The electromotive force for the following cel | vedclass

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(C) For the given concentration cell,the cell reaction is: $Cu_{(s)} + Cu^{+2} (0.04 \ M) \rightarrow Cu^{+2} (0.01 \ M) + Cu_{(s)}$.Applying the Nern

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$ - \frac{RT}{F} \ln (0.5) $

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(C) For the given concentration cell,the cell reaction is: $Cu_{(s)} + Cu^{+2} (0.04 \ M) ightarrow Cu^{+2} (0.01 \ M) + Cu_{(s)}$.Applying the Nernst equation: $E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q$.Here,$E^0_{cell} = 0$ for a concentration cell,$n = 2$,and $Q = \frac{[Cu^{+2}]_{anode}}{[Cu^{+2}]_{cathode}} = \frac{0.01}{0.04} = 0.25$.Substituting these values: $E_{cell} = 0 - \frac{RT}{2F} \ln (0.25)$.Since $0.25 = (0.5)^2$,we have $\ln (0.25) = 2 \ln (0.5)$.Therefore,$E_{cell} = - \frac{RT}{2F} 	imes 2 \ln (0.5) = - \frac{RT}{F} \ln (0.5)$.

The electromotive force for the following cell $Cu_{(s)} | Cu^{+2} (0.01 \ M) || Cu^{+2} (0.04 \ M) | Cu_{(s)}$ is

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