Chemistry - Electrochemistry Question with Solution | TestHub

The relation between standard emf of the cell and equilibrium constant at 298 K is given by:

$E_{\text{cell}}^{\circ }=\frac{0.0592}{n}{\log }_{10}K$

From the given cell reaction, the number of electrons transferred is $n=2$.

Given $K=10^4$, substituting the values we get:

$E_{\text{cell}}^{\circ }=\frac{0.0592}{2}{\log }_{10}(10^4)$

$E_{\text{cell}}^{\circ }=\frac{0.0592}{2}\times 4$

$E_{\text{cell}}^{\circ }=0.0592\times 2=0.1184\text{ V}$

Answer: 0.1184 V