First slide

Solubility product(KSP)

Question

$\begin{array}{l} Pb (s) + {Hg}_{2} {SO}_{4} (s) ⇌ {PbSO}_{4} (s) + 2 Hg (1) E_{cell}^{o} = 0 .92 V \\ K_{sp} ({PbSO}_{4}) = 2 \times 10^{- 8}, K_{sp} ({Hg}_{2} {SO}_{4}) = 1 \times 10^{- 6} \end{array}$
Hence, E cell for a cell containing saturated solutions of the two salts in their respective electrodes would be:
Given [antilog (0.141) = 1.38]

Difficult

A

0.92 V
B

0.89 V
C

1.04 V
D

0.96 V

Solution

${Pb}_{(s)} + {Hg}_{2} {SO}_{4 (s)} ⇌ {PbSO}_{4 (s)} + 2 Hg$
Can be visualized as
$\begin{array}{l} {Pb}_{(s)} \to {Pb}_{aq}^{2 +} + 2 e^{-} \\ {SO}_{4}^{2 -} + {Pb}^{2 +} \to {PbSO}_{4} \\ {Hg}_{2} {SO}_{4} ⇌ {Hg}_{2}^{2 +} + {SO}_{4}^{2 -} \\ {Hg}_{2} {SO}_{4} ⇌ {Hg}_{2}^{2 +} + {SO}^{2 -} \\ {Hg}_{2}^{2 +} + 2 e^{-} \to 2 Hg \end{array}$
$∴$ same cell can be written as:
$\begin{array}{l} Pb + {Hg}_{2}^{2 +} \to {Pb}^{2 +} + 2 Hg \\ E_{cell} = E_{cell}^{o} (E_{{Hg}_{2}^{2 +} | Hg}^{0} - E_{{Pb}^{2 +} | Pb}^{0}) \\ - \frac{0 .059}{2} \log \frac{({Pb}^{2 +})}{({Hg}^{2 +})} \\ = 0 .92 + \frac{0 .06}{4} \times 1 .7 \\ = 0 .92 - 0 .0255 \\ ≅ 0 .89 V \end{array}$
Now, ${Pb}^{2 +} = \frac{K_{sp} {PbSO}_{4}}{{SO}_{4}^{2 -}}$
$\begin{array}{l} {Hg}_{2}^{2 +} = \frac{K_{sp} {Hg}_{2} {SO}_{4}}{{SO}_{4}^{2 -}} \\ ∴ E_{cell}^{0} = 0 .92 + \frac{0 .059}{2} \log \frac{K_{sp} {PbSO}_{4}}{K_{sp} {Hg}_{2} {SO}_{4}} \end{array}$
$∴ {SO}_{4}^{2 -}$ cancels for separate cell
$\begin{array}{l} E_{cell} = E_{cell}^{0} - \frac{0 .059}{2} \log \frac{({Pb}^{2 +})}{({Hg}_{2}^{2 +})} \\ {Pb}^{2 +} = \sqrt{K_{sp} {PbSO}_{4}} \\ {Hg}^{2 +} = \sqrt{K_{sp} {HgSO}_{4}} \\ E_{cell} = E_{cell}^{0} - \frac{0 .059}{2} \log \sqrt{\frac{K_{sp} {PbSO}_{4}}{K_{sp} {Hg}_{2} {SO}_{4}}} \end{array}$
$\begin{array}{l} = 0 .92 + \frac{0 .059}{2} \log 2 \times 10^{- 2} \\ - \frac{0 .059}{4} \log 2 \times 10^{- 2} \\ = 0 .92 + \frac{0 .059}{4} [2 - \log 2] \end{array}$
=0.96V

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