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The emf of the cell, `Zn|Zn^(2+)(0.01M)|| Fe^(2+)(0.001M) | Fe` at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is:A. `e^(0.32/0.0295)`B. `1-^(0.32/0.0295)`C. `10^(0.26/0.0295)`D. `10^(0.32/0.0591)` |
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Answer» Correct Answer - B |
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