By how much will the potential of the half-ce | vedclass

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(B) The Nernst equation for the half-cell $Cu^{2+} + 2e^- \rightarrow Cu$ is:$E = E^0 - \frac{0.0591}{2} \log \frac{1}{[Cu^{2+}]}$After dilution to $1

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Decreases by $59 \ mV$

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(B) The Nernst equation for the half-cell $Cu^{2+} + 2e^- ightarrow Cu$ is:$E = E^0 - \frac{0.0591}{2} \log \frac{1}{[Cu^{2+}]}$After dilution to $100$ times,the concentration $[Cu^{2+}]$ becomes $\frac{[Cu^{2+}]}{100}$.New potential $E' = E^0 - \frac{0.0591}{2} \log \frac{1}{[Cu^{2+}]/100} = E^0 - \frac{0.0591}{2} (\log \frac{1}{[Cu^{2+}]} + \log 100)$$E' = E - \frac{0.0591}{2} 	imes 2 = E - 0.0591 \ V = E - 59 \ mV$.Therefore,the potential decreases by $59 \ mV$.

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

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