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The cell $Zn | {Zn}^{2 +} (1 M) | | {Cu}^{2 +} (1 M) | Cu (E_{cell}^{o} = 1 .10 V)$ , was allowed to be completely discharged at 298 K. The relative concentration of $Z n^{2 +}$ to ${Cu}^{2 +}, [\frac{{Zn}^{2 +}}{{Cu}^{2 +}}]$ is:

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$10^{37.3}$

antilog (24.08)

$9.65 \times 10^{4}$

37.3

answer is C.

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Detailed Solution

$\begin{array}{l} E_{cell} = E_{cell}^{o} - \frac{0 .0591}{n} \log Q \\ Where, Q= \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \end{array}$

For complete discharge $E_{cell} = 0$

So $E_{cell}^{o} = \frac{0 .0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$

$∴ [\frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}] = 10^{37 .3}$

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