Hey Syed!

The answer to your question is b) ie -46.32 kj/mol. The solution to the answer can be explained as follows:-

We have E°= 0.24 V

F= 96500 C/mol

and by writing the reduction reactions at two half cells occurring as follows  2Fe3+  +   2e-    =  2 Fe2+

and                                       I2          +2e-     = 2I-

We get the value of n= 2 ( no. of electrons involved in reduction and oxidation of species in two half cells)

Now, we know that Standard Gibbs Free energy, G= -nFE°

PUTTING THE VALUES OBTAINED, G= -2×96500×0.24 ====>  46.32 kj

Hello,

(a) Given:

Reaction: 2Fe ³⁺ (aq) + 2I ^– (aq) → 2Fe ²⁺ (aq) + I ₂ (s)

Standard electrode potential of the cell (E ^° _cell ) = 0.24V

1 F = 96,500 C mol ^–1 )

To calculate the standard Gibb’s free energy, we apply the formula given below:

ΔG ^° = -nFE ^° _cell

Where n is the no. of transferred electrons

F is the quantity of electricity flowing,

E ^° _cell is the standard cell potential

Reduction reaction:

Oxidation reaction:

In the above reactions, the number of electrons transfer(n) = 2e-

∴ ΔG ^° = -2× 96,500 × 0.24

⇒ ΔG ^° = -46320 J/mol

⇒ ΔG ^° = -46.320 KJ/mol

Thus, the standard Gibbs energy of the cell reaction is (b) -46.32 KJ/mol.