Question
Calculate the standard free energy change for the reaction occuring in the cell:
Zn(s) | Zn2+ (1 M)|| Cu2+ (1M) | Cu(s)
[Given E°Zn2+/Zn = – 0.076 V, E°Cu2+/Cu = + 0.34 V, F = 96500 C mol–1]
Solution
Given:
= – 0.076 V
= + 0.34 V
F = 96500 C mol–1
from the reaction
n=2
= -
=0.34 -(-0.076)
= 0.416
We know that
= -2 x 96500 x 0.416
=–802.88 kJ mol–1
= – 0.076 V
= + 0.34 V
F = 96500 C mol–1
from the reaction
n=2
= -
=0.34 -(-0.076)
= 0.416
We know that
= -2 x 96500 x 0.416
=–802.88 kJ mol–1
