The cell in which the following reaction occurs:
2Fe3+ (aq) + 2I- (aq) → 2Fe2+ (aq) + I2 (s)
Has Eocell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction (Given: 1F = 96,500 C/mol)
The standard Gibbs free energy (∆G) can be obtained from the equation,
∆G = -nFEcell
Where,
n is the number of electrons taking part in the reaction
Ecell is the EMF of the cell Electrode potential
F is Faraday’s Constant (1F= 96500 C/mol)
On Splitting the cell reaction into two half reactions, we get,
Reduction half-reaction
Fe3+ + e- → Fe2+ (here n=1)
So, for 2 moles of Fe3+
2Fe3+ + 2e- → 2Fe2+ (n =2)
Similarly, for the Oxidation Half-reaction
I- → ½ I2 + e- (n=1)
For 2 moles of I-
2I- → I2 + 2e- (n=2)
Using the obtained values in the equation, we get,
-∆G = nFE = 2 x 96500 x 0.236 = -45548 kJ/mol
Solving the equation using the log table,
Log x= Log (2 x 96500 x 0.236) = Log 2 + Log 96500 + log (2.36 x 10-1)
= Log 2 + Log 96500 + log 2.36 + log 10-1 = 0.3010 + 4.9845 + 0.3729 -1
Separating the characteristic and the mantissa and solving further,
= 0.3010 + 4 + 0.9845 + 0.3729 -1
= 3 + 1.6584
Log (x) = 4.6584
x= Antilog (4.6584) = 45540
-∆G= 45540 kJ/mol or ∆G= -45540 kJ/mol
The reaction is exothermic and spontaneous.