Can I get help on this

Question

Consider the following reaction and its Δ𝐺∘ at 25.00 °C.

Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s) Δ𝐺∘=−408.0 kJ/mol

Calculate the standard cell potential, 𝐸∘cell, for the reaction.

𝐸∘cell=V?

Calculate the equilibrium constant, 𝐾, for the reaction.

𝐾=

Answer 1

∆Gº = -nFEº = -RT ln K

To find Eº, we first need to find n (number of electrons transferred)

Mg(s) ==> Mg2+ + 2e- ... oxidation

Ni2+ + 2e- ==> Ni(s) ... reduction

Therefore n = 2 mols electrons transferred

∆Gº = -nFEº

-408.0 kJ/mol = -(2 mol e-)(96,500 Coulombs/mol)(Eº)

1 kJ = 1000 coulomb volts, so -408.0 kJ = -408,000 CV

-408,000 CV = -193,000 C (Eº)

Eº = -408,000 C V / -193,000 C

Eº = +2.11 V

To find K, we can use ∆Gº = -RT ln K

∆Gº = -408 kJ/mol

R = 0.008314 kJ/Kmol

T = 298 K

K = ?

-408 = -(0.008314)(298) ln K

-408 = -2.478 ln K

ln K = 164.6

K = 3.1x10⁷¹