What will be the electrode potential of Cu el | vedclass

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(B) The electrode reaction for $Cu$ is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.Using the Nernst equation at $298 \ K$:$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^

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$0.293$

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(B) The electrode reaction for $Cu$ is: $Cu^{2+}(aq) + 2e^- ightarrow Cu(s)$.Using the Nernst equation at $298 \ K$:$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$.Here,$E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$,$n = 2$,and $[Cu^{2+}] = 0.025 \ M$.$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \log \frac{1}{0.025}$.$E_{Cu^{2+}/Cu} = 0.34 - 0.02955 	imes \log(40)$.Since $\log(40) = \log(4 	imes 10) = 0.602 + 1 = 1.602$.$E_{Cu^{2+}/Cu} = 0.34 - 0.02955 	imes 1.602 \approx 0.34 - 0.0473 = 0.2927 \ V$.Rounding to three decimal places,we get $0.293 \ V$.

What will be the electrode potential of $Cu$ electrode dipped in $0.025 \ M$ $CuSO_4$ solution at $298 \ K$? Given that the standard reduction potential of $Cu^{2+}/Cu$ is $0.34 \ V$.

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