Calculate the equilibrium constant for the following reaction: $Ni_{(s)} + Cu_{(aq)}^{2+} \to Cu_{(s)} + Ni_{(aq)}^{2+}$. Given: $E_{Ni^{2+}|Ni}^o = -0.25 \ V$ and $E_{Cu^{2+}|Cu}^o = 0.34 \ V$.
(N/A) The standard cell potential is calculated as:
$E_{\text{cell}}^o = E_{\text{cathode}}^o - E_{\text{anode}}^o = E_{Cu^{2+}|Cu}^o - E_{Ni^{2+}|Ni}^o$
$E_{\text{cell}}^o = 0.34 \ V - (-0.25 \ V) = 0.59 \ V$
The relationship between standard cell potential and equilibrium constant $K_C$ is given by:
$E_{\text{cell}}^o = \frac{0.0591}{n} \log K_C$
Here,$n = 2$ (number of electrons transferred).
$0.59 = \frac{0.0591}{2} \log K_C$
$\log K_C = \frac{0.59 \times 2}{0.0591} \approx 20$
$K_C = 10^{20}$
$E_{\text{cell}}^o = E_{\text{cathode}}^o - E_{\text{anode}}^o = E_{Cu^{2+}|Cu}^o - E_{Ni^{2+}|Ni}^o$
$E_{\text{cell}}^o = 0.34 \ V - (-0.25 \ V) = 0.59 \ V$
The relationship between standard cell potential and equilibrium constant $K_C$ is given by:
$E_{\text{cell}}^o = \frac{0.0591}{n} \log K_C$
Here,$n = 2$ (number of electrons transferred).
$0.59 = \frac{0.0591}{2} \log K_C$
$\log K_C = \frac{0.59 \times 2}{0.0591} \approx 20$
$K_C = 10^{20}$
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