(a) The cell in which the following reaction occurs :

2Fe³⁺(aq) + 2I^–(aq) → 2Fe²⁺(aq) + I₂(s)

has E°_cell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given : 1 F = 96,500 C mol^–1)

(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given : 1 F = 96,500 C mol^–1)

(a) Given:

Reaction: 2Fe³⁺(aq) + 2I^–(aq) → 2Fe²⁺(aq) + I₂(s)

Standard electrode potential of the cell (E^°_cell) = 0.236V

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 × 0.236

⇒ ΔG^° = -45548 J/mol

⇒ ΔG^° = -45.548 KJ/mol

Thus, the standard Gibbs energy of the cell reaction is -45.548KJ/mol.

(b) Given : 1 F = 96,500 C mol^–1)

Current (I) = 0.5 A

Time (t) = 2 hours = 2 × 3600 = 7200 sec. (∵ 1 hr=3600 seconds)

To calculate charge, we apply the formula Q= It

∴ Q = 0.5 × 7200

⇒ Q = 3600 coulombs

As we know that, electrons flowing through the wire on passing charge of one faraday (96500C) = 6.023 × 10²³

Electrons flowing through the wire on passing a charge of 3600 coulombs =

⇒ 2.246 × 10²² electrons

Thus, 2.246 × 10²² electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours.