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Solubility Product

A silver elec...

Question

A silver electrode is immersed in saturated $A g_{2} S O_{4} (a q .)$ . The potential difference between silver and the standard hydrogen electrode is found to be $0.711 V$ . Determine $K_{s p} (A g_{2} S O_{4})$ . (Given, $E_{A g^{+} / A g}^{\circ} = 0.799 V)$ .

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Solution

The cell may be represented as:
$P t (H_{2} 1 a t m) | H^{+} (1 M) | | A g^{+} (s a l t) | A g (s)$
$E_{c e l l}^{\circ} = E_{A g^{+} / A g}^{\circ} - E_{H^{+} / H_{2}}^{\circ}$
$= 0.799 - 0 = 0.799 V$
Given, emf of the cell $= 0.711 V$
$H_{2} + 2 A g^{+} \to 2 A g + 2 H^{+}$
$Q = \frac{[A g]^{2} [H^{+}]^{2}}{[H_{2}] [A g^{+}]^{2}} = \frac{1^{2} \times 1^{2}}{1 \times [A g^{+}]^{2}} = \frac{1}{[A g^{+}]^{2}}$
Applying Nernst equation,
$E = E^{\circ} - \frac{0.0591}{n} {log}_{10} Q$
or $0.711 = 0.799 - \frac{0.0591}{2} {log}_{10} \frac{1}{[A g^{+}]^{2}}$
$[A g^{+}] = 0.03243 m o l L^{- 1}$
$A g_{2} S O_{3} ⇌ 2 A g^{+} 0.03243 + S O_{4}^{2 -} 0.016215$
$K_{s p} [A g^{+}]^{2} [S O_{4}^{2 -}] = (0.03243)^{2} \times (0.016215) = 1.705 \times 10^{- 2}$ .

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