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(A) The cell reaction is spontaneous in the direction where $E^o_{cell} > 0$. Given $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \

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$1.122 	imes 10^3$

Vedclass · Answer

(A) The cell reaction is spontaneous in the direction where $E^o_{cell} > 0$. Given $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$. Since $E^o_{(Sn^{2+}|Sn)} > E^o_{(Ni^{2+}|Ni)}$,$Sn^{2+}$ will be reduced and $Ni$ will be oxidized. The cell reaction is: $Ni(s) + Sn^{2+}(aq) ightarrow Ni^{2+}(aq) + Sn(s)$. $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{(Sn^{2+}|Sn)} - E^o_{(Ni^{2+}|Ni)} = -0.14 - (-0.23) = 0.09 \ V$. The relationship between $E^o_{cell}$ and $K_C$ is given by: $E^o_{cell} = \frac{0.0591}{n} \log K_C$. Here $n = 2$ (number of electrons transferred). $0.09 = \frac{0.0591}{2} \log K_C$. $\log K_C = \frac{0.09 	imes 2}{0.0591} = \frac{0.18}{0.0591} \approx 3.0457$. $K_C = 10^{3.0457} \approx 1.111 	imes 10^3$. Rounding to the provided option,$K_C = 1.122 	imes 10^3$.

Calculate the equilibrium constant $(K_C)$ for the cell obtained by connecting two electrodes with standard electrode potentials $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$ at $298 \ K$.

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