The Cell in which the following reaction occurs: 2Fe 3+ (aq) + 2I – (aq) → 2Fe 2+ (aq) + I 2 (s) has E 0 cell = 0.236V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Question

Philoid · Accepted Answer

Given:

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

E⁰_cell = 0.236V

n = moles of e^- from balanced redox reaction = 2

F = Faraday's constant = 96,485 C/mol

T = 298 K.

Using the formula, we get,

∆_rG⁰ = – nFE⁰_cell

⇒ ∆_rG⁰ = – 2 × FE⁰_cell

⇒ ∆_rG⁰ = −2 × 96485 C mol^-1 × 0.236 V

⇒ ∆_rG⁰ = −45540 J mol⁻¹

⇒ ∆_rG⁰ = −45.54 kJ mol⁻¹

Now,

∆_rG⁰ = −2.303RT log K_c

Where, K is the equilibrium constant of the reaction.

R is the gas constant; R = 8.314 J-mol-C^-1

⇒ −45540 J mol⁻¹ = –2.303× (8.314 J-mol-C^-1)× (298 K) × (log Kc)

Solving for K_c we get,

⇒ log Kc = 7.98

Taking antilog both side, we get

⇒ Kc = Antilog (7.98)

⇒ K_c = 9.6 × 107

Trick to remember:

The Cell in which the following reaction occurs:2Fe3+(aq) + 2I– (aq) → 2Fe2+(aq) + I2(s) has E0cell = 0.236V at 298 K.Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

The Cell in which the following reaction occurs:
2Fe³⁺(aq) + 2I^– (aq) → 2Fe²⁺(aq) + I₂(s) has E⁰_cell = 0.236V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.