Q. 125.0( 2 Votes )

# (a) The cell in w

Answer :

(a) GivenReaction- 2Fe3+(aq) + 2I(aq) 2Fe2+(aq) + I2(s)

(1 F = 96,500 C mol–1)

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

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.

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