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The following reaction takes place at $298 \mathrm{~K}$ in an electrochemical cell involving two metals $\mathrm{A}$ and B,
$\mathrm{A}^{2+}(\mathrm{aq} .)+\mathrm{B}(\mathrm{s}) \rightarrow \mathrm{B}^{2+}($ aq $)+\mathrm{A}(\mathrm{s})$
With $\left[\mathrm{A}^{2+}\right]=4 \times 10^{-3} \mathrm{M}$ and $\left[\mathrm{B}^{2+}\right]=2 \times 10^{-3} \mathrm{M}$ in the respective half-cells, the cell EMF is $1.091 \mathrm{~V}$. The equilibrium constant of the reaction is closest to
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$\mathrm{A}^{2+}(\mathrm{aq} .)+\mathrm{B}(\mathrm{s}) \rightarrow \mathrm{B}^{2+}($ aq $)+\mathrm{A}(\mathrm{s})$
With $\left[\mathrm{A}^{2+}\right]=4 \times 10^{-3} \mathrm{M}$ and $\left[\mathrm{B}^{2+}\right]=2 \times 10^{-3} \mathrm{M}$ in the respective half-cells, the cell EMF is $1.091 \mathrm{~V}$. The equilibrium constant of the reaction is closest to
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Verified Answer
The correct answer is:
$2 \times 10^{37}$
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{.0591}{2} \log \frac{2 \times 10^{-3}}{4 \times 10^{-3}}$
$1.091=\mathrm{E}^{\circ}{ }_{\text {cell }}-\frac{.0591}{2} \log (.5)$
$\mathrm{E}_{\text {cell }}^{\circ}=1.099$
$\mathrm{E}_{\text {cell }}^{\circ}=-\frac{.0591}{2} \log \mathrm{k}$
$\log \mathrm{k}=-\frac{1.099 \times 2}{.0591}=-37.22$
$\therefore \mathrm{k}=2 \times 10^{37}$
$1.091=\mathrm{E}^{\circ}{ }_{\text {cell }}-\frac{.0591}{2} \log (.5)$
$\mathrm{E}_{\text {cell }}^{\circ}=1.099$
$\mathrm{E}_{\text {cell }}^{\circ}=-\frac{.0591}{2} \log \mathrm{k}$
$\log \mathrm{k}=-\frac{1.099 \times 2}{.0591}=-37.22$
$\therefore \mathrm{k}=2 \times 10^{37}$
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