Pb(s)PbSO4saturated solutionPbI2saturated solutionPb(s). Which of the following expressions represent the EMF of the above cell at 25°C?

PbII2+→PbI2+; Ecell0=0∴Ecell=(0.0591/n)log⁡Pb2+II/Pb2+I

Pb(s)PbSO4saturated solutionPbI2saturated solutionPb(s). Which of the following expressions represent the EMF of the above cell at 25°C?

PbII2+→PbI2+; Ecell0=0∴Ecell=(0.0591/n)log⁡Pb2+II/Pb2+I

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${Pb}_{(s)} |\underset{saturated solution}{{PbSO}_{4}}| |\underset{saturated solution}{{PbI}_{2}}| {Pb}_{(s)}$ . Which of the following expressions represent the EMF of the above cell at 25°C?

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$E = \frac{0.0592}{2} \log \frac{{[a_{{Pb}^{+ 2}}]}_{RHS}}{{[a_{{Pb}^{2 +}}]}_{LHS}}$

$E = \frac{0.0592}{2} \log \frac{{[a_{{Pb}^{2 +}}]}_{LHS}}{{[a_{{Pb}^{2 +}}]}_{RHS}}$

$E = \frac{0.0592}{2} \log {[K_{sp ({PbI}_{2})}]}^{1 / 2}$

$E = \frac{0 .0592}{2} \log \frac{K_{sp [{PbI}_{2}]}}{K_{sp [{PbSO}_{4}]}}$

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$\begin{array}{l} {Pb}_{II}^{2 +} \to {Pb}_{I}^{2 +}; E_{cell}^{0} = 0 \\ ∴ E_{cell} = (0 .0591 / n) \log {[{Pb}^{2 +}]}_{II} / {[{Pb}^{2 +}]}_{I} \end{array}$

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Pb(s)PbSO4saturated solutionPbI2saturated solutionPb(s). Which of the following expressions represent the EMF of the above cell at 25°C?

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${Pb}_{(s)} |\underset{saturated solution}{{PbSO}_{4}}| |\underset{saturated solution}{{PbI}_{2}}| {Pb}_{(s)}$ . Which of the following expressions represent the EMF of the above cell at 25°C?