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Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at...

Calculate the equilibrium constant for each of the reactions at 25 ∘C.

Standard Electrode Potentials at 25 ∘C
Reduction Half-Reaction E∘(V)
Pb2+(aq)+2e− →Pb(s) -0.13
Zn2+(aq)+2e− →Zn(s) -0.76
Br2(l)+2e− →2Br−(aq) 1.09
Cl2(g)+2e− →2Cl−(aq) 1.36
MnO2(s)+4H+(aq)+2e− →Mn2+(aq)+2H2O(l) 1.21
Pb2+(aq)+2e− →Pb(s) -0.13

Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g)

Express your answer using two significant figures.

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Answer #1

from data table:
Eo(Cl2(g)/2Cl-) = 1.36 V
Eo(Br2(l)/2Br-) = 1.09 V

As per given reaction/cell notation,
cathode is (Br2(l)/2Br-)
anode is (Cl2(g)/2Cl-)

Eocell = Eocathode - Eoanode
= (1.09) - (1.36)
= -0.27 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
-0.27 = (0.0592/2)*log Kc
log Kc = -9.1216
Kc = 7.558*10^-10
Answer: 7.6*10^-10

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