Explanation

(a)
$Cr/{Cr}^{3 +}\ (0.1\ M)\ ||\ {Fe}^{2 +}\ (0.01\ M)\ |\ Fe$
Oxidation half-cell; $Cr\ \rightarrow \ {Cr}^{3 +} + \ {3e}^{-}\ \times 2$
Reduction half-cell; ${Fe}^{2 +} + \ {2e}^{-}\ \rightarrow \ Fe\ \times \ 3$
Net cell reaction;
$2Cr\ + \ 2{Fe}^{2 +}\ \rightarrow \ 2{Cr}^{3 +}\ + \ 3Fe\ \ \ (n\ = \ 6)$
$E_{cell}^{{^\circ}} = \ E_{oxidation}^{{^\circ}}\ - \ E_{reduction}^{{^\circ}}$
= 0.72 – 0.42
= 0.30 V
$E_{cell}^{{^\circ}} = \ E_{cell}^{{^\circ}}\ - \ \frac{0.0591}{n}\log\frac{\lbrack{Cr}^{3 +}\rbrack^{2}}{\lbrack{Fe}^{2 +}\rbrack^{3}}$
$\ \ \ \ \ = \ 0.30\ - \ \frac{0.0591}{6}\log\frac{{(0.1)}^{2}}{{(0.01)}^{3}}$
$\ \ \ \ = \ 0.30\ - \ \frac{0.0591}{6}\log\frac{10^{- 2}}{10^{- 6}}$
$= \ 0.30\ –\ \frac{0.0591}{6}\log 10^{4}$
$\ E_{cell}\ = \ 0.2606\ V$