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(D) Given $n = 2$ and $E_{	ext{cell}}^{\circ} = 0.3 	ext{ V}$. At $298 	ext{ K}$,the relation between $E_{	ext{cell}}^{\circ}$ and equilibrium con

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$10^{10}$

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(D) Given $n = 2$ and $E_{	ext{cell}}^{\circ} = 0.3 	ext{ V}$. At $298 	ext{ K}$,the relation between $E_{	ext{cell}}^{\circ}$ and equilibrium constant $K_c$ is given by the Nernst equation: $E_{	ext{cell}}^{\circ} = \frac{0.0591}{n} \log K_c$. Substituting the values: $0.3 = \frac{0.0591}{2} \log K_c$. $\log K_c = \frac{0.3 	imes 2}{0.0591} = \frac{0.6}{0.0591} \approx 10.15$. $K_c = 	ext{antilog}(10.15) \approx 1.41 	imes 10^{10}$. Among the given options,the closest value is $10^{10}$.

For a cell reaction involving two electron changes,$E_{\text{cell}}^{\circ} = 0.3 \text{ V}$ at $25^{\circ}\text{C}$. The equilibrium constant of the reaction is:

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