Find the equilibrium constant for the reaction,
$I n^{2 +} + C u^{+} ⇌ I n^{3 +} + C u^{+}$ at $298 K$
Given: $E_{C u^{2 +} / C u^{+}} = 0.15 V; E_{I n^{2 +} / I n^{+}}^{\circ} = - 0.4 v o l t$
$E_{I n^{3 +} / I n^{+}}^{\circ} = - 0.42 v o l t$ .

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Solution

The given reversible reaction can be obtained as,
$C u^{2 +} + e^{-} \to C u^{+}; △ G^{\circ} = - 0.15 F ∵ △ G^{\circ} - n F E^{\circ}$
$I n^{2 +} + e^{-} \to I n^{+}; △ G^{\circ} = + 0.40 F$
$I n^{+} \to I n^{3 +} + 2 e^{-}; △ G^{\circ} = - 0.84 F - ------------------------------------ -$
$C u^{2 +} + I n^{2 +} ⇌ I n^{3 +} + C u^{+}; △ G^{\circ} = - 0.59 F - ------------------------------------------------ -$
$- n F E^{\circ} = - 0.59 F$
$E^{\circ} = 0.59 v o l t$
Equilibrium constant can be calculated as,
$K = a n t i l o g [\frac{n E^{\circ}}{0.059}]$
$= a n t i l o g [\frac{1 \times 0.59}{0.059}] = 10^{10}$ .

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Q. The equilibrium constant for the reaction:

I n^{2 +} + C u^{2 +} ⇌ I n^{3 +} + C u^{+}

at 298 K
Given

E_{C u^{2 +} / C u}^{o} = 0.15 V; E_{I n^{3 +} / I n^{+}}^{o} = 0.42 V

and

E_{I n^{2 +} / I n^{+}}^{o} = - 0.40 V

Find the equilibrium constant for the reaction $I n^{2 +} + C u^{2 +} \to I n^{3 +} + C u^{+}$ at 298 K. Given, $E {^{o}}_{C u^{2 +} / C u^{+}} = 0.15 V, E {^{o}}_{I n^{3 +} / I n^{+}} = - 0.42 V$ $E {^{o}}_{I n^{2 +} / I n^{+}} = - 0.40 V$ is approximately: