ElectrochemistryMedium

Question

For the cell reaction: 4Br− + O2 + 4H+ $\rightleftharpoons$ 2Br2 + 2H2O; E° = 0.18 V. The value of (log KC) at 298 K is [2.303 RT/F = 0.06]

Options

A.12
B.6
C.18
D.3

Solution

$E_{cell}^{o} = \frac{0.06}{n}.\log K_{eq} \Rightarrow 0.18 = \frac{0.06}{4}.\log K_{eq}$

$\therefore\log K_{c} = 12$

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