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By how much will the potential of half-cell $Cu^{2+}|Cu$ change if the solution is diluted to 100 times at 298K?

$\begin{array}{1 1}\text{Increases by 59mV}\\\text{Decreases by 59mV}\\\text{Increases by 29.5mV}\\\text{Decreases by 29.5mV}\end{array} $

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Answer : Decreases by 59mV
For the reaction $Cu^{2+}+2e^-\rightarrow Cu$,Nernst equation is
$E=E^{\large\circ}-\large\frac{RT}{2F}$$\ln \large\frac{1}{[Cu^{2+}]}$
If the solution is diluted 100 times,then we have
$E_2-E_1=-\large\frac{RT}{2F}$$\ln \large\frac{[Cu^{2+}]}{(0.01)[Cu^{2+}]}$
$\Rightarrow \large\frac{RT}{2F}$$\ln(0.01)=-\large\frac{2\times 0.059V}{2}$$=-0.059V$
answered Jul 21, 2014 by sreemathi.v

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