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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 $E_{A g^{+} / A g}^{0} = 0.799 V$ ]

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Solution

$p H H_{2} | H^{+} 1 M | A g_{2} S O_{4} (a q)$ saturated $| A g$
The reaction is,
$H_{2} ⟶ 2 H^{+} + 2 e^{-}$
$2 A g^{-} + 2 e^{-} ⟶ 2 A g$
$F_{c e l l} = F_{O p H} + F_{p p A g}$
$0.711 = 0.799 + \frac{0.059}{2} L o g [A g^{+}]^{2}$
$L o g ⎡ ⎣ \frac{1}{{[{A g}^{+}]}^{2}} ⎤ ⎦ = \frac{(0.799 - 0.711)}{2} \times 0.059 = 3$
$[A g^{+}]^{2} = 10^{- 3}$
$[A g^{+}] = 3.2 \times 10^{- 2}$
Solubility Equilibrium,
$A g_{2} S O_{4} ⟶ 2 A g^{+} + S O_{4}^{2 -}$
$K s p = [A g^{+}]^{2} [S O_{4}^{2 -}]$
$= (3.2 \times 10^{- 2})^{2} \frac{3.2 \times 10^{- 2}}{2} = 1.6 \times 10^{- 5}$

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