Consider the following hypothetical redox reaction
$ \text{A}(s)+2\text{B}^{+}(aq)\rightarrow\text{A}^{2+}(aq)+2\text{B}(s):E_{\text{cell}}^0=0.295\ \text{V} $
The equilibrium constant of the reaction at 298 K will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $ 10^{10} $
For the reaction
$\text{A}(s)+2\text{B}^{+}(aq)\rightarrow\text{A}^{2+}(aq)+2\text{B}(s)\ E_{\text{cell}}^0=0.295$
This reaction shows that solid A is oxidized to $A^{2+}$ and $B^+$ ions are reduced to solid B
Total electrons transferred $n = 2$
The relationship between standard cell potential and equilibrium constant is:
$\Delta G^{\circ}=-nFE_{\text{cell}}^{\circ}$
$\Delta G^{\circ}=-RT\ln K$
Equating both:
$-nFE_{\text{cell}}^{\circ}=-RT\ln K\Rightarrow\ln K=\frac{nFE_{\text{cell}}^{\circ}}{RT}$
Substituting Values:
$n = 2$
$F=96500\ \text{C}/\text{mol}$
$E_{\text{cell}}^{\circ}=0.295$
$R=8.314\ \text{J}/\text{mol}\cdot\text{K}$
$T=298\ \text{K}$
$\ln K=(2\times96500\times0.295)/(8.314\times298)=22.98$
$K=10^{10}$