Calculate the equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$
(N/A) The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation at $298 \ K$:
$E^{\Theta}_{cell} = \frac{0.059 \ V}{n} \log K_C$
Here,$n = 2$ (number of electrons transferred in the reaction).
Substituting the values:
$0.46 \ V = \frac{0.059 \ V}{2} \log K_C$
$\log K_C = \frac{0.46 \times 2}{0.059} = \frac{0.92}{0.059} \approx 15.593$
$K_C = \text{antilog}(15.593) = 3.92 \times 10^{15}$
$E^{\Theta}_{cell} = \frac{0.059 \ V}{n} \log K_C$
Here,$n = 2$ (number of electrons transferred in the reaction).
Substituting the values:
$0.46 \ V = \frac{0.059 \ V}{2} \log K_C$
$\log K_C = \frac{0.46 \times 2}{0.059} = \frac{0.92}{0.059} \approx 15.593$
$K_C = \text{antilog}(15.593) = 3.92 \times 10^{15}$
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Consider the following $4$ electrodes:
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| $C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ | $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$ |
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