Verified answer

Answer

Part 1

∆G° = –RTlnK = –(8.314 J/mol-K)*(298 K)*ln(5.7 × 10-2) =

T = 298 k

R= 8.314 J/mol-K

this equals to, ∆G° = [ -8.314*298*2.303*(-2)*log57 ] / 1000 = 20.03 kj/mol

Part 2

ΔG = -nFEcell

where

ΔG is the free energy of the reaction

n is the number of moles of electrons exchanged in the reaction

F is Faraday's constant (96484.56 C/mol)

E0 is the cell potential.

E0 = - ∆G°/2F = - 20.03 kj/mol /2* (96,485 C/mol )

this equals to - 0.1036 ( j / coulomb) = - 0.1036 v

1 faraday = 96 485 coulombs

1 joule per coulomb is equivalent to 1 volt.