Question

The free energy change for the following reaction at 25 °C, when [Fe2+1 - 6.16*10' M and (Ag - 1.19 M, is 252 kJ: Fe2+(6.16*109 M) + 2Ag(s) >Fe(s) + 2Ag*(1.19 M) AG = 252 kJ What is the cell potential for the reaction as written under these conditions? Would this reaction be spontaneous in the forward or the reverse direction?

Answer 1

Cell potential (E_cell) and free energy change (∆G) are related by the following equation:

AG = -nFE cell

n=Change in number of transferred electrons and F= Faraday constant (96,485 C)

For Reaction 1

Fe2+(6.16 x 10-3M) + 2Ag(s) → Fe(s) + 2Ag+ (1.19 M)

And n=2

AG = 252 KJ

Then

252 x 103 = -2 x 96,485 x Ecell

Ecell = - 1.30V

Reaction is non spontaneous and doesn’t proceed in forward direction.

For Reaction 2

Pb2+ (4.82 x 10-3M) + 2Ag(s) + Pb(s) + 2Ag+ (1.17 M)

And n= 2

AG = 192 KJ

Then

192 x 103 = -2 x 96,485 x Ecell

Ecell = -0.99 V

Reaction is non spontaneous and doesn’t proceed in forward direction.

For Reaction 3

3Pb2+ (1.19M) + 2Cr(s) + 3 Pb(s) + 2 Cr3+(9.96 x 10-3M)

And n=3

AG = -367 KT

Then

-367 x 103 = -3 x 96,485 x Ecell

Ecell = 1.26 V

Reaction is spontaneous and proceed in forward direction.