Question

In the electrochemical cell, Zn | ZnSO₄ (0.01 M) || CuSO₄ (1.0 M) | Cu, the emf of this Daniell cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ changed to 0.01 M, the emf changes to E₂. From the following, which one is the relationship between E₁ and E₂?

`("Given", (RT)/F = 0.059)`

Options

• E₁ = E₂

• E₁ < E₂

• E₁ > E₂

• E₂ = 0 ≠ E₁

Answer 1

E₁ > E₂

Explanation:

For the Daniell cell:

Zn | ZnSO₄ (0.01 M) || CuSO₄ (1.0 M) | Cu

The Nernst equation for the cell is:

\[\ce{E = E{^{\circ}_{cell}} - \frac{0.0591}{n} log \frac{[Zn^{2+}]}{[Cu^{2+}]}}\]

Case 1 (E₁):

[Zn²⁺] = 0.01 

[Cu²⁺] = 1.0

\[\ce{E_1 = E{^{\circ}} - \frac{0.0591}{2} log \frac{0.01}{1}}\]

= \[\ce{E{^{\circ}} - \frac{0.059}{2} × (-2)}\]

E₂ = E° + 0.059

Case 2 (E₂):

[Zn²⁺] = 1.0

[Cu²⁺] = 0.01

\[\ce{E_2 = E^{\circ} - \frac{0.0591}{2} log \frac{1}{0.01}}\]

= \[\ce{E^{\circ} - \frac{0.059}{2} × (2)}\]

E₂ = E° − 0.059

∴ E₁ > E₂    ...[From case (1) and (2)]