Question

Calculate E^o, E, and delta G for the following cell reaction

3Zn(s) + 2Cr3+(aq) ⇌ 3Zn2+(aq) + 2Cr(s) where [Cr3+] = 0.090 M and [Zn2+] = 0.0095 M

I found E^o to be 0.02 V

I'm having trouble finding E and delta G

Answer 1

1)

Lets find Eo 1st

from data table:

Eo(Zn2+/Zn(s)) = -0.7618 V

Eo(Cr3+/Cr(s)) = -0.74 V

As per given reaction/cell notation,

cathode is (Cr3+/Cr(s))

anode is (Zn2+/Zn(s))

Eocell = Eocathode - Eoanode

= (-0.74) - (-0.7618)

= 0.0218 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[Zn2+]^3/[Cr3+]^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 {[Zn2+]^3/[Cr3+]^2}

E = 2.18*10^-2 - (0.0591/6) log (0.0095^3/0.09^2)

E = 2.18*10^-2-(-3.918*10^-2)

E = 6.098*10^-2 V

Answer: E = 0.061 V

2)

number of electrons being transferred, n = 6

F = 96500.0 C

we have below equation to be used:

deltaG = -n*F*E

= -6*96500.0*0.06098

= -35307.42 J/mol

= -35.3 KJ/mol

Answer: delta G = -35.3 KJ/mol