Question

Consider the cell
$\left. Pt ( s )\left| H _2( g )(1 atm )\right| H ^{+} \text {(aq, }\left[ H ^{+}\right]=1\right) \| Fe ^{3+}( aq ), Fe ^{2+}( aq ) \mid Pt ( s )$
Given $E ^{\circ} Fe ^{3+} Fe ^{2+}=0.771 V ^{2+}$ and $E ^{\circ} H ^{+} 1/ 2 H _2=0 V , T =298 K$
If the potential of the cell is $0.712 V$, the ratio of concentration of $Fe ^{2+}$ to $Fe ^{3+}$ is

Answer 1

$ \frac{1}{2} H _2( g )+ Fe ^{3+}( aq .) \longrightarrow H ^{+}( aq )+ Fe ^{2+}( aq .) $
$ E = E ^{ o }-\frac{0.059}{1} \log \frac{\left[ Fe ^{2+}\right]}{\left[ Fe ^{3+}\right]} $
$ \Rightarrow 0.712=(0.771-0)-\frac{0.059}{1} \log \frac{\left[ Fe ^{2+}\right]}{\left[ Fe ^{3+}\right]}$
$ \Rightarrow \log \frac{\left[ Fe ^{2+}\right]}{\left[ Fe ^{3+}\right]}=\frac{(0.771-0712)}{0.059}=1$
$ \Rightarrow \frac{\left[ Fe ^{2+}\right]}{\left[ Fe ^{3+}\right]}=10$