic medium


In Basic and neutral media


In lab, benzoic acid is prepared by alkaline KMnO4 oxidation of toluene. In alcoholic KMnO4 is preferred over acidic or

alkaline KMnO4 because reaction occurs faster in homogenous medium than in heterogeneous medium,

so alcohol helps in mixing two reactants and toluene.

Also, cost of adding acid or base is avoided because in the neutral medium, base are produced in the reaction.

         When concentrated sulphuric acid is added to an inorganic mixture containing chloride,

          a pungent smelling gas HCl is produced.


  • This is because a stronger acid displaces a weaker acid from its salt.

HCl is a very weak reducing agent, it cannot reduce sulphuric acid to SO2.  Hence, HCl is not oxidized to Cl2.

When the mixture contains bromide ions, initially produced HBr being a strong reducing agent than HCl.

It reduces sulphuric acid to SO2 and itself gets oxidized to produce Br2.


Question 8.13

Identify the substance oxidized reduced, oxidising agent and reducing agent for each of the following reactions:

  • 2AgBr (s) + C6H6O2 (aq) --> 2Ag(s) + 2HBr (aq) + C6H4O2 (aq)
  • HCHO (l) + 2 [Ag(NH3)2]+ (aq) + 3OH- (aq) --> 2Ag(s) + HCOO- (aq) + 4NH3 (aq) + 2H2O(l)
  • HCHO (l) + 2Cu2+ (aq) + 5OH- (aq) --> Cu2O (s) + HCOO- (aq) + 3H2O (l)
  • N2H4 (l) + 2H2O2 (s) + 2H2SO4 (aq) -->2PbSO4 (s) + 2H2O(l)
  • Pb(s) + PbO2(s) + H2SO4 (aq)--> 2PbSO4 (s) + 2H2O(l)


             Oxidised         Reduced       Oxidising agent        Reducing agent

  •   C6H6O2                 AgBr                AgBr                 C6H6O
  • HCHO                [Ag(NH3)2]+   [Ag(NH3)2]+               HCHO
  • HCHO                 Cu2+                   Cu2+                          HCHO
  • N2H4                  H2O2                    H2O2                   N2H4
  • Pb                     PbO2                     PbO2                          Pb



Question 8.14

Consider the reactions:

2S2O32- (aq) + I2 (s) --> S4O62- (aq) + 2I-(aq)

S2O32- (aq) + 2Br2 (l) + 5H2O (l) --> 2SO43- (aq) + 4Br- (aq) + 10H+ (aq)

Why does the same reactant, thiosulphate react differently with iodine and bromine?


The average oxidation number of S in S2O32- is +2 while in S4O62-, it is +2.5.

Oxidation number of S in SO42- = + 6

Br2 is a stronger oxidizing agent than I2. It oxidises S to S2O32- to a

higher oxidation state of +6 and hence forms SO42- ions.

I2 being weaker oxidizing agent oxidizes S to S2O32- ion to lower oxidation of +2.5 in S4O62- ion.

This happens because thiosulphate reacts differently with Br2 and I2.


Question 8.15

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds,

hydroiodic acid is the best reductant.


Halogens have a strong tendency to accept electrons. They are strong oxidizing agents.

Their relative oxidizing power is measured in terms of their electrode potentials.

Electrode potential of halogens decreases in order


F2 oxidises Cl- to Cl2, Br- to Br2 , I- to I2.

Cl2 oxidises Br- to Br2 and F to I2 but not F- to F2.

Br2 oxidises F to I2 but F- to F2 and Cl- to Cl2.


Hence, F2 is the best oxidant.

Halide ion has tendency to lose electrons and hence can act as reducing agents.

Since, electrode potentials of halide ions decrease in the order:


So, reducing power of the halide ions decreases in the same order. HI > HBr > HCl > HF.

Hydroiodic acid is the best reductant.

Let us take some examples:

HI and HBr reduce sulphuric acid to SO2 while HCl and HF do not.


Question 8.16

Why does the following reaction occur?

XeO64- (aq) + 2F- (aq) + 6H+ (aq)--> XeO3 (g) + F2 (g) + 3H2O(l)

What conclusion about the compound Na4XeO6 (of which XeO64- is a part) can be drawn from the reaction?



Oxidation number of Xe decreases from +8 in XeO64- to +6 in XeO3 while that of F increases from –I in F- to 0 in F2.

XeO64- is reduced while F- is oxidized. This reaction occurs because Na2XeO64- is a stronger oxidizing agent than F2.


Question 8.17

Consider the reactions:

  • H3PO2 (aq) + 4AgNO3 (aq) + 2H2O (l) --> H3PO4 (aq) + 4Ag (s) + 4 HNO3(aq)
  • H3PO2 (aq) + 2CuSO4 (aq) + 2H2O (l) --> H3PO4 (aq) + 2Cu(s) + H2SO4 (aq)
  • C6H5CHO (l) + 2[Ag(NH3)2]+ (aq) + 3OH- --> C6H5COO- (aq) + 2Ag(s) + 2NH3 (aq) + 2H2O (l)
  • C6H5CHO (l) + 2Cu2+ (aq) + 5OH- (aq) --> No change observed

What inference do you draw about the behavior of Ag+ and Cu2+ from these reactions?


Reaction (a) and (b) indicates that H3PO4 is a reducing agent and this reduced AgNO3 and CuSO4 to Ag and Cu respectively.

AgNO3 and CuSO4 acts as oxidizing agent and H3PO2 to H3PO4.

In reaction (c), [Ag(NH3)2]+ oxidizes C6H5CHO to C6H5COO- .

In reaction (d) Cu2+ ions cannot oxidize C6H5CHO to C6H5COO- .

Hence, Ag+ ion is a strong deoxidizing agent than Cu2+ ion.


Question 8.18

Balance the following redox reactions by ion-electron method.

  • MnO4- (aq) + I- (aq) --> MnO2 (s) + I2 (s) (in basic medium)
  • MnO4- (aq) + SO2 (g) --> Mn2+ (aq) + HSO4- (in acidic solution)
  • H2O2 (aq) + Fe2+ (aq) --> Fe3+ (aq) + H2O (l) (in acidic solution)
  • Cr2O72- + SO2 (g) --> Cr3+ (aq) + O2 (g) + H+ (in acidic solution)


  • Skeleton equation is




Question 8.19

Balance the following equations in basic medium by ion-electron method and oxidation number

methods and identify agent and the reducing agent.

  • P4 (s) + OH- (aq) --> PH3 (g) + HPO2- (aq)
  • N2H4 (l) + ClO3- (aq) --> NO (g) + Cl- (g)
  • Cl2O7 (g) + H2O2 (aq) --> ClO2- (aq) + O2 (g) + H+



P4 acts both as an oxidizing as well as a reducing agent.

Oxidation number method:

Total decrease in oxidation number of P4 in PH3 = 3 x 4 = 12

Total increase in oxidation number of P4 in H2PO4 = 1 x 4 = 4

Balance the equation,

P4(s) + OH-(aq) ---> PH3(g) + 3H2PO2-(aq)

Balance O atoms, multiply OH- by 6

P4(s) + 6OH-(aq) ---> PH3(g) + 3H2PO2-(aq)

Balance H atoms,

 P4(s) + 6OH-(aq) +3H2O(l) ---> PH3(g) + 3H2PO2-(aq) + 3OH-(aq)

P4(s) + 3OH-(aq) +3H2O(l) --> PH3(g) + 3H2PO2-(aq) (1)

Ion electron method. There are two half reactions –

Oxidation half reactions:

Balance P atoms,


Balance oxidation number by adding electrons

P4(s) --> 4H2PO2-(aq) + 4e-

Balance charge now,

 Class_11_Chemistry_Redox_Reactions_Equation_48  ----(3)

Reduction half reaction


Balance P atoms

P4(s) --> 4PH3(g)

Now, balance Oxidation number

P4(s) +12e- --> 4PH3(g)

Balance charge

P4(s) +12H2O(l) +12e- ---> 4PH3(g) + 12OH-(aq)

Balance O atoms

P4(s) +12H2O(l) +12e- ---> 4PH3(g) + 12OH-(aq)


Multiply equation (3) and (5)


This is a balanced equation.

  • Reaction is as follows:

 N2H4(l) + ClO3-(aq)  ---> NO(g) + Cl-(aq)

Oxidation number method

Total increase in Oxidation number of N = 2 x 4 = 8

Total decrease in Oxidation number of Cl = 1 x 6 = 6

To enhance increase or decrease in oxidation number multiply

 3N2H4(l) + 4ClO3-(aq)  ---> NO(g) + Cl-(aq)

Balance N and Cl atoms,

 3N2H4(l) + 4ClO3-(aq)  ---> 6NO(g) + 4Cl-(aq)

Balance O

3N2H4(l) + 4ClO3-(aq)  ---> 6NO(g) + 4Cl-(aq) +6H2O(l)   -----(1)

Ion electron method

Oxidation half reaction


Balance Oxidation number

N2H4(l) --> 2 NO(g) + 8e-

Balance charge,

N2H4(l) + 8OH-(aq) --> 2 NO(g) + 8e-

Balance O atoms,


Reduction half reaction


Balance Oxidation number

4ClO3-(aq) +6e- -->Cl-(aq) 

Balance charge,

4ClO3-(aq) +6e- -->Cl-(aq) + 6OH-(aq)

Balance O atoms

4ClO3-(aq) + 3H2O(l) + 6e- --> Cl-(aq) + 6OH-(aq)  (3)

This is a correct balanced reaction half equation.

3N2H4(l) + 4ClO3-(aq) ---> 6NO(g) + 4Cl-(aq) + 6H2O (l)  (4)

This is a correct balanced equation.



Cl2O7 acts as oxidizing agent while H2O2 as the reducing agent.

Oxidation number method

Total decrease in Oxidation number of Cl2O7 = 4 x 2 = 8

Total increase in Oxidation number of H2O2 = 2 x 1 = 2

Balance oxidation number


Balance Cl atoms


Balance O atoms


Balance H atoms


This represents balanced redox reactions.

Ion electron method

Oxidation half reaction


Balance Oxidation number

H2O2(aq) ---> O2(g) +  2e-

Balance charge

H2O2(aq) + 2OH-(aq) ---> O2(g)  + 2e-  (1)

Balance O atoms

H2O2(aq) + 2OH-(aq) ---> O2(g) + 2H2O(l) + 2e-  (1)

Reduction half reaction


Balance Cl atoms

Cl2O7 (g) --> 2 ClO2-(aq)

Balance Oxidation number

Cl2O7 (g) + 8e- ---> 2 ClO2-(aq)

Balance charge

Cl2O7 (g) + 8e- --> 2 ClO2-(aq) + 6OH-

Balance O atoms


Using equation 1 and 2, we get



Question 8.20

What sorts of information can you draw from the following reaction?

(CN)2 (g) + 2OH- (aq) --> CN- (aq) + CNO-(aq) + H2O(l)


Let Oxidation number of C is ‘x’.

Oxidation number of C in cynogen = 2 (x – 3) = 0

x = + 3

Oxidation number of C in cyanide ion,CN- = x- 3 = -1

x = 2

Oxidation number of C in cyanate ion, CNO = x – 3 -2 =- 1

x  = + 4


  • This reaction involves decomposition of cynogen on alkaline medium to cyanide ion and cyanate ion.
  • Oxidation number of C decreases from + 3 in (CN)2 to + 2 in CN- ion and increases from + 3 in (CN)2 to + 4
  • in CNO- Cynogen is reduced to cyanide ion and oxidized to cyanate ion.
  • Cyanogen in s pseudohalogen while cyanide ion is a pseudohalide ion.
  • It is a redox reaction and a disproportionation reaction.


Question 8.21

The Mn3+ ion is unstable in sodium and undergoes disproportionation to give Mn2+, MnO2 and H+ ion.

Write a balanced ionic equation for the reaction.




Question 8.22

Consider the elements:

Cs, Ne, I and F

  • Identify the element that exhibits only negative oxidation state.
  • Identify the element that exhibits only positive oxidation state.
  • Identify the element that exhibit both positive and negative oxidation states.
  • Identify the element which exhibit neither the negative nor does the positive oxidation state.


(i) F exhibits only negative oxidation no. That is -1.

(ii) Cs exhibits only positive oxidation no. That is +1.

(iii) I exhibit both negative and positive oxidation no. That is -1, +1, +3, +5 and +7.

(iv)  Ne exhibits neither negative nor positive oxidation no. That is 0.​


Question 8.23

Chlorine is used to purify drinking water. Excess of chlorine is harmful.

The excess of chlorine is removed by treating with sulphur dioxide.

Present a balanced equation for this redox change taking place in water.


Skeletal equation is



Question 8.24

Refer to the periodic table given in your book and now answer the following questions.

  • Select the possible non metals that can show disproportionation reaction.
  • Select three metals that can show disproportionation reaction.



Question 8.25

In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas

by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide

that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?


Balanced equation for the reaction is


Amount of oxygen available = 20 g (which is less than the amount needed).

So, oxygen is the limiting reagent.

Calculations are made on the amount of oxygen is taken and not on the amount of ammonia taken.

160 g of oxygen produce NO = 120 g

20 g of oxygen will produce NO = (120/160) x 20 = 15 g of NO


Question 8.26

Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

  • Fe3+ (aq) and I- (aq)
  • Ag+ (aq) and Cu (s)
  • Fe3+ and Cu (s)
  • Ag (s) and Fe3+ (aq)
  • Br2 (aq) and Fe2+ (aq)



2Fe3+(aq) + 2I-(aq) -à 2Fe2+(aq) + I2(s); E0 = +0.23V

Since, the EMF is positive, the reaction above is feasible.


Since, EMF is positive, above reaction is feasible.


EMF of the above reaction is positive, above reaction is feasible.

If we take it alternatively,

2Cu(s) + 2Fe3+(aq) --> 3Cu2+(aq) + 2Fe(s)

EMF of the reaction is negative (-0.376 V). Hence, the reaction is not feasible.


EMF of the reaction is negative, above reaction is not feasible.

Alternatively, reaction that can occur is

 3Ag(s) + Fe3+(aq) ----> 3 Ag+(aq) + Fe(s)

EMF of this reaction is also negative (-0.836 V). Hence, this redox reaction is not feasible.


EMF of the reaction is positive, therefore, reaction is feasible.


Question 8.27

Predict the products of electrolysis in each of the following:

  • An aqueous solution of AgNO3 with silver electrodes
  • An aqueous solution AgNO3 with platinum electrodes
  • A dilute solution of H2SO4 with platinum electrodes
  • An aqueous solution of CuCl2 with platinum electrodes


  • AgNO3 ionises to give Ag+ (aq) and NO3- (aq) ions. When electricity is passed,
  • Ag+ ions move towards the cathode while NO3- moves towards the anode.

At cathode, either Ag+ ions or water molecules may get reduced.

Ag+(aq) + e- --> Ag(s); E0 =+0.08V

2H2O + 2e- --->H2(g) + 2OH-(aq); E0 = -0.83 V

Their electrode potential are

Ag(s) ---> Ag+(aq) + e-; E0 = -0.80V

2H2O(l) ---> O2(g) + 4H+(aq) + 4e-; E0 =-1.23V

Oxidation potential of Ag is higher than that of water.

Hence, at anode, it is Ag of silver anode which gets oxidized and not water molecules.

Electrode potential of NO3- is even lower than that of water since, more bonds are broken during reduction of NO3 ions.

When an aqueous solution of AgNO3 is electrolysed, silver from Ag anode dissolves while Ag+ (aq) ions

present in the solution get reduced and deposited on cathode.

  • During electrolysis of AgNO3 solution with platinum electrodes, oxidation of water occurs at anode since,
  • platinum is a noble metal it does not undergo oxidation easily. As a result, oxygen is liberated at the anode.

When an aqueous solution of AgNO3 is electrolysed using platinum electrodes, Ag+ ions from the solution

gets deposited on the cathode while oxygen is liberated at the anode.

  • In aqueous solution, H2SO4 ionises to give H+ and SO42-

When electricity is passed H+ ions move towards cathode while SO42- ions move towards anode.

Their electrode potentials are

2H+(aq) + 2e- --> H2(g) ; E0 = 0.0V

H2O (aq) + 2e- ---> H2(g) + 2OH-(aq); E0 = -0.83 V

Electrode potential of H+ ions is higher than water. At cathode, H+ ions are reduced to evolve hydrogen gas.

CuCl2(aq) ---> Cu2+(aq) + 2Cl- (aq)

On passing electricity, Cu2+ ions move towards cathode and Cu2+ ions move towards anode.

At cathode, Cu2+ or water molecules are reduced.

Electrode potentials are

Cu2+ + 2e- --->Cu(s); E0 =+0.34 V

H2O(l) + 2e- ---> H2(g) + 2OH-; E0 = -0.83 V

Since, electrode potential of Cu2+ ions is much higher than that of water, at cathode Cu2+ is reduced and not water molecules.

At anode, either Cl- ions or water molecules are oxidized.

Their oxidation potentials are

2Cl-(aq) ---> Cl2(g) + 2e-; ∆E0 = -1.36V

2H2O(l) ---> O2(g) + 4H+(aq) + 4e- ; ∆E0 = -1.23V

When an aqueous solution of CuCl2 is electrolysed, Cu metal is liberated at the cathode while Cl2 gas is evolved at the anode.                                


Question 8.28

Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn


A metal with stronger reducing power displaces another metal with weaker reducing power from its solution of salt.

The order of the increasing reducing power of the given metals is as given below:

Cu < Fe < Zn < Al < Mg

Therefore, Mg can displace Al from its salt solution, but Al cannot displace Mg.

Thus, the order in which the given metals displace each other from the solution of their salts is as given below: Mg >Al>Zn> Fe >Cu


Question 8.29

Given the standard electrode potentials,

K+/K = -2.93V, Ag+/Ag = 0.80 V

Hg2+/Hg = 0.79 V

Mg2+/Mg = -2.37 V

Cr3+/Cr = -0.74 V

Arrange these metals in their increasing order of reducing power.


Lower the electrode potential, better is the reducing agent.

Electrode potential increase as

K+/K = -2.93V, Mg2+/Mg = -2.37 V, Cr3+/Cr = -0.74 V, Hg2+/Hg = 0.79 V, Ag+/Ag = 0.80 V.

Reducing power of metals decreases in same order is K, Mg, Cr, Hg, Ag.


Question 8.30

Depict the galvanic cell in which the reaction Zn(s) + 2Ag+ (aq) --> Zn2+ (aq) + 2Ag (s) takes place. Further show:

  • Which of the electrode is negatively charged
  • The carriers of the current in the cell, and
  • Individual reaction at each electrode.


Given reaction:

Zn(s) + 2Ag+ (aq) --> Zn2+ (aq) + 2Ag (s)

Zn gets oxidized to Zn2+ ions and Ag+ gets reduced to Ag metal.

Oxidation occurs at zinc electrode and reduction occurs at silver electrode.

  • Oxidation occurs at zinc electrode and accumulate on zinc electrode. Hence, zinc electrode is negatively charged.
  • Ions are the carriers of the current in the cell.
  • Reactions which occurs at the two electrodes:

    Zn(s) --> Zn2+(aq) + 2e-

    Ag+(aq) + e- --> Ag(s)

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