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(A) The cell reaction is:Anode: $Cu_{(s)} \rightarrow Cu^{2+}(0.1 \ M) + 2e^{-}$Cathode: $Cu^{2+}(1 \ M) + 2e^{-} \rightarrow Cu_{(s)}$Overall reactio

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$0.0295$

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(A) The cell reaction is:Anode: $Cu_{(s)} ightarrow Cu^{2+}(0.1 \ M) + 2e^{-}$Cathode: $Cu^{2+}(1 \ M) + 2e^{-} ightarrow Cu_{(s)}$Overall reaction: $Cu^{2+}(1 \ M) ightarrow Cu^{2+}(0.1 \ M)$Standard cell potential $E_{cell}^o = E_{cathode}^o - E_{anode}^o = 0.34 \ V - 0.34 \ V = 0.0 \ V$Using the Nernst equation at $298 \ K$:$E_{cell} = E_{cell}^o - \frac{0.0591}{n} \log \frac{[Cu^{2+}]_{anode}}{[Cu^{2+}]_{cathode}}$Here $n = 2$,$[Cu^{2+}]_{anode} = 0.1 \ M$,and $[Cu^{2+}]_{cathode} = 1 \ M$$E_{cell} = 0.0 - \frac{0.0591}{2} \log \left( \frac{0.1}{1} ight)$$E_{cell} = -0.02955 	imes \log(10^{-1})$$E_{cell} = -0.02955 	imes (-1) = +0.02955 \ V$Thus,the cell potential is $0.02955 \ V$.

Calculate the cell potential at $298 \ K$ for the following cell:
$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$
Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)

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Calculate the cell potential at $298 \ K$ for the following cell:$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)