NCERT solution Class 12 electrochemistry

NCERT Solutions

Class 12 Chemistry

Electrochemistry

NCERT Questions

Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Q.7
Q.8
Q.9
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Q.12
Q.13
Q.14
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Q.18

Q.4

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

2Cr(s) + 3Cd²⁺ (aq) → 2Cr³⁺ (aq) + 3Cd
Fe²⁺ (aq) + Ag⁺ (aq) → Fe³⁺ (aq) + Ag(s)

Calculate the Δ_rG^(-) and equilibrium constant of the reactions

(i) E⁰ Cr³⁺ / Cr = - 0.74 V E⁰ Cd²⁺ / Cd = - 0.40 V

The galvanic cell of the given reaction is depicted as:

Cr(s) |Cr³⁺ (aq) ||Cd²⁺ (aq) |Cd(s)

Now, the standard cell potential is

Ecell (-) = (E R (-) – EL (-))

= (0.40 – (-0.74))

=+0.34 V

ΔrG (-) = -nFE cell (-)

In the given equation, n =6

F = 96487 C mol⁻¹

Ecell (-) =+0.34 V

Then Δ_rG^(-) = (−6 × 96487 C mol⁻¹ × 0.34 V)

= −196833.48 CV mol⁻¹

= −196833.48 J mol⁻¹

= −196.83 kJ mol⁻¹

Again,

Δ_rG^(-) = -RT ln K

=> Δ_rG^(-) =-2.303 Rt ln K

=> log K = - (Δ_rG^(-) /2.303 RT)

=-(196.83 x10³)/ (2.303 x 8.314 x298)

= 34.496

K = antilog (34.496) = 3.13 × 10³⁴

(ii) E⁰ _Fe³⁺ _/Fe²⁺ = 0.77 V E0 Ag+ / Ag = 0.80 V

The galvanic cell of the given reaction is depicted as:

Fe²⁺(aq) |Fe³⁺ (aq) ||Ag⁺ (aq) |Ag(s)

Now, the standard cell potential is

Ecell (-) = (E R (-) – EL (-))

= (0.80-0.77)

= 0.03 V Here, n = 1.

Then, ΔrG (-) = -nFE cell (-)

= (−1 × 96487 C mol⁻¹ × 0.03 V)

= -2894.61 Jmol^-1

= -2.89 KJ mol^-1

Again, Δ_rG^(-) =-2.303 Rt ln K

 log K = - (Δ_rG^(-) /2.303 RT)

=-(2.894.61)/ (2.303 x 8.314 x298)

= 0.5073

Therefore, K = antilog (0.5073)

= 3.2 (approximately)

Video solution:

Electrochemistry Class 12 Chemistry NCERT Chapter 2 NCERT Solutions

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NCERT solutions of Electrochemistry Class 12 Chemistry - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 12 Chemistry Electrochemistry NCERT solutions.