Class 12 Chemistry Electrochemistry NCERT Solutions: Chapter 3 Questions with Answer PDF

A 3.1 To determine the standard electrode potential of the system Mg²⁺|Mg, connect it to the standard hydrogen electrode (SHE). Keep the Mg²⁺|Mg system as cathode and SHE as cathode. This is represented as shown below.
Pt(s) | H₂(g, 1 bar)| H⁺ (aq, 1 M) ||Mg²⁺ (aq, 1M)| Mg
The electrode potential of a cell is given by
E^⊖ = E^⊖R – E^⊖L
Where, 
E^⊖R- Potential of the half-cell in the right side of the above representation
E^⊖L- Potential of the half-cell in the left side of the above representation
It is to be noted that the potential of the standard hydrogen electrode is zero.
Therefore, E^⊖L = 0
E^⊖ = E^⊖R – 0 
⇒ E^⊖ = E^⊖R

A 3.2 NO, because Zn is very reactive with Cu. It reacts with copper sulphate to form zinc sulphate i.e., Zn displaces Cu and metallic Cu is also formed.
The reaction is given as:
Zn + CuSO₄ ⇒ ZnSO₄ + Cu

A 3.3 :For a substance to oxidise Fe²⁺ to Fe³⁺ ion, it must have high reduction potential than Fe³⁺. The reduction potential of Fe³⁺ to Fe²⁺ reaction is 0.77V, the substances which have reduction potentials higher than this value will oxidise Fe²⁺ ions. Comparing the values, from the table:

A 3.4 Given: 
For hydrogen electrode, pH = 10
n = 1
(n = moles of e- from balanced redox reaction)
On using the formula [H+] = 10– pH

⇒ [H+] = 10 − 10 M
We know,

 

A 3.5 Given: 
[Ag⁺] = 0.002 M
[Ni²⁺] = 0.160 M
n = 2
(n = moles of e- from balanced redox reaction)
E⁰cell= 1.05 V
Now, using the Nernst equation we get,

A 3.6 Given: 
2Fe³⁺(aq) + 2I^- (aq) → 2Fe²⁺(aq) + I₂(s)
E^0cell = 0.236V
n = moles of e^- from balanced redox reaction = 2
F = Faraday's constant = 96,485 C/mol
T = 298 K.
Using the formula, we get
∆_rG⁰ = – nFE⁰_cell
⇒ ∆_rG⁰ = – 2 × FE⁰_cell
⇒ ∆_rG⁰ = −2 × 96485 C mol^-1 × 0.236 V
⇒ ∆_rG⁰ = −45540 J mol^-1
⇒ ∆_rG⁰ = −45.54 kJ mol^-1
Now,
∆_rG⁰ = −2.303RT log K_c
Where, K is the equilibrium constant of the reaction

R is the gas constant; R = 8.314 J-mol-C^-1
⇒ −45540 J mol^-1 = –2.303× (8.314 J-mol-C^-1)× (298 K) × (log K_c)
Solving for K_c we get,
⇒ logK_c = 7.98
Taking antilog both side, we get
⇒ K_c = Antilog (7.98)
⇒ K_c = 9.6 × 107
Trick to remember

A 3.7 The conductivity of a solution depends on the amount of ions present per volume of the solution. When diluted, the concentration of the ions decreases which implies that the number of ions per volume decreases thus, in turn, conductivity decreases.

A 3.8

A 3.9 C = 0.025 mol L^-1
A_m = 46.1 Scm² mol L^-1
λ⁰ (H⁺) = 349.6 Scm² mol L^-1
λ⁰ (HCOO^-) = 54.6 Scm² mol L^-1
Λ⁰m (HCOOH) = Λ⁰(H⁺) + Λ⁰(HCOO^-) 
= 349.6 + 54.6
= 404.2 S cm² mol L^-1

^{Now, the degree of dissociation:}

A 3.10 Current I = 0.5A
Time t = 2hrs = 2×60×60 = 7200 seconds
Charge Q = I × t
Q = 0.5×7200 = 3600 C
Charge carried by 1 mole of electrons (6.023×10²³electrons) is equal to 96487C.
No of electrons = 6.023×10²³ × 3600/96487
No of electrons = 2.25×10²² electrons

A 3.11 Metals with greater reactivity can be extracted electrolytically. Sodium, potassium, calcium, lithium, magnesium, aluminium which are present in the top of the reactivity series are extracted electrolytically. 

Cr₂O₇^2– + 14H⁺ + 6e^–⇒ 2Cr³⁺ + 7H₂O

A 3.12 

Cr₂O₇^2– + 14H⁺ + 6e^–⇒ 2Cr³⁺ + 7H₂O

For reducing one mole of Cr₂O₇^2–, 6 mole of electrons are required. Hence, 6 faraday charge is needed. Hence, 6F = 6×96487 = 578922 C. Thus, the quantity of electricity is needed is 578922 C.

A 3.13

Anode: Lead (Pb)

Cathode: a grid of lead packed with lead oxide (PbO₂)

Electrolyte: 38% solution of sulphuric acid (H₂SO₄)

The cell reactions are as follows :

Pb (s) + SO₂^-4(aq) ⇒ PbSO₄ (s) + 2e^- (anode)

PbO₂(s) + SO₂^-4(aq) + 4H⁺(aq) +2e^-⇒ PbSO₄(s) +2H₂O(l) (cathode)

Pb(s) + PbO₂(s) +2H₂SO₄(aq)⇒ 2PbSO₄(s) +2H₂O(l)

(overall cell reaction)

On charging, all these reactions will be reversed.

A 3.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

A 3.15 In the corrosion reaction, due to the presence of air and moisture, oxidation takes place at a particular point of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

Fe(s) ⇒ Fe²⁺ (aq) + 2e-

Electrons released at the anodic spot move through the metal and go to another spot of the object, wherein presence of H⁺ ions, the electrons reduce oxygen. This spot behaves as the cathode. These H⁺ ions come either from H₂CO₃, which are formed due to the dissolution of carbon dioxide from the air into water. The cathodic reaction is given by

O_2(air) + 4H_aq⁺+4e^-⇒ 2H O

The overall reaction is given by,

2Fe(s) + O_2(air) + 4H_(aq)⁺⇒ 2Fe²⁺+2H₂O

A 3.1 The order in which the given metals displace each other from the solution of their salts is given by,
Mg>Al> Zn> Fe> Cu
A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt. The order of increasing the reducing power of given metals is Cu< Fe< Zn< Algiven metals displace each other from the solution of their salts is given by, Mg>Al> Zn> Fe> Cu. This is hence arranged in decreasing order of its reactivity

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

A 3.2 Reducing power of metals increase with the decrease of reduction potential. Hence, the increasing order of reducing power will be as,
Ag < Hg < Cr < Mg < K
When the reduction potential is lower, the element has more tendency to get oxidized and thus more will be reducing power. The metal that has more negative electrode potential will be the one with more reducing power. Thus, here potassium (K) has the highest reducing power among the given elements.

A 3.3 The galvanic cell corresponding to the given redox reaction can be represented as:

Zn|Zn²⁺_(aq)||Ag ⁺_(aq)|Ag

• 1) Zn electrode (anode) is negatively charged because, at this electrode, Zn is oxidized to Zn²⁺, causing electron accumulation at the
• 2) Electrons (ions) are the carriers of the current in the cell and in the external circuit, current flows from Ag (cathode) to Zn(anode) which is normally opposite to the electron flow which is from anode to cathode.
• 3) At anode:

Zn_(s)⇒ Zn2 + (aq) + 2e– At cathode:

Ag + (aq) + e –⇒ Ag(s)

A 3.4 (1) Known - E⁰_Cr³⁺_/Cr = - 0.74 V

E⁰ cd²⁺ = - 0.40 V

∆_rG⁰ = ? K = ?

The galvanic cell of the given reaction is written as - Cr_(s)|Cr³⁺_(aq)|| Cd²⁺_(aq)|Cd_(s)→ Reaction 1

Hence, the standard cell potential is given as, E⁰ = E_R⁰ - E_L⁰

= - 0.40 - (- 0.74)

∴ E⁰ = + 0.34 V

To calculate the standard Gibb’s free energy, ∆_rG⁰, we use,

∆_rG⁰ = - nE⁰F → Equation 1

wherenF is the amount of charge passed and E⁰ is the standard reduction electrode potential. Substituting n = 6 (no. of e ^- involved in the reaction 1), F = 96487 C mol^-1,

E⁰ = + 0.34 V in Equation 1, we get,l

∆_rG⁰ = - 6×0.34V×96487 C mol^-1

= - 196833.48 CV mol^-1

= - 196833.48 J mol^-1

∴ ∆_rG⁰ = - 196.83348 kJ mol^-1

To find out the equilibrium constant, K, we use the formula,

log K =n E⁰ / 0.0591

=6 X 0.3 V / 0.0591

log K = 34.5177

K = antilog 34.5177

∴ K = 3.294 × 10³⁴

The standard Gibb’s free energy, ∆_rG⁰ is - 196.83348 kJ mol ^–1 and equilibrium constant, K is 3.294 × 10³⁴

(2) Known -

E⁰ Fe³⁺ / Fe²⁺ = 0.77V

E⁰ Ag⁺ / Ag = 0.80 V

∆_rG⁰ = ? K = ?

The galvanic cell of the given reaction is written as - Fe²⁺_(aq)|Fe³⁺_(aq)|| Ag ⁺_(aq)|Ag_(s)→ Reaction 1

Hence, the standard cell potential is given as, E⁰ = E_R⁰ - E_L⁰

= 0.80 - (0.77)

∴ E⁰ = + 0.03 V

To calculate the standard Gibb’s free energy, ∆_rG⁰, we use,

∆_rG⁰ = - nE⁰F → Equation 1

wherenF is the amount of charge passed and E⁰ is the standard reduction electrode potential.

Substituting n = 1 (no. of e ^- involved in the reaction 1), F = 96487 C mol^-1, E⁰ = + 0.03V in Equation 1, we get,

∆_rG⁰ = - 1×0.03V×96487 C mol^-1

= - 2894.61 CV mol ^-1

= - 2894.61 J mol^-1

∴ ∆_rG⁰ = - 2.89461 kJ mol^-1

To find out the equilibrium constant, K, we use the formula,

log K =n E⁰ / 0.0591

=1 X 0.03 V / 0.0591

log K = 0.5076

K = Antilog 0.5076

∴ K = 3.218

The standard Gibb’s free energy, ∆_rG⁰ is - 2.89461 kJ mol ^–1 and equilibrium constant, K is 3.218

(i) Mg(s)|Mg ²⁺(0.001M)||Cu²⁺(0.0001 M)|Cu(s) (ii) Fe(s)|Fe ²⁺(0.001M)||H⁺ (1M)|H2 (g)(1bar)| Pt(s) (iii) Sn(s)|Sn²⁺(0.050 M)||H⁺ (0.020 M)|H2 (g) (1 bar)|Pt(s) (iv) Pt(s)|Br ^– (0.010 M)|Br2 (l )||H⁺ (0.030 M)| H2 (g) (1 bar)|Pt(s).

A 3.5 Ecell = ?

(i) Mg + Cu²⁺ → Mg²⁺ + Cu (n = 2)

E⁰ Cu²⁺ / Cu⁺ = 0.34V

E⁰ Mg  ²⁺ / Mg = - 2.37 V

E_cell⁰ = E_R⁰-E_L⁰

E_cell⁰  = 0.34 - ( - 2.37) → Equation 1

Using Nernst equation, we get,

 Substituting Equation 1 in equation 2, we get,

∴ E_cell = 0.34 - ( - 2.37) - 0.0591 / 2 log 10 ^-3/ 10^-4

= 2.71 - 0.0591 / 2 log 10

= 2.71 - 0.02955

∴ E_cell = 2.68 V

The e.m.f of the cell, E_cell is 2.68 V

ii) Fe + 2H ⁺ → Fe²⁺ + H₂ (n = 2)

E⁰ H⁺ / H2 = 0V

E⁰ Fe  ²⁺ / Fe = - 0.44 V

E_cell⁰ = E_R⁰-E_L⁰

E_cell⁰  = 0 - ( - 0.44) → Equation 1

Using Nernst equation, we get,

∴ E_cell = 0.5287 V

The e.m.f of the cell, E_cell is 0.5287 V

iii) Sn + 2H ⁺ → Sn²⁺ + H₂ (n = 2)

E_cell⁰ = 0 - ( - 0.14) → Equation 1

Using Nernst equation, we get,

A 3.6 Given - Zn → Zn²⁺ + 2e ^- , E⁰ = 0.76V (anode)

Ag₂O + H₂O + 2e ^- →2Ag + 2OH ^- , E⁰ = 0.344V (cathode), n = 2

Δ_rG⁰ = ?

E_cell⁰= ?

Zn is oxidized and Ag₂O is reduced.

Hence, the standard cell potential, E_cell⁰ is given as,

E_cell⁰ = ER⁰ - E_L⁰

E⁰ _cell = 0.344 + 0.76

∴ E⁰_cell = 1.104 V

To calculate the standard Gibb’s free energy, ∆_rG⁰, we use,

∆_rG⁰ = - nE⁰F → Equation 1

= - 2×96487×1.104 J

= - 213043.296 J

∴ ∆_rG⁰ = - 2.13×10⁵ J

The standard cell potential, E⁰_cell is 1.104 V and the standard Gibb’s free energy, ∆_rG⁰ is - 2.13×10⁵ J

A 3.7 The conductivity of a solution is defined as the conductance of one unit volume of solution kept between two platinum electrodes with a unit area of cross-section and at a distance of unit length.The molar conductivity of the solution is defined as the conducting power of all the ions produced by 
one gram mole of an electrolyte in a solution. It is denoted by ∧m. 

The conductivity of a solution (both for strong and weak electrolytes) always decreases with the decrease in concentration of the electrolyte i.e., on dilution. This pattern is seen because the number of ions per unit volume that carry the current in the solution decreases on dilution. The molar conductivity of the solution increases with the decrease in concentration of the electrolyte. This is because both the number of ions as well as mobility of ions increases with dilution.

A 3.8 Given - 
Molarity, C = 0.20 M
Electrolytic conductivity of a solution, κ = 0.0248 S cm^-1
Molar conductivity = ?
Molar conductivity, ∧_m = K/C X 1000 S cm² mol^-1
=0.0248S cm^-1 X 1000 Cm³L^-1 / 0.20 mol L^-1

∴ ∧_m = 124 S cm² mol^-1
Molar conductivity(∧_m) of 0.20 M solution of KCl at 298 K is 124 S cm² mol^-1

A 3.9 Given - 
Resistance of a conductivity cell, R = 1500 Ω
Electrolytic conductivity of a solution, κ = 0.146 × 10^-3 S cm^-1
Cell constant = ?
Conductivity, κ = cell constant/resistance
Cell constant = κ × R
= 0.146 × 10^-3 S cm^-1×1500 Ω
Cell constant = 0.219 cm^-1
The cell constant of the cell containing 0.001M KCl solution at 298 K is 0.219 cm^-1

A 3.10

?⁰ R = Intercept on the ? axis = 124.0 S cm² mol^-1, which is obtained by extrapolation to zero concentration.

A 3.11 Given -

Molarity, C = 0.00241 M

Conductivity, κ = 7.896 × 10^–5 S cm^–1

Molar conductivity, ?_m = ?

?⁰_m for acetic acid = 390.5 S cm²mol^–1

Molar conductivity, ?_m = k/c X 1000 S cm² mol^-1

 = 7.896 X 10^–5 S cm^–1X 1000 cm³ L^-1  / 0.00241 mol L^-1

∴ ?_m = 32.76 S cm² mol^-1

To calculate the dissociation constant, K_a, we use

K_a =  → Equation 1

Here, we need to find the value of α (degree of dissociation), by the formula,

The molar conductivity, ?_m is 32.76 S cm² mol^-1 and the dissociation constant, K_a is 1.86×10^-5

A 3.12 The electrode reaction is given as, 

Al³⁺ (aq) + 3e^- → Al(s)

∴ The quantity of charge required for the reduction of 1 mol of Al³⁺ = 3F

= 3×96487 C

= 289461 C

The electrode reaction is given as,

Cu²⁺ (aq) + 2e^- → Cu(s)

∴ The quantity of charge required for the reduction of 1 mol of Cu²⁺ = 2F

= 2×96487 C

= 192974 C

The electrode reaction is given as, MnO₄→ Mn²⁺

i.e., Mn⁷⁺ + 5e ^- → Mn²⁺

∴ The quantity of charge required for the reduction of 1 mol of Mn⁷⁺ = 5F

= 5×96487 C

= 482435 C

A 3.13 (i) Ca²⁺ + 2e^- → Ca
⇒ Here, 1 mole of Ca, i.e., 40g of Ca requires = 2 F electricity (F if Faraday)
∴ 20g of Ca requires = 20X2/40
= 1 F of electricity
Electricity in terms of Faraday required to produce 20.0 g of Ca from molten CaCl₂ is 1 F of electricity.
(ii) Al³⁺ + 3e^-  → Al
⇒ 1 mole of Al, i.e., 27g of Al requires = 3 F electricity (F if Faraday)
∴ 40.0 g of Al will require = 3/27 X 40
= 4.44 F of electricity
Electricity in terms of Faraday required to produce 40.0 g of Al from molten Al₂O₃ is 4.44 F of electricit

A 3.14 The electrode reaction for 1 mole of H₂O is given as,

H₂O → H₂ + 1/2O₂

i.e., O^2- →1/2 O₂ + 2e ^-

∴ The quantity of electricity required = 2F

= 2×96487 C

= 192974 C

The quantity of electricity required in coulomb for the oxidation of 1 mol of H₂O to O₂ is 192974 C

• (ii) The electrode reaction for 1 mole of FeO is

FeO + 1/2 O₂ → 1/2 Fe₂O₃

i.e., Fe²⁺ → Fe³⁺ + e ^-

∴ The quantity of electricity required = 1F

= 1×96487 C

= 96487 C

The quantity of electricity required in coulomb for the oxidation of 1 mol of FeO to Fe₂O₃ is 96487 C

A 3.15 Quantity of electricity passed = 5 A × (20 × 60 sec)
= 6000 C ⇒ Equation 1
The electrode reaction is written as,
Ni²⁺ + 2e → Ni
Thus, the quantity of electricity required = 2F
= 2×96487 C
= 192974 C
∵ 192974 C of electricity deposits 1 mole of Ni, which is 58.7 g ⇒ Equation 2
Thus, equating equations 1 and 2, we get

192974 C of electricity deposits = 58.7 g
6000 C of electricity will deposit = 58.7 X 6000 / 192974
= 1.825g of Ni
The mass of Ni deposited at the cathode is 1.825g of Ni

A 3.16 Equivalent weight is Ag, E_Ag = 180/1  = 180

Equivalent weight is Cu, E_Cu = 63.5 / 2 = 31.75

Equivalent weight is Zn, E_Zn= 65/2 = 32.5

Using Faraday’s second law of electrolysis, to find the mass of Cu and Zn, we use Equation 1,

∴ W_Zn = 0.436 g

To find the time of current flow, using Faraday’s first law of electrolysis we get,

M = Z ×I ×t ⇒ Equation 2

∵ Z = Equivalent Weight / 96487 , Equation 2 becomes,

M = 108 / 96487 X 1.5 X t

t = 1.45 X 96487 / 108X 1.5

t = 864 seconds.

The time of current flow, t = 864 seconds, the mass of Cu is 0.426 g and mass of Zn is 0.436 g

A 3.17 The electrode reaction is written as,

2Fe³⁺ + 2I ^- → 2Fe²⁺ + I₂

= 0.54V - 0.77V

∴ E⁰_cell = - 0.23 V

It is not feasible, as E⁰_cell is negative, ∴ ∆G⁰ is positive.

• The electrode reaction is written as,
• 2Ag⁺ (aq) + Cu(s)→ Cu²⁺ (aq) + Ag(s)

= + 0.80V - 0.34V

∴ E⁰_cell = 0.46V

It is feasible, as E_cell ⁰ is positive, ∴ ∆G⁰ is negative.

• (iii) The electrode reaction is written as,
• 2Fe³⁺ (aq) + 2Br^- (aq)→ 2Fe²⁺ (aq) + Br₂

k

= 0.77V - 1.09V

∴ E⁰_cell = - 0.32 V

It is not feasible, as E⁰_cell is negative, ∴ ∆G⁰ is positive.

• (iv) The electrode reaction is written as,
• Ag(s) + Fe³⁺ (aq) → Fe²⁺ (aq) + Ag⁺ (aq)

= 0.77V - 0.80V

∴ E⁰_cell = - 0.03

It is not feasible, as E⁰_cell is negative, ∴ ∆G⁰ is positive.

• (v) The electrode reaction is written as,
• Br₂ + 2Fe²⁺ (aq) → 2Br^- (aq) + 2Fe³⁺ (aq)

= 1.09V - 0.77V

∴ E⁰_cell = 0.32 V

It is feasible, as E_cell⁰ is positive, ∴ ∆G⁰ is negative.

A 3.18 Given -

All the ions are in aqueous state.

Reaction in solution:

AgNO_3(s) + aq → Ag ⁺ + NO^3-

H₂O → H ⁺ + OH ^-

At cathode:

Ag + (aq) + e - →Ag(s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference to H ⁺ ions.

At anode:

Ag(s)→ Ag + (aq) + e -

As Ag anode is attacked by NO^3- ions, Ag of the anode will dissolve to form Ag ⁺ ions in the aqueous solution.

(ii) Reaction in solution:

AgNO_3(s) + aq → Ag ⁺ + NO^3-

H₂O óH ⁺ + OH ^-

At cathode:

2Ag ⁺_(aq) + 2e ^- →2Ag_(s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference to H ⁺ ions.

At anode:

2OH^- (aq) → O₂(g) + 2H⁺ (aq) + 4e -

As anode is not attackable, out of OH ^- and NO^3- ions, OH ^- having lower discharge potential, will be discharged in preference to NO^3- ions. These then decompose to give out O₂.

(iii) Reaction in solution

H₂SO₄(aq) → 2H⁺ (aq) + SO^2-₄  (aq)

At cathode:

2H + (aq) + 2e - →H₂(g)

At anode:

2OH^- (aq) → O₂(g) + 2H⁺ (aq) + 4e -

∴ H₂ gas is evolved at cathode and O_2(g) is evolved at anode.

(iv) Reaction in solution: CuCl₂(s) + aq → Cu²⁺ (aq) + Cl^- (aq)

H₂O óH ⁺ + OH ^-

At cathode:

Cu²⁺ (aq) + 2e - →Cu(g)

At anode:

2Cl^- (aq) - 2e^- → Cl₂(g)

∴ Cu will be deposited at cathode and Cl₂ gas will be liberated at anode.