Question

The equilibrium constant for the reaction $Zn(s)+S{n}^{2+}(aq)\rightleftharpoons Z{n}^{2+}(aq)+Sn(s)$ is $1\times 10^{20}$ at $298K$. The magnitude of standard electrode potential of $Sn/S{n}^{2+}$ if $E_{Z{n}^{2+}/Zn}^0=-0.76V$ is ___$\times 10^{-2}V$. (Nearest integer)
Given: $\frac{2.303RT}{F}=0.059V$

Answer 1

Answer: 17

$Zn(s)+S{n}^{2+}(aq)\rightleftharpoons Z{n}^{2+}(aq)+Sn(s)$
$\Delta G^{\circ }=-2.303RT{\log }_{10}Keq$
$-nF\left(E_{\text{ cell }}^0\right)=-2.303RT{\log }_{10}Keq$
$E_{Zn/Z{n}^{2+}}^0+E_{S{n}^{2+}/Sn}^0=\frac{0.059}{2}{\log }_{10}Keq$
$0.76+E_{S{n}^{2+}/Sn}^0=\frac{0.059}{2}{\log }_{10}10^{20}$
$0.76+E_{S{n}^{2+}/Sn}^0=\frac{0.059\times 20}{2}$
$E_{S{n}^{2+}/S{n}^0}=0.59-0.76=-0.17$
$E_{Sn/S{n}^{2+}}^0=17\times 10^{-2}V$
$=17$