The standard cell potential of the following cell Zn | Zn2+(aq) || Fe2+(aq) | Fe is 0.32 V. Calculate the standard Gibbs energy change for the reaction: Zn(s) + Fe2+(aq) Zn2+(aq) + Fe(s) (Given : 1 F = 96487 C)
- -61.75 kJ mol-1
- +5.006 kJ mol-1
- -5.006 kJ mol-1
- +61.75 kJ mol-1
Answer
(1) -61.75 kJ mol-1
Solution
r G = -nFE cell, with n = 2
r G = -2 96487 0.32
= -61751.7 J mol-1
-61.75 kJ mol-1
Answer: (1)