Calculate the equilibrium constant for each of the reactions at 25∘C∘C. Standard Electrode Potentials at 25...
Calculate the equilibrium constant for each of the reactions at 25∘C∘C.
| Standard Electrode Potentials at 25 ∘C |
| Reduction Half-Reaction |
| Cl2(g)+2e ---> 2CI- 1.36 |
I2(s)+2e --> 2I- 0.54
Part A
Cl2(g)+2I−(aq)→2Cl−(aq)+I2(s)
K= ?
Homework Answers
from data table:
Eo(I2(s)/2I-) = 0.54 V
Eo(Cl2(g)/2Cl-) = 1.36 V
As per given reaction,
cathode is (Cl2(g)/2Cl-)
anode is (I2(s)/2I-)
Eocell = Eocathode - Eoanode
= (1.36) - (0.54)
= 0.82 V
here, number of electrons being transferred, n = 2
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
0.82 = (0.0592/2)*log Kc
log Kc = 27.7027
Kc = 5.043*10^27
Answer: 5.04*10^27
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Calculate the equilibrium constant for each of the reactions at
25∘C∘C.
Standard Electrode Potentials at 25...
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