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!!!

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles