Question

Calculate E^o , E, and ΔG for the following cell reaction

3Zn(s) + 2Cr³⁺(aq) ⇌ 3Zn²⁺(aq) + 2Cr(s)
where [Cr³⁺] = 0.090 M and [Zn²⁺] = 0.0085 M

Answer 1

1)

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

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

Answer: 0.0218 V

2)

Number of electron being transferred in balanced reaction is 6

So, n = 6

use:

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 (0.0085^3/0.09^2)

E = 2.18*10^-2-(-4.06*10^-2)

E = 6.24*10^-2 V

Answer: 6.24*10^-2 V

C)

number of electrons being transferred, n = 6

F = 96500 C

use:

ΔG = -n*F*E

= -6*96500*0.0624

= -36129.6 J/mol

= -36.1 KJ/mol

Answer: -36.1 KJ/mol