for a cell reaction 2Fe3+ +2I- --------> 2Fe2+ +I2 E=0.24 V at 298K. The standard Gibbs energy of the cell reaction is? (F= 96500 C/mol) a) 23.16kJ/mol. B) -46.32kJ/mol c) -23.16kJ/mol. D) 46.32kJ/mol
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.