Question

For cell reaction
2fe³⁺(aq) + 2 I⁻ (aq) → 2fe²⁺(aq) + I₂(aq)
E⁻cₑₗₗ = 0.24 V at 298K. The standard Gibbs energy (ΔᵣG⁻) of the cell reaction is :
[Given that Faraday constant F = 96500 C mol⁻¹}
(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 1

The Gibbs free energy of the given reaction is -46.32kJ/mol

Step by step explanation:

The given chemical reaction is as follows.

$\bold{2Fe^{3+}(aq)+2I^{-}(aq)\rightarrow 2Fe^{2+}(aq)+I_{2}(aq)}$

$E_{cell}^{o}=0.24$

Total number of electrons transferred in the reaction "n" = 2

Gibbs free energy is calculated by the following formula.

$\bold{\Delta\,G^{0}=-nFE_{cell}^{o}}$

Substitute the given values.

$=-2\times 96500\times 0.24\,=-46.32kJ\,mol^{-1}$

Therefore, The Gibbs free energy of the given reaction is -46.32kJ/mol

Hence, correct option is -1