The EMF of the cell, Zn | Zn 2+ (0.01 M) || Fe 2+ (0.001M) | Fe at 298 K is 0.2905 then…
Chemistry · JEE Advanced · NTA Exams — Electrochemistry
The EMF of the cell,
Zn | Zn2+ (0.01 M) || Fe2+ (0.001M) | Fe
at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is
Zn | Zn2+ (0.01 M) || Fe2+ (0.001M) | Fe
at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is
Answer
(B)
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