ElectrochemistryHard

Question

We have an oxidation−reduction system: [Fe(CN)6]3− + e− $\rightleftharpoons$ [Fe(CN)6]4−; Eo = +0.36 V.

The ratio of concentrations of oxidized and reduced from at which the potential of the system becomes 0.24 V, is [Given: 2.303 RT/F = 0.06)

Options

A.2:1
B.1:2
C.1:20
D.1:100

Solution

$E = E^{o} - \frac{0.06}{n}.\log\frac{\lbrack R\rbrack}{\lbrack O\rbrack}$

$\Rightarrow 0.24 = 0.36 - \frac{0.06}{1}.\log\frac{\lbrack R\rbrack}{\lbrack O\rbrack} $$$\therefore\frac{\lbrack O\rbrack}{\lbrack R\rbrack} = \frac{1}{100}$$

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