Question

Calculate the free energy change (ΔG) for the following cell: Cu/Cu+2 (0.010M) // Ag+ (0.050M)/Ag Ag+ + e-  Ag Eo = 0.799V Cu+2 + 2e-  Cu Eo = 0.337V

Answer 1

Lets find Eo 1st

from data table:

Eo(Cu2+/Cu(s)) = 0.337 V

Eo(Ag+/Ag(s)) = 0.7996 V

As per given reaction/cell notation,

cathode is (Ag+/Ag(s))

anode is (Cu2+/Cu(s))

Eocell = Eocathode - Eoanode

= (0.799) - (0.337)

= 0.462 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

Use:

E = Eo - (2.303*RT/nF) log {[Cu2+]^1/[Ag+]^2}

Here:

2.303*R*T/n

= 2.303*8.314*298.0/F

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Cu2+]^1/[Ag+]^2}

E = 0.462 - (0.0591/2) log (0.01^1/0.05^2)

E = 0.462-(1.78*10^-2)

E = 0.444 V

number of electrons being transferred, n = 2

F = 96500.0 C

use:

ΔG = -n*F*E

= -2*96500.0*0.444

= -85692 J/mol

= -85.69 KJ/mol

Answer: -85.69 KJ/mol