Explanation

(a)
$E_{cell}^{{^\circ}} = E_{C} - \ E_{A}$
= - 0.45 – (-2.37)
= -1.92 V
$Mg\ (s)\ |\ {Mg}^{2 +}(aq)\ x\ M\ ||\ {Fe}^{2 +}(aq)\ 0.01\ M\ |\ Fe\ (s)$
The cell reaction is
$Mg\ + \ {Fe}^{2 +}\ \rightarrow \ {Mg}^{2 +}\ + \ Fe$
$E_{cell} = \ E_{cell}^{{^\circ}} - \ \frac{0.059}{n}\log\frac{\lbrack{Mg}^{2 +}\rbrack}{\lbrack{Fe}^{2 +}\rbrack}$
$1.92\ = \ 1.92\ –\ \frac{0.059}{n}\ \ \log\ \frac{x}{0.01}$
$0\ = \ \frac{- 0.059}{2}\ \log\ \frac{x}{0.01}$
∴ $x = 0.01\ M$