ElectrochemistryHard
Question
The standard reduction potentials in acidic conditions are 0.77 V and 0.53 V, respectively, for Fe3+|Fe2+ and I3−|I− couples. The equilibrium constant for the reaction: 2Fe3+ + 3I− $\rightleftharpoons$2Fe2+ + I3−, is (2.303 RT/F = 0.06)
Options
A.2 × 108
B.108
C.104
D.10−8
Solution
$E_{cell}^{o} = \frac{0.06}{n}.\log K_{eq}$
$\Rightarrow (0.77 - 0.53) = \frac{0.06}{2}.\log K_{eq} $$$\therefore K_{eq} = 10^{8}$$
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