Question

Consider a galvanic cell that uses the half reactions: Fe3+(aq) + 3e → Fe(s) Ered=-0.04V Co2+ (aq) + 2e → Co(s) Ered - -0.28 V • 1) Calculate AGº for this reaction in kJ/mol. • 2) Calculate the equilibrium constant, K, for this reaction at 298 K • 3) What is the cell potential at 25°C when [Fe3+] - 1.09 M and [Co2+] = 0.057 M?

Answer 1

1)

from data table:

Eo(Co2+/Co(s)) = -0.28 V

Eo(Fe3+/Fe(s)) = -0.04 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Fe3+/Fe(s))

anode is (Co2+/Co(s))

The chemical reaction taking place is

2 Fe3+(aq) + 3 Co(s) --> 2 Fe(s) + 3 Co2+(aq)

Eocell = Eocathode - Eoanode

= (-0.04) - (-0.28)

= 0.24 V

number of electrons being transferred, n = 6

F = 96500 C

use:

ΔGo = -n*F*Eo

= -6*96500.0*0.24

= -138960 J/mol

= -139 KJ/mol

Answer: -139 KJ/mol

2)

here, number of electrons being transferred, n = 6

Eo = (2.303*R*T)/(n*F) log Kc

At 25 oC or 298 K, R*T/F = 0.0592

So, Eo = (0.0592/n)*log Kc

0.24 = (0.0592/6)*log Kc

log Kc = 24.3243

Kc = 2.11*10^24

Answer: 2.11*10^24

3)

Number of electron being transferred in balanced reaction is 6

So, n = 6

use:

E = Eo - (2.303*RT/nF) log {[Co2+]^3/[Fe3+]^2}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Co2+]^3/[Fe3+]^2}

E = 0.24 - (0.0591/6) log (0.057^3/1.09^2)

E = 0.24-(-3.752*10^-2)

E = 0.2775 V

Answer: 0.277 V