For the cell Zn_{(s)} | Zn^{2+}_{(1,M)} || Cu | vedclass

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(C) The Nernst equation is given by: $E_{cell} = E^o_{cell} - \frac{0.0592}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.When the cell is fully discharged,$E_

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$10^{37.3}$

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(C) The Nernst equation is given by: $E_{cell} = E^o_{cell} - \frac{0.0592}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.When the cell is fully discharged,$E_{cell} = 0$ and $n = 2$.Substituting the values: $0 = 1.10 - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.$\frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} = 1.10$.$\log \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{1.10 	imes 2}{0.0592} \approx 37.16$.Rounding to the nearest provided option,the ratio is $10^{37.3}$.

For the cell $Zn_{(s)} | Zn^{2+}_{(1\,M)} || Cu^{2+}_{(1\,M)} | Cu_{(s)}$,the $E^o_{cell} = 1.10 \, V$. When the cell is fully discharged at $298 \, K$,the ratio of the concentrations $[Zn^{2+}] / [Cu^{2+}]$ is:

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