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MHT CET Chemistry question

ChemistryElectrochemistryMHT CET

The e.m.f. of the cell \mathrm{Zn|Zn^{2+}(0.01\,M)||Fe^{2+}(0.001\,M)|Fe} at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is

Answer: (B)

Solution

E = E° − (0.0591/2) log(0.01/0.001) gives E° = 0.2905 + 0.0296 = 0.32 V, and log K = nE°/0.0592 = 2 × 0.32/0.0592 = 0.32/0.0296,

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Question ID VK-066766