NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 3 – Free PDF Download

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry play a pivotal role in the CBSE Class 12 Chemistry board examination. NCERT Solutions for Class 12 Chemistry are comprehensive materials that have answers to the exercise present in the NCERT Textbook. These solutions are developed by subject experts at BYJU’S, following the latest CBSE Syllabus for 2023-24 and its guidelines.

By studying these NCERT Solutions for Class 12 Chemistry, students will be able to solve different kinds of questions that might appear in the board examination and entrance examinations. These solutions are presented in a clear and step-wise format for ease of understanding. To download the NCERT Solutions for Class 12 Chemistry Chapter 3 PDF, click the link given below.

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

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Class 12 Chemistry NCERT Solutions Chapter 3 Electrochemistry – Important Questions

Q 3.1:

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.

Answer:

According to their reactivity, the given metals replace the others from their salt solutions in the said order: Mg, Al, Zn, Fe, and Cu.

Mg: Al: Zn: Fe: Cu

Q 3.2:

Given the standard electrode potentials.
K+/K = –2.93V

Ag+/Ag = 0.80V

Hg2+/Hg = 0.79V

Mg2+/Mg = –2.37 V

Cr3+/Cr = – 0.74V
Arrange these metals in their increasing order of reducing power.

Ans:

The reducing power increases with the lowering of the reduction potential. In order of given standard electrode potential (increasing order): K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag

Thus, in the order of reducing power, we can arrange the given metals as Ag< Hg < Cr < Mg < K

Q 3.3 :

Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) →Zn2+(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Ans :

The galvanic cell in which the given reaction takes place is depicted as

\(\begin{array}{l}Zn_{ ( s ) } | Zn^{ 2+ }_{( aq )}||Ag^{ + }_{( aq )}|Ag_{( s )}\end{array} \)

(i) The negatively charged electrode is the Zn electrode (anode).

(ii) The current carriers in the cell are ions. Current flows to zinc from silver in the external circuit.

(iii) Reaction at the anode is given by

\(\begin{array}{l}Zn_{ ( s ) }\rightarrow Zn^{ 2+ }_{( aq )} + 2 e^-\end{array} \)

Reaction at the anode is given by

\(\begin{array}{l}Ag^{+}_{ ( aq ) } + e^- \rightarrow Ag_{( s )}\end{array} \)

Q 3.4:

Calculate the standard cell potentials of the galvanic cell in which the following
reactions take place.
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ∆rGJ and equilibrium constant of the reactions.

Ans :

(i)

\(\begin{array}{l}E^{\Theta}_{Cr^{3+}/Cr}\end{array} \)
= 0.74 V

\(\begin{array}{l}E^{\Theta}_{Cd^{2+}/Cd}\end{array} \)
= -0.40 V

The galvanic cell of the given reaction is depicted as

\(\begin{array}{l}Cr_{ ( s ) }|Cr^{ 3+ }_{ ( aq ) }||Cd^{ 2+ }_{ aq }|Cd_{ ( s ) }\end{array} \)

Now, the standard cell potential is

\(\begin{array}{l}E^{\Theta }_{cell} = E^{\Theta }_{g}-E^{\Theta }_{L}\end{array} \)

= – 0.40 – ( -0.74 )

= + 0.34 V

In the given equation, n = 6

F = 96487 C mol−1

\(\begin{array}{l}E^{\Theta }_{cell}\end{array} \)
= + 0.34 V

Then,

\(\begin{array}{l}\Delta_rG^{\Theta}\end{array} \)
= −6 × 96487 C mol−1 × 0.34 V

= −196833.48 CV mol−1

= −196833.48 J mol−1

= −196.83 kJ mol−1

Again,

\(\begin{array}{l}\Delta_rG^{\Theta} = – R T ln K\end{array} \)
\(\begin{array}{l}\Delta_rG^{\Theta} = – 2.303 R T ln K\end{array} \)
\(\begin{array}{l}log {k} = \frac{\Delta_rG}{ 2.303 R T }\end{array} \)
\(\begin{array}{l}= \frac{-196.83\times10^{3}}{ 2.303 \times8.314\times 298 }\end{array} \)

= 34.496

K = antilog (34.496) = 3.13 × 1034

The galvanic cell of the given reaction is depicted as

\(\begin{array}{l}Fe^{ 2+ }_{( aq )}|Fe^{ 3+ }_{( aq )}|| Ag^+_{( aq )}|Ag_{( s )}\end{array} \)

Now, the standard cell potential is

\(\begin{array}{l}E^{\Theta }_{cell} = E^{\Theta }_{g}-E^{\Theta }_{L}\end{array} \)

Here, n = 1

Then,

\(\begin{array}{l}\Delta_t G^0 = -nFE^0_{cell}\end{array} \)

= −1 × 96487 C mol−1 × 0.03 V

= −2894.61 J mol−1

= −2.89 kJ mol−1

Again,  

\(\begin{array}{l}\Delta_t G^0 = -2.303 RT \; ln K\end{array} \)
\(\begin{array}{l}ln K = \frac{\Delta_t G}{ 2.303 RT }\end{array} \)
\(\begin{array}{l}= \frac{-2894.61 }{ 2.303 \times 8.314 \times 298 }\end{array} \)

= 0.5073

K = antilog (0.5073)

= 3.2 (approximately)

Q 3.5:

Write the Nernst equation and emf of the following cells at 298 K.
(i) Mg(s)|Mg2+(0.001M)||Cu2+(0.0001 M)|Cu(s)
(ii) Fe(s)|Fe2+(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)

Answer

(i) For the given reaction, the Nernst equation can be given as

\(\begin{array}{l}E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{[Mg^{2+}]}{[Cu^{2+}]}\end{array} \)
\(\begin{array}{l}= 0.34 – (-2.36) – \frac{0.0591}{2} log \frac{0.001}{0.0001}\end{array} \)
\(\begin{array}{l}2.7 -\frac{0.0591}{2}log10\end{array} \)

= 2.7 − 0.02955

= 2.67 V (approximately)

(ii) For the given reaction, the Nernst equation can be given as

\(\begin{array}{l}E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{[Fe^{2+}]}{[H ^{+}]^2}\end{array} \)

= 0 – ( – 0.14) –

\(\begin{array}{l}\frac{0.0591}{n}log\frac{0.050}{(0.020)^{2}}\end{array} \)

= 0.52865 V

= 0.53 V (approximately)

(iii) For the given reaction, the Nernst equation can be given as

\(\begin{array}{l}E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{[Sn^{2+}]}{[H ^{+}]^2}\end{array} \)

= 0 – ( – 0.14) –

\(\begin{array}{l}\frac{0.591}{2}log\frac{0.050}{(0.020)^2}\end{array} \)

= 0.14 − 0.0295 × log125

= 0.14 − 0.062

= 0.078 V

= 0.08 V (approximately)

(iv) For the given reaction, the Nernst equation can be given as

\(\begin{array}{l}E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{1}{[Br^{-}]^2[H ^{+}]^2}\end{array} \)

= 0 – 1.09 –

\(\begin{array}{l}\frac{0.591}{2}log\frac{1}{(0.010)^2(0.030)^2}\end{array} \)

= -1.09 – 0.02955 x

\(\begin{array}{l}log\frac{1}{0.00000009}\end{array} \)

= -1.09 – 0.02955 x

\(\begin{array}{l}log\frac{1}{9\times 10^{-8}}\end{array} \)

= -1.09 – 0.02955 x

\(\begin{array}{l}log{ (1.11 \times 10^{7} )}\end{array} \)

= -1.09 – 0.02955 x (0.0453 + 7)

= -1.09 – 0.208

= -1.298 V

Q 3.6:

In the button cells widely used in watches and other devices, the following reaction takes place:

Reaction taking place in the button cells

 

Determine ∆r GJ and EJ for the reaction.

Ans:

\(\begin{array}{l}E^0\end{array} \)
= 1.104 V

We know that,

\(\begin{array}{l}\Delta_r G^\Theta = -nFE^\Theta\end{array} \)

= −2 × 96487 × 1.04

= −213043.296 J

= −213.04 kJ

Q 3.7:

Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Answer

The conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. Specific conductance is the inverse of resistivity, and it is represented by the symbol κ. If ρ is resistivity, then we can write

\(\begin{array}{l}k = \frac{1}{\rho}\end{array} \)

At any given concentration, the conductivity of a solution is defined as the unit volume of solution kept between two platinum electrodes with the unit area of the cross-section at a distance of unit length.

\(\begin{array}{l}G = k \frac{a}{l} = k \times 1 = k\end{array} \)
    [Since a = 1 , l = 1]

When concentration decreases, there will be a decrease in Conductivity. It is applicable for both weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity –

The molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte, kept between two electrodes with the area of cross-section A and distance of unit length.

\(\begin{array}{l}\Lambda_m = k \frac{A}{l}\end{array} \)

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte)

\(\begin{array}{l}\Lambda_m = k V\end{array} \)

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution. The variation of

\(\begin{array}{l}\Lambda_m\end{array} \)
with
\(\begin{array}{l}\sqrt{c}\end{array} \)
for strong and weak electrolytes is shown in the following plot :

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry Q 3.7 Plot

Q  3.8:

The conductivity of the 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.

Ans :

Given, κ = 0.0248 S cm−1 c

c = 0.20 M

Molar conductivity,

\(\begin{array}{l}\Lambda_m = \frac{k \times 1000}{c}\end{array} \)
\(\begin{array}{l}= \frac{0.0248 \times 1000}{0.2}\end{array} \)

= 124 Scm2mol-1

Q 3.9:

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1

Answer

Given,

Conductivity, k = 0.146 × 10−3 S cm−1

Resistance, R = 1500 Ω

Cell constant = k × R

= 0.146 × 10−3 × 1500

= 0.219 cm−1

Q 3.10:

The conductivity of sodium chloride at 298 K has been determined at different concentrations, and the results are given below.

Concentration/M            0.001     0.010     0.020     0.050     0.100

102 × k/S m−1                      1.237     11.85     23.15     55.53     106.74

Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0 Λ m.

Ans:

Given,

κ = 1.237 × 10−2 S m−1, c = 0.001 M

Then, κ = 1.237 × 10−4 S cm−1, c1⁄2 = 0.0316 M1/2

\(\begin{array}{l}\Lambda_m =\frac{k}{c}\end{array} \)
\(\begin{array}{l}=\frac{1.237 \times 10^{ -4 } S\;cm^{-1} }{0.001 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\end{array} \)

= 123.7 S cm2 mol−1

Given,

κ = 11.85 × 10−2 S m−1, c = 0.010M

Then, κ = 11.85 × 10−4 S cm−1, c1⁄2 = 0.1 M1/2

\(\begin{array}{l}\Lambda_m =\frac{k}{c}\end{array} \)
\(\begin{array}{l}=\frac{11.85 \times 10^{ -4 } S\;cm^{-1} }{0.010 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\end{array} \)

= 118.5 S cm2 mol−1

Given,

κ = 23.15 × 10−2 S m−1, c = 0.020 M

Then, κ = 23.15 × 10−4 S cm−1, c1/2 = 0.1414 M1/2

\(\begin{array}{l}\Lambda_m =\frac{k}{c}\end{array} \)
\(\begin{array}{l}=\frac{23.15 \times 10^{ -4 } S\;cm^{-1} }{0.020 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\end{array} \)

= 115.8 S cm2 mol−1

Given,

κ = 55.53 × 10−2 S m−1, c = 0.050 M

Then, κ = 55.53 × 10−4 S cm−1, c1/2 = 0.2236 M1/2

\(\begin{array}{l}\Lambda_m =\frac{k}{c}\end{array} \)
\(\begin{array}{l}=\frac{106.74 \times 10^{ -4 } S\;cm^{-1} }{0.050 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\end{array} \)

= 111.1 1 S cm2 mol−1

Given,

κ = 106.74 × 10−2 S m−1, c = 0.100 M

Then, κ = 106.74 × 10−4 S cm−1, c1/2 = 0.3162 M1/2

\(\begin{array}{l}\Lambda_m =\frac{k}{c}\end{array} \)
\(\begin{array}{l}=\frac{106.74 \times 10^{ -4 } S\;cm^{-1} }{0.100 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\end{array} \)

= 106.74 S cm2 mol−1

Now, we have the following data:

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry Q 3.10 Data & Graph

Since the line interrupts

\(\begin{array}{l}\Lambda_m\end{array} \)
at 124.0 S cm2 mol−1,
\(\begin{array}{l}\Lambda^0_m\end{array} \)
=  124.0 S cm2 mol−1

Q 3.11:

The conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0 Λ m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?

Ans:

Given, κ = 7.896 × 10−5 S m−1 c

= 0.00241 mol L−1

Then, molar conductivity,

\(\begin{array}{l}\Lambda_m = \frac{k}{c}\end{array} \)

=

\(\begin{array}{l}\frac{7.896 \times 10^{-5} S cm^{-1}}{0.00241 \; mol \; L^{-1}}\times \frac{1000 cm^3}{L}\end{array} \)

= 32.76S cm2 mol−1

\(\begin{array}{l}\Lambda^0_m =\end{array} \)
390.5 S cm2 mol−1

Again,

\(\begin{array}{l}\alpha =\frac{\Lambda_m }{\Lambda^0_m }\end{array} \)

=

\(\begin{array}{l}= \frac{32.76 \; S\; cm^2 \; mol^{-1} }{390.5 \; S\; cm^2 \; mol^{-1} }\end{array} \)

Now,

= 0.084

Dissociation constant,

\(\begin{array}{l}K_a = \frac{c\alpha^2}{(1-\alpha)}\end{array} \)

=

\(\begin{array}{l}\frac{ ( 0.00241 \; mol \; L^{-1} )( 0.084 )^2}{ ( 1 – 0.084 ) }\end{array} \)

= 1.86 × 10−5 mol L−1

Q 3.12:

How much charge is required for the following reductions?
(i) 1 mol of Al3+ to Al
(ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of MnO4– to Mn2+

Ans :                  

(i)

\(\begin{array}{l}Al^{3+} + 3e^- \rightarrow Al\end{array} \)

Required charge = 3 F

= 3 × 96487 C

= 289461 C

(ii)

\(\begin{array}{l}Cu^{2+} + 2e^- \rightarrow Cu\end{array} \)

Required charge = 2 F

= 2 × 96487 C

= 192974 C

(iii)

\(\begin{array}{l}MnO^-_4 \rightarrow Mn^{2+}\end{array} \)

i.e

\(\begin{array}{l}Mn^{7+} + 5e^-\rightarrow Mn^{2+}\end{array} \)

Required charge = 5 F

= 5 × 96487 C

= 482435 C

Q 3.13:

How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2?
(ii) 40.0 g of Al from molten Al2O3?

Ans:

(i)  From the given data,

\(\begin{array}{l}Ca^{2+} + 2e^- \rightarrow Ca\end{array} \)

Electricity required to produce 40 g of calcium = 2 F

Therefore, electricity required to produce 20 g of calcium = (2 x 20 )/ 40 F

= 1 F

(ii) From the given data,

\(\begin{array}{l}Al^{3+} + 3e^- \rightarrow Al\end{array} \)

Electricity required to produce 27 g of Al = 3 F

Therefore, electricity required to produce 40 g of Al = ( 3 x 40 )/27 F

= 4.44 F

Q 3.14:

How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2?
(ii) 1 mol of FeO to Fe2O3?

Ans :

(i) From the given data,

\(\begin{array}{l}H_2O\rightarrow H_2 + \frac{1}{2}O_2\end{array} \)

We can say that

\(\begin{array}{l}O^{2-}\rightarrow \frac{1}{2}O_2 + 2e^-\end{array} \)

Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F

= 2 × 96487 C

= 192974 C

(ii) From the given data,

\(\begin{array}{l}Fe^{2+}\rightarrow Fe^{3+} + e^-\end{array} \)

Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F

= 96487 C

Q 3.15:

A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Ans :

Given,

Current = 5A

Time = 20 × 60 = 1200 s

Charge = current × time

= 5 × 1200

= 6000 C

According to the reaction,

\(\begin{array}{l}Ni^{2+} + 2e^-\rightarrow Ni_{ (s) } + e^-\end{array} \)

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C =

\(\begin{array}{l}\frac{58.71 \times 6000}{2 \times 96487}g\end{array} \)

= 1.825 g

Hence, 1.825 g of nickel will be deposited at the cathode.

Q 3.16:

Three electrolytic cells, A, B, and C, containing solutions of ZnSO4, AgNO3 and CuSO4, respectively, are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Ans :

According to the reaction,

\(\begin{array}{l}Ag^+_{(aq)} +e^- \rightarrow Ag_{(s)}\end{array} \)

i.e., 108 g of Ag is deposited by 96487 C.

Therefore, 1.45 g of Ag is deposited by =

\(\begin{array}{l}\frac{96487\times 1.45}{107}C\end{array} \)

= 1295.43 C

Given,

Current = 1.5 A

Time = 1295.43/ 1.5 s

= 863.6 s

= 864 s

= 14.40 min

Again,

\(\begin{array}{l}Cu^{2+}_{(aq)} +2e^-\rightarrow Cu_{(s)}\end{array} \)

i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1295.43 C of charge will deposit

\(\begin{array}{l}\frac{63.5 \times 1295.43}{2 \times 96487}\end{array} \)

= 0.426 g of Cu

\(\begin{array}{l}Zn^{2+}_{(aq)} +2e^-\rightarrow Zn_{(s)}\end{array} \)

i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn

Therefore, 1295.43 C of charge will deposit

\(\begin{array}{l}\frac{65.4 \times 1295.43}{2 \times 96487}\end{array} \)

= 0.439 g of Zn

Q 3.17:

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible.
(i) Fe3+(aq) and I–(aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br– (aq)
(iv) Ag(s) and Fe 3+ (aq)
(v) Br2 (aq) and Fe2+ (aq)

Ans :

(i)

Standard electrode potential calculation

 

(ii)

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry Q 3.17(ii)

E0  is positive; hence, the reaction is feasible.

 

(iii)

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry Q 3.17(iii)

E0  is negative; hence, the reaction is not feasible.

 

(iv)

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry Q 3.17(iv)

E0  is negative; hence, the reaction is not feasible.

 

(v)

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry Q 3.17(v)

E0  is positive; hence, the reaction is feasible.

Q  3.18:

Predict the products of electrolysis in each of the following.
(i) An aqueous solution of AgNO3 with silver electrodes
(ii) An aqueous solution of AgNO3 with platinum electrodes
(iii) A dilute solution of H2SO4 with platinum electrodes
(iv) An aqueous solution of CuCl2 with platinum electrodes

Ans:

(i) At the cathode,

The following reduction reactions compete to take place at the cathode.

\(\begin{array}{l}Ag^+_{(aq)}+e^- \rightarrow Ag_{(s)}\end{array} \)
; E0 = 0.80 V
\(\begin{array}{l}H^+_{(aq)}+e^- \rightarrow \frac{1}{2}H_{2(g)}\end{array} \)
;E0 = 0.00 V

The reaction with a higher value of E0 takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At the anode,

The Ag anode is attacked by

\(\begin{array}{l}NO^+_3\end{array} \)
 ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.

(ii) At the cathode,

The following reduction reactions compete to take place at the cathode.

\(\begin{array}{l}Ag^+_{(aq)}+e^- \rightarrow Ag_{(s)}\end{array} \)
; E0 = 0.80 V
\(\begin{array}{l}H^+_{(aq)}+e^- \rightarrow \frac{1}{2}H_{2(g)}\end{array} \)
;E0 = 0.00 V

The reaction with a higher value of E0 takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At the anode,

Since Pt electrodes are inert, the anode is not attacked by

\(\begin{array}{l}NO^+_3\end{array} \)
ions. Therefore, OH− or
\(\begin{array}{l}NO^+_3\end{array} \)
ions can be oxidised at the anode. But OH− ions have a lower discharge potential and get preference and decompose to liberate O2.

\(\begin{array}{l}OH^-\rightarrow OH + E^-\end{array} \)
\(\begin{array}{l}4OH^-\rightarrow 2H_2O + O_2\end{array} \)

(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.

\(\begin{array}{l}H^+_{(aq)}+e^-\rightarrow \frac{1}{2}H_{2(g)}\end{array} \)

At the anode, the following processes are possible.

\(\begin{array}{l}2H_2O_{(l)}\rightarrow O_{2(g)} + 4H^+_{(aq)}+4e^-\end{array} \)
; E0 = +1.23 V             —–(i)
\(\begin{array}{l}2SO^{2-}_{4(aq)}\rightarrow S_2O^{2-}_{6(aq)} + 2e^-\end{array} \)
; E0 = +1.96 V          —–(ii)

For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.

(iv) At the cathode,

The following reduction reactions compete to take place at the cathode.

\(\begin{array}{l}Cu^{2+}_{(aq)}+2e^- \rightarrow Cu_{(s)}\end{array} \)
; E0 = 0.34 V
\(\begin{array}{l}H^+_{(aq)}+e^- \rightarrow \frac{1}{2}H_{2(g)}\end{array} \)
;E0 = 0.00 V

The reaction with a higher value takes place at the cathode. Therefore, the deposition of copper will take place at the cathode.

At the anode,

The following oxidation reactions are possible at the anode.

\(\begin{array}{l}Cl^{-}_{(aq)} \rightarrow \frac{1}{2} Cl_{2(g)}+e^-\end{array} \)
; E0 = 1.36 V
\(\begin{array}{l}2H_20_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} +e^-\end{array} \)
; E0 = +1.23 V

At the anode, the reaction with a lower value of E0 is preferred. But due to the overpotential of oxygen, Cl− gets oxidised at the anode to produce Cl2 gas.

Also Access 
NCERT Exemplar for Class 12 Chemistry Chapter 3
CBSE Notes for Class 12 Chemistry Chapter 3

Electrochemistry is the branch of chemistry that deals with the relationship between chemical energy and electrical energy produced in a redox reaction and how they can be converted into each other. NCERT Solutions for Class 12 are prepared by the best subject experts. In essence, these solutions can be useful for those students preparing for Class 12 exams, JEE Advance and other medical entrance exams. Students can successfully answer the numerical problems based on electrochemistry by downloading the free PDF.

Class 12 NCERT Solutions for Chemistry Chapter 3 Electrochemistry

Chapter 3 Electrochemistry of Class 12 Chemistry, is prepared as per the CBSE Syllabus for 2023-24. Class 12 Chemistry NCERT Solutions for Chapter 3 – Electrochemistry have been designed to help the students prepare well and score good marks in the CBSE Class 12 Chemistry exam. Further, the solutions consist of well thought and structured questions, along with detailed explanations, to help students learn and remember concepts easily.

Subtopics for Class 12 Chemistry Chapter 3 – Electrochemistry

  1. Electrochemical Cells
  2. Galvanic Cells
    1. Measurement of Electrode Potential
  3. Nernst Equation
    1. Equilibrium Constant from Nernst Equation
    2. Electrochemical Cell and Gibbs Energy of Reaction
  4. The conductance of Electrolytic Solutions
    1. Measurement of the Conductivity of Ionic Solutions
    2. Variation of Conductivity and Molar Conductivity with Concentration
  5. Electrolytic Cells and Electrolysis
    1. Products of Electrolysis
  6. Batteries
    1. Primary Batteries
    2. Secondary Batteries
  7. Fuel Cells
  8. Corrosion

After studying Electrochemistry Class 12 important textbook questions solutions, students will be able to describe an electrochemical cell and differentiate between galvanic and electrolytic cells. They will also study the application of the Nernst equation for calculating the emf of a galvanic cell and define the standard potential of the cell.

This chapter has derivations of the relation between the standard potential of the cell, Gibbs energy of cell reaction and its equilibrium constant. This solution will give the definition of resistivity (p), conductivity (K) and molar conductivity ( Am) of ionic solutions; differentiate between ionic (electrolytic) and electronic conductivity; describe the method for measurement of conductivity of electrolytic solutions and calculation of their molar conductivity; justify the variation of conductivity and molar conductivity of solutions with change in their concentration and define Aom (molar conductivity at zero concentration or infinite dilution); enunciate Kohlrausch law and learn its applications; understand quantitative aspects of electrolysis; describe the construction of some primary and secondary batteries and fuel cells and explain corrosion as an electrochemical process.

About BYJU’S NCERT Solutions

BYJU’S provides a comprehensive set of NCERT Solutions for students that have been designed to offer several benefits to the users.  BYJU’S solutions offer the opportunity to learn NCERT Class 12 Chemistry syllabus including Chapter 3 from anywhere and within the comfort zone of the students. Students can learn, practise and revise the different chemistry chapters’ topics right from their homes or from any place. NCERT Solutions for Class 12 Chemistry Chapter 3 are easily accessible, and students can view the solutions right on the website, or they can download and use BYJU’S – The Learning App for a more enhanced learning experience.

With these NCERT Solutions for Class 12, students can easily customise how they learn. Apart from these NCERT Solutions, BYJU’S also has the best subject experts who can guide students to learn the concepts in a more simple and precise manner. Further, if students come across any doubts or queries while going through the Class 12 Chemistry NCERT Solutions, they can always approach BYJU’S responsive support team to clear all their doubts. Besides, BYJU’S keeps track of all the progress that students make and offers feedback, as well as counselling via periodic assessments. Moreover, students can bring in all their queries regarding Chemistry, and other subjects, including Physics, Biology and Maths.

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Frequently Asked Questions on NCERT Solutions for Class 12 Chemistry Chapter 3

Q1

Explain the concept of Electrochemistry discussed in Chapter 3 of NCERT Solutions for Class 12 Chemistry.

Electrochemistry is the concept that deals with both electricity and Chemistry. It explains how electricity can be produced by chemical changes. The reactions which take place in a battery between an anode and a cathode with an electrolyte between them are discussed here. Electrochemical reactions help students to understand the chemical reaction in a battery that produces an electric current. This chapter might be difficult for the students to understand and might need lots of practice to get a grip on the concepts efficiently.
Q2

How should I prepare Chapter 3 of NCERT Solutions for Class 12 Chemistry for the board exams?

Studying this chapter on a daily basis will help you to get a grip on the concepts. Taking notes about complex topics and difficult numericals will boost your confidence during revision. After completing the questions from the NCERT textbook, you can refer to the solutions prepared by the expert faculty at BYJU’S to understand the other method of answering the questions. The solutions contain detailed explanations for minute topics, which are important from the exam point of view.
Q3

Why should I download the NCERT Solutions for Class 12 Chemistry Chapter 3 PDF from BYJU’S?

Students can download the NCERT Solutions for Class 12 Chemistry Chapter 3 PDF from BYJU’S. The solutions created are 100% accurate based on the latest CBSE Syllabus and its guidelines. Students can find the PDF format of solutions, which are free for download on BYJU’S. Both chapter-wise and exercise-format solutions are available to help students boost their exam preparation.

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