The E^{circ} of M mid M^{2+} parallel Cu^{2+} | vedclass

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(C) The cell reaction is: $M + Cu^{2+} ightarrow M^{2+} + Cu$ According to the Nernst equation: $E_{	ext{cell}} = E^{\circ}_{	ext{cell}} - \frac{0

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$10^{-11}$

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(C) The cell reaction is: $M + Cu^{2+} ightarrow M^{2+} + Cu$ According to the Nernst equation: $E_{	ext{cell}} = E^{\circ}_{	ext{cell}} - \frac{0.06}{n} \log \frac{[M^{2+}]}{[Cu^{2+}]}$ Given: $E^{\circ}_{	ext{cell}} = 0.3 \ V$ $n = 2$ $[M^{2+}] = 0.1 \ M$ $E_{	ext{cell}} = 0$ Substituting the values: $0 = 0.3 - \frac{0.06}{2} \log \frac{0.1}{[Cu^{2+}]}$ $0.3 = 0.03 \log \frac{0.1}{[Cu^{2+}]}$ $10 = \log \frac{0.1}{[Cu^{2+}]}$ $\frac{0.1}{[Cu^{2+}]} = 10^{10}$ $[Cu^{2+}] = \frac{0.1}{10^{10}} = 10^{-11} \ mol \ L^{-1}$

The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

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The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero? $\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$