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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

SbistudyPosted oninchemistry

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 

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