Question

The emf of the cell
Zn | Zn²⁺ (0.01 M) | | Fe²⁺ (0.001 M) | Fe
at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is
(a) $e^{\frac{0.32}{0.0295} }$
(b) $10^{\frac{0.32}{0.0295} }$
(c) $10^{\frac{0.26}{0.0295} }$
(d) $10^{\frac{0.26}{0.0591} }$

Answer 1

Answer:

the ans is c becoz

Explanation: