Question

8H^(+) + MnO4^(-) + 5Fe^(2+) → Mn^(2+)+ 5Fe^(3+) + 4H2O Calculate the standard Gibbs energy change for the reaction. (E=+0.74V)

Calculate the equilibrium constant.

Answer 1

+7 +2

8H⁺ + MnO₄^- + 5Fe²⁺ → Mn²⁺+ 5Fe³⁺ + 4H₂O

We know that G = - nFE

Where

n = number of electrons transferred = change in oxidation state of Mn = 5

F = Faraday = 96500 C

E = potential of the cell = +0.74 V

Plug the values we get

G = - nFE

       = -(5x96500x0.74) J

      = -357x10³ J

      = -357 kJ

Therefore the standard Gibbs energy change of the reaction is 357 kJ

Also we know that G = - RT lnK

Where

R = gas constant = 8.314 J/(mol-K)

T = Temperature = 25 ^oC = 25+273 = 298 K

K = Equilibrium constant =?

Plug the values we get

ln K = -(G / RT)

      = 144

   K = e ¹⁴⁴

      = 3.45x10⁶²

Therefore the equilibrium constant is 3.45x10⁶²