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Calculate the equilibrium constant K_c for the reaction at 298 K.
$3 {Sn}^{4 +} + 2 Cr ⇋ 3 {Sn}^{2 +} + 2 {Cr}^{3 +}$
$(Given E °_{{Sn}^{4 +} / {Sn}^{2 +}} = 0.15 V; E °_{{Cr}^{3 +} / Cr} = - 074 V) . = 0.34 V;$

Solution

E °_{cell} = E °_{{Sn}^{4 +} / {Sn}^{2 +}} - E °_{{Sn}^{4 +} / {Sn}^{2 +}} - E °_{{Cr}^{3 +} / Cr} = 0.15 - (- 0.74) = 0.89 V

It is clear from the given equation that electrons involved in the reaction (n) = 6

\log K_{C} = \frac{E °_{cell} \times n}{0.059}

or

\log K_{C} = \frac{0.89 \times 6}{0.059} = 90.508

or

K_{C} = antilog 90.508

or

K_{C} = 3.221 \times 10^{90} .

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