Question

Calculate the non standard Gibbs Free Energy change, ΔG at 250C for the following reaction with the indicated concentrations.

Zn + 2Ag+ (0.30M) → 2Ag + Zn2+ (0.50M)

Remember you'll need to calculate both the standard and non standard cell potential Ecell.

Remember ΔG0 = − nFE0cell  and a similar equation for the NONstandard value ΔG = −nFEcell

Here are the standard reduction potentials:

Zn2+/ Zn: - 0.763V
Ag+/ Ag: + 0.799V

Answer in kJ to 3SF including signs as needed

Part 2

Calculate the value of lnK where K is the equilibrium constant at 250C for the following reaction with the indicated concentrations.

Zn + 2Ag+ (0.30M) → 2Ag + Zn2+ (0.50M)

You should have already calculated both the standard and non standard cell potentials.

Remember ΔG0 = − nFE0cell  and ΔG0 = - RTlnK

You must use the standard value of ΔG to find K!!!

Think about what value of Ecell you need to use!!

Give your answer to lnK to 3SF

Answer 1

Zn + 2Ag+ (0.30M) ----> 2Ag + Zn2+ (0.50M)

E0cell = Ecathode - E0anode

       = 0.799 -(-0.763)

       = 1.562 v

DG0 = -nFE0cell

    = -2*96500*1.562

    = -301.5 kj

    = -3.01*10^2 kj

Ecell = E0cell- (0.0591/n)log([Zn2+]/[Ag+]^2

      = 1.562-(0.0591/2)log(0.5/0.3^2)

      = 1.54 v

DG = - nFEcell

   = -2*96500*1.54

   = -297.22 kj

   = -2.97*10^2 kj

part 2

DG0   = -3.01*10^2 kj

DG0 = - RTlnK

-3.01*10^5 = -8.314*298lnK

lnK = 121.5

K = 5.79*10^52