Answer:
The equilibrium constant is [tex]8.83\times 10^{20}[/tex].
The value of ΔG° is -120. kJ/mol.
Explanation:
[tex]Cu^{2+}(aq)+Co(s)\rightarrow Cu(s)+Co^{2+}(aq)[/tex]
[tex]E^o_{Cu^{2+}/Cu}=0.34 V[/tex]
[tex]E^o_{Co^{2+}/Co}=-0.28 V [/tex]
[tex]E^o_{cell}=0.34 V-(-0.28 V)=0.62 V[/tex]
[tex]E^o_{cell}=\frac{0.592}{n}\log[K_c][/tex]
n = 2
[tex]0.62 V=\frac{0.0592}{2}\log[K_c][/tex]
[tex]K_c=8.83\times 10^{20}[/tex]
The equilibrium constant is [tex]8.83\times 10^{20}[/tex].
Relation between standard Gibbs free energy and equilibrium constant follows:
[tex]\Delta G^o=-RT\ln K_c[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy =
R = Gas constant = [tex]8.3145J/K mol[/tex]
T = temperature =
[tex]K_c[/tex] = Equilibrium constant
We have :
T = 298 K
[tex]K_c=8.83\times 10^{20}[/tex]
[tex]\Delta G^o=?[/tex]
[tex]\Delta G^o=-8.314 J/mol K\times 298 K\ln (8.83\times 10^{20})[/tex]
[tex]\Delta G^o=-119,492.86 J/mol=-119.49286 kJ/mol\approx -120. kJ/mol[/tex]
The value of ΔG° is -120. kJ/mol.
The equilibrium constant is 8.8 × 10^20 and the change in free energy is -119.5 kJ/mol.
The balanced equation of the reaction is; [tex]Cu^{2+}(aq) + Co(s)------> Cu(s) + Co^{2+} (aq)[/tex], we can see that 2 electrons were transferred in the process.
From;
E°cell = 0.0592/n logK
E°cell = E°cathode - E°anode
E°cell = 0.34V - (-0.28 V)
E°cell = 0.62 V
Hence;
0.62 = 0.0592/2 log K
K = Antilog (0.62 × 2/0.0592)
Keq = 8.8 × 10^20
From;
ΔG = -RTlnK
R = 8.314 J/K
T =298 K
ΔG = -(8.314 J/K.mol × 298 K × ln 8.8 × 10^20)
ΔG = -119.5 kJ/mol
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