Calculate the cell potential for a Cu plate k | vedclass

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(A) The electrode reaction is: $Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$.Given: $[Cu^{2+}] = 0.2 \ M$,$n = 2$,and $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$.Us

Vedclass · Accepted Answer

$0.3194$

Vedclass · Answer

(A) The electrode reaction is: $Cu^{2+}(aq) + 2e^{-} ightarrow Cu(s)$.Given: $[Cu^{2+}] = 0.2 \ M$,$n = 2$,and $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$.Using the Nernst equation for the electrode potential:$E_{Cu^{2+} \mid Cu} = E_{Cu^{2+} \mid Cu}^o - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$Substituting the values:$E_{Cu^{2+} \mid Cu} = 0.34 - \frac{0.0591}{2} \log \frac{1}{0.2}$$E_{Cu^{2+} \mid Cu} = 0.34 - 0.02955 	imes \log(5)$Since $\log(5) \approx 0.6990$:$E_{Cu^{2+} \mid Cu} = 0.34 - (0.02955 	imes 0.6990)$$E_{Cu^{2+} \mid Cu} = 0.34 - 0.02065$$E_{Cu^{2+} \mid Cu} \approx 0.3194 \ V$.

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)

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