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By how much will the potential of a copper electrode change if the solution of ${CuSO}_{4}$ in which it is immersed is diluted to 10 times at 298 K?

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a

Increases by 30 mV

b

Decreases by 30 mV

c

Increases by 60 m V

d

Decreases by 60 m V

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detailed solution

Correct option is B

$\begin{array}{l} E_{1} = E^{\circ} - \frac{R T}{2 F} \ln \frac{1}{{[{Cu}^{2 +}]}_{1}} \\ \begin{aligned} E_{2} & = E^{\circ} - \frac{R T}{2 F} \ln \frac{1}{{[{Cu}^{2 +}]}_{2}} = E^{\circ} - \frac{R T}{2 F} \ln \frac{1}{0.1 {[{Cu}^{2 +}]}_{1}} \\ E_{2} - E_{1} & = - \frac{R T}{2 F} \ln \frac{1}{0.1 {[{Cu}^{2 +}]}_{1}} + \frac{R T}{2 F} \ln \frac{1}{{[{Cu}^{2 +}]}_{1}} = \frac{R T}{2 F} \ln 0.1 = - \frac{R T}{2 F} = - \frac{0.059 V}{2} = - 0.0295 V \\ = - 29.5 mV \end{aligned} \end{array}$

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