Question

$Zn|Z{n}^{2+}(a=0.1M)||F{e}^{2+}(a=0.01M)|Fe$.
The EMF of the above cell is $0.2905$. The equilibrium constant for the cell reaction is:

• A. ${10}^{0.32/0.0591}$
• B. ${10}^{0.32/0.0295}$
• C. ${10}^{0.26/0.0295}$
• D. $e^{0.32/0.0295}$

Answer 1

The correct option is D $e^{0.32/0.0295}$

Ans $\left(D\right)$

$E=E^o-=\frac{0.059}{n}{\log }_{10}\frac{Z{n}^{2+}}{F{e}^{2+}}$

$\therefore 0.2905=E^o-\frac{0.059}{2}{\log }_{10}\left(\frac{0.1}{0.01}\right)$

$\therefore 0.2905+\frac{0.059}{2}=E^o$

$\therefore E^o=0.32$

But $-nF{E}^o=-RT\ln Keq$

$\therefore +\frac{2\times 96500\times 0.32}{8.314\times 298}=\ln keq$

$\therefore Keq=e^{0.32/0.0295}$