Calculate the equilibrium constant (K_c) of t | vedclass

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Question

(A) The relationship between the standard cell potential $(E_{cell}^{\circ})$ and the equilibrium constant $(K_c)$ is given by the formula: $E_{cell}^

Vedclass · Accepted Answer

$10^{35}$

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(A) The relationship between the standard cell potential $(E_{cell}^{\circ})$ and the equilibrium constant $(K_c)$ is given by the formula: $E_{cell}^{\circ} = \frac{0.0591}{n} \log K_c$. Given $\frac{2.303 \ RT}{F} = 0.06$,the formula becomes $E_{cell}^{\circ} = \frac{0.06}{n} \log K_c$. In the reaction $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$,the number of electrons transferred $(n)$ is $2$. Substituting the values: $1.05 = \frac{0.06}{2} \log K_c$. $1.05 = 0.03 \log K_c$. $\log K_c = \frac{1.05}{0.03} = 35$. $K_c = 10^{35}$.

Calculate the equilibrium constant $(K_c)$ of the reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$; $E_{cell}^{\circ} = 1.05 \ V$. (Given: $\frac{2.303 \ RT}{F} = 0.06$)

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