Question : For the cell reaction
2Fe3+ (aq) + 2I– (aq) → 2Fe2+ (aq) + I2 (aq)
Eo (cell) = 0.24V at 298 K. The standard Gibbs energy (△r GO) of the cell reaction is
(1) -46.32 kJ mol-1
(2) -23.16 kJ mol-1
(3) 46.32 kJ mol-1
(4) 23.16 KJ mol-1
answer : We have AG° =-nFE°
For the given reaction,
n = 2, therefore,
AG° = -2 x 96500 x 0.24
= -46320J mol-1
= -46.320 kJ mol-1
that is why the correct Answer is option (1) -46.32 kJ mol-1
Answer explanation : as we know that The reduction process of two half cell is occurring here it means two half cell reduction is take place here as given below –
2Fe3+ + 2e-1 = 2 Fe2+
and
I2 +2e-1 = 2I-1
here 2 electrons take place in the reaction so the n = 2
as we know that the Standard Gibbs Free energy is calculated by ,
G = -nFE°
F = 96500 , Eo (cell) = 0.24V
so
G° = -2 x 96500 x 0.24
= -46320J mol-1
= -46.320 kJ mol-1
correct Answer is option (1) -46.32 kJ mol-1