Calculate the equilibrium constant, k, at 25°c for the galvanic cell reaction shown below: 8 h+(aq) + 5 fe2+(aq) + mno4-(aq) → mn2+(aq) + 5 fe3+(aq) + 4 h2o(l) e° = +0.74 v enter your answer in exponential format (sample 1.23e-4) with two decimal places and no units.

Equilibrium Constant = (3.60 × 10⁶²)

Explanation:

The change in Gibb's free energy for a galvanic cell is given as

Δ = -nFE°

where n = number of electrons transferred = 5 for this reaction

F = Faraday's constant = 96500 C

E° = cell potential = +0.74 V

But the change in Gibb's free energy for galvanic cell reaction is also given as

ΔG = -RT In K

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature in Kelvin = 25 + 273.15 = 298.15 K

K = Equilibrium constant.

Equating these two expressions

-nFE° = - RT In K

RT In K = nFE°

In K = (nFE°) ÷ (RT)

In K = (5 × 96500 × 0.74) ÷ (8.314 × 298.15)

In K = 144.04

K = e^(144.04)

K = (3.60 × 10⁶²)

Hope this Helps!!!

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