ElectrochemistryHard

Question

What is the equilibrium constant of the reaction: 2Fe3+ + Au+→ 2Fe2+ + Au3+?

Given $E_{Au^{+}|Au}^{o}$= 1.68 V, $E_{Au^{3 +}|Au}^{o}$= 1.50 V, $E_{Fe^{3 +}|Fe^{2 +}}^{o}$= 0.75 V and 2.303 RT/F = 0.06.

Options

A.1 × 1022
B.1 × 10−22
C.1 × 10−11
D.1 × 10−72

Solution

$E_{cell}^{o} = \frac{0.06}{2}.\log K_{eq}$

$\Rightarrow 0.75 - \frac{1.50 \times 3 - 1.68 \times 1}{3 - 1} = 0.03\log K_{eq} $$$\therefore K_{eq} = 10^{- 22}$$

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