Electrochemistry Question 7
Question: The equilibrium constant for the reaction
$Zn(s)+Sn^{2+}(aq) \rightarrow Zn^{2+}(aq)+Sn(s)$ is $1 \times 10^{20}$
at $298 K$. The magnitude of standard electrode potential of $Sn / Sn^{2+}$ if $E _{Zn^{2+} / Zn}^{o}=-0.76 V$ is $\times 10^{-2} V$. (Nearest integer)
Given : $\frac{2.303 RT}{F}=0.059 V$
Answer: (17)
Solution:
$ \begin{aligned} & Zn(s)+Sn^{2+}(aq) \rightarrow Zn^{2+}(aq)+Sn(s) \\ & \Delta G^{\circ}=-2.303 RT \log _{10} Keq \\ & -nF(E_cell^{0})=-2.303 RT \log _{10} Keq \\ & E _{Zn / Zn^{2+}}^{0}+E _{Sn^{2+} / Sn}^{0}=\frac{0.059}{2} \log _{10} Keq \\ & 0.76+E _{Sn^{2+} / Sn}^{0}=\frac{0.059}{2} \log _{10} 10^{20} \\ & 0.76+E _{Sn^{2+} / Sn}^{0}=\frac{0.059 \times 20}{2} \\ & E _{Sn^{2+} / Sn}^{0}=0.59-0.76=-0.17 \\ & E _{Sn / Sn^{2+}}^{0 \text{ natho }}=17 \times 10^{-2} V \end{aligned} $
