For the cell Zn | Zn^{2+}(0.01 , M) || Fe^{2+ | vedclass

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(B) The cell reaction is: $Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)$.Here,$n = 2$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac

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$10^{\frac{0.32}{0.0295}}$

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(B) The cell reaction is: $Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)$.Here,$n = 2$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Fe^{2+}]}$.$0.2905 = E^0_{cell} - \frac{0.0591}{2} \log \frac{0.01}{0.001}$.$0.2905 = E^0_{cell} - 0.02955 \log(10)$.$E^0_{cell} = 0.2905 + 0.02955 = 0.32005 \, V \approx 0.32 \, V$.At equilibrium,$E_{cell} = 0$ and $E^0_{cell} = \frac{0.0591}{n} \log K_c$.$0.32 = \frac{0.0591}{2} \log K_c$.$0.32 = 0.02955 \log K_c$.$\log K_c = \frac{0.32}{0.02955} \approx \frac{0.32}{0.0295}$.$K_c = 10^{\frac{0.32}{0.0295}}$.

For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

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