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Electrode Potential

The standard ...

Question

The standard oxidation potential of zinc of $0.76 v o l t$ and of silver is $- 0.80 v o l t$ . Calculate the emf of the cell $Z n | Z n (N O_{3})_{2} 0.25 M ∥ A g N O_{3} 0.1 M | A g$ at $25^{\circ} C$ .

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Solution

The cell reaction is
$Z n + 2 A g^{+} \to 2 A g + Z n^{2 +}$
$E_{O X}^{\circ}$ of $Z n = 0.76 v o l t$
$E_{r e d}^{\circ}$ of $A g = 0.80 v o l t$
$E_{c e l l}^{\circ} = E_{O X}^{\circ}$ of $Z n + E_{r e d}^{\circ}$ of $A g = 0.76 + 0.80 = 1.56 v o l t$
We know that, $E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.0591}{n} log \frac{[P r o d u c t s]}{[R e a c t a n t s]}$
$= E_{c e l l}^{\circ} - \frac{0.0591}{2} log \frac{0.25}{0.1 \times 0.1}$
$= 1.56 - \frac{0.0591}{2} \times 1.3979$
$= (1.56 - 0.0413) v o l t = 1.5187 v o l t$
Alternative method: First of all, the single electrode potentials of both the electrodes are determined on the basis of given concentrations.
$E_{O X (Z i n c)} = E_{O X}^{\circ} - \frac{0.0591}{2} log 0.25$
$= 0.76 + 0.0177 = 0.7777 v o l t$
$E_{r e d (S i l v e r)} = E_{r e d}^{\circ} + \frac{0.0591}{1} log 0.1$
$= 0.80 - 0.0591$
$- 0.7409 v o l t$
$E_{c e l l} = E_{O X (Z i n c)} + E_{r e d (S i l v e r)}$
$= 0.7777 + 0.7409$
$= 1.5186 v o l t$ .

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