Calculate Eo , 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
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
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