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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

$P t$ $H_{2} ∣ ∣ H^{+} 1 M ∣ ∣ A g_{2} S O_{4} (a q)$ Saturated $| A g$ the reaction is
$H_{2} ⟶ 2 H^{+} + 2 e^{-}$
$2 A g^{+} + 2 e^{-} ⟶ 2 A g$
or, $0.711 = 0.799 + (0.059 / 2) l o g [A g^{+}]^{2}$
$l o g (\frac{1}{[A g^{+}]^{2}}) = \frac{[0.799 - 0.711] \times 2}{0.059} = 3$
$[A g^{+}]^{2} = 10^{- 3}$
$[A g^{+}] = 3.2 \times 10^{- 2}$
Now, Solubility equilibrium is
$A g_{2} S O_{4} ⟷ 2 A g^{+} + S O_{4}^{2 -}$
$K_{s p} = [A g^{+}]^{2} [S O_{4}^{2 -}]$
$= (3.2 \times 10^{- 2})^{2} (3.2 \times 10^{- 2}) / 2$
$= 1.6 \times 10^{- 5}$
$K_{s p} [A g_{2} S O_{4}] = 16 \times 10^{- 5}$

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