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

Solution

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