For the redox change; Zn(s)+Cu2+(1M)→ Zn2+(0.1M)+Cu(s)In a cell Eo cell is 1.10 volt. Ecell for the cell would be

A) E=Ecell0+0.0592log[Cu2+][Zn2+]Since, Ecell=Ecell0−0.05912log[Zn2+][Cu2+] =1.10−0.0592log0.11=1.13V

For the redox change; Zn(s)+Cu2+(1M)→ Zn2+(0.1M)+Cu(s)In a cell Eo cell is 1.10 volt. Ecell for the cell would be

A) E=Ecell0+0.0592log[Cu2+][Zn2+]Since, Ecell=Ecell0−0.05912log[Zn2+][Cu2+] =1.10−0.0592log0.11=1.13V

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For the redox change;
$Z n (s) + C u^{2 +} (1 M) \to Z n^{2 +} (0.1 M) + C u (s)$
In a cell $E^{o}$ cell is 1.10 volt. $E_{c e l l}$ for the cell would be

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2.14 V

180 V

1.13 V

0.82 V

answer is A.

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Detailed Solution

A) $E = E_{c e l l}^{0} + \frac{0.059}{2} \log \frac{[C u^{2 +}]}{[Z n^{2 +}]}$

Since,

$\begin{array}{l} E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} \log \frac{[Z n^{2 +}]}{[C u^{2 +}]} \\ = 1.10 - \frac{0.059}{2} \log \frac{0.1}{1} = 1.13 V \end{array}$

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For the redox change; Zn(s)+Cu2+(1M)→ Zn2+(0.1M)+Cu(s)In a cell Eo cell is 1.10 volt. Ecell for the cell would be

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For the redox change;
$Z n (s) + C u^{2 +} (1 M) \to Z n^{2 +} (0.1 M) + C u (s)$
In a cell $E^{o}$ cell is 1.10 volt. $E_{c e l l}$ for the cell would be