Explanation

(b)
For all $Zn\ |\ {Zn}^{2 +}\ (A = \ 0.1\ M)\ ||\ {Fe}^{2 +}\ (A = \ 0.01\ M)\ |\ Fe.$
The cell reaction,
$Zn\ (s) \rightarrow \ {Zn}^{2 +}\ (aq) + \ {2e}^{-}$
${Fe}^{2 +}\ (aq) + \ {2e}^{-}\ \rightarrow Fe\ (s)$
[table] $Zn\ (s) + {Fe}^{2 +}\ (aq)\ \rightarrow \ {Zn}^{2 +}\ (aq) + \ Fe\ (s)$ [/table]
On applying Nernst equation,
$E_{cell} = \ E_{cell}^{{^\circ}}\ - \ \frac{0.0591}{n}\ \log_{10}\ \frac{\lbrack{Zn}^{2 +}\rbrack}{\lbrack{Fe}^{2 +}\rbrack}$
$0.2905 = \ E_{cell}^{{^\circ}}\ - \ - \ \frac{0.0591}{2}\ \log_{10}\ \frac{0.1}{0.01}$
$0.2905 = \ E_{cell}^{{^\circ}} - \ 0.0295\ \times \ \log_{10}\ 10$
$0.2905\ = \ E_{cell}^{{^\circ}}\ - \ 0.0295\ \times \ 1$
$\therefore E_{cell}^{{^\circ}}\ = \ 0.2905\ + \ 0.0295\ = \ 0.32\ V$
At equilibrium ($E_{cell}\ = \ 0$)
$E_{cell} = \ E_{cell}^{{^\circ}}\ - \ \frac{0.0591}{n}\ \log_{10}\ K_{c}$
∴ $\ 0\ = \ E_{cell}^{{^\circ}}\ - \ \frac{0.0591}{n}\ \log_{10}\ K_{c}$
or$\ \ E_{cell}^{{^\circ}} = \ \ \frac{0.0591}{2}\ \log_{10}\ K_{c}$
$0.32\ = \ \ \frac{0.0591}{2}\ \log_{10}\ K_{c}$
or $K_{c}\ = \ 10^{0.32/0.0295}\$