M.Sc. Chemistry (PU-CET P.G) 2021 Panjab University Entrance Exam With Answers

Practice Mode:
53.

What is the equilibrium constant for the reaction given below at 298 K, if Ecell = 0.2905V at 298 K?

Zn(s) + Fe2+ (aq) -->Zn2+ (aq)(0.01M) + Fe(s)
A: e^0.32/0.0295
B: 10^0.595/0.76
C: 10^0.0250/0.32
D: 10^0.32/0.0295

The answer is: D