Explanation

(b)
Half-cell reaction is
${Cr}_{2}O_{7}^{2 -} + 14H^{+} + 6e^{-} \rightarrow 2{Cr}^{3 +} + 7H_{2}O$
$E_{{Cr}_{2}O_{7}^{2 -}/{Cr}^{3 +}} - E_{{Cr}_{2}O_{7}^{2 -}/\ {Cr}^{3 +}}^{{^\circ}}$
$= - \frac{0.0591}{n}\log\frac{\left\lbrack {Cr}^{3 +} \right\rbrack^{2}}{\left\lbrack Cr_{2}O_{7}^{2 -} \right\rbrack\left\lbrack H^{+} \right\rbrack^{14}}$
$= \frac{0.0591}{6}\log{\left\lbrack 10^{- 3} \right\rbrack^{14} = - 0.414\ V}$