Determine the equilibrium constant of the reaction at 298 K.
$2{\mathrm{Fe}}^{3+}+{\mathrm{Sn}}^{2+} \leftrightharpoons 2{\mathrm{Fe}}^{2+}+{\mathrm{Sn}}^{4+}$
From the obtained value of the equilibrium constant, predict whether Sn²⁺ ions can reduce Fe³⁺ to Fe²⁺ quantitatively or not.
$\mathrm{E}{°}_{{\mathrm{Sn}}^{4+}/{\mathrm{Sn}}^{2+}} = 0.15 \mathrm{V} \mathrm{and} \mathrm{E}{°}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}} = +0.77 \mathrm{V}$

The cell is
                 $\mathrm{Pt}\left(\mathrm{s}\right)| {\mathrm{Sn}}^{2+}(\mathrm{aq}\left)\right| {\mathrm{Sn}}^{4+}| (\mathrm{aq}\left)\right|| {\mathrm{Fe}}^{3+}(\mathrm{aq}) | {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\left|\mathrm{Pt}\right(\mathrm{s})$
and                            $$\begin{gathered} \mathrm{E}{°}_{\mathrm{cell}} = \mathrm{E}{°}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}} - \mathrm{E}{°}_{{\mathrm{Sn}}^{4+}/{\mathrm{Sn}}^{2+}} \\ = + 0.77 - (+0.15) = +0.62 \mathrm{V} \end{gathered}$$
and                                n = 2
∴              $\mathrm{K}= \mathrm{antilog}\left(\frac{\mathrm{nE}°}{0.059}\right) = \mathrm{antilog} \left(\frac{2\times 0.62}{0.059}\right) = \mathrm{antilog} (21.0169)$
                                      $\mathrm{K} = 1.039 \times {10}^{21}$
As K is very high, the reaction is favoured in the forward direction, so, Sn²⁺ can easily reduce Fe³⁺ ion to Fe²⁺ ion.

t = 600 s
Charge = current x time
= 1.5 A x 600s = 900 C
According to the reaction : Cu²⁺(aq) + 2e^– → Cu(s)
We require 2F or 2 x 96487 C to deposit 1 mol or 63 g of Cu
For 900 C, the mass of Cu deposited  = $\frac{(63 \mathrm{g} {\mathrm{mol}}^{-1} \times 900 \mathrm{C})}{2 \times 96487 \mathrm{C} {\mathrm{mol}}^{-1}} = 0.2938 \mathrm{g}.$

The cathode reaction is Cu²⁺ + 2e^– → Cu
I = 1.5 amperes, t = 10 x 60 = 600 s
∴ Q = It = 1.5 x 600 = 900 C
The reaction states that 2 x 96500 C are required to deposit 63.5 g Cu at cathode
$\therefore$  900 C will deposit 

$\frac{63.5\times 900}{2\times 96500}\times 900 = 0.296 \mathrm{g} \mathrm{Cu} \mathrm{cathode}.$