Question

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
Mg2+(aq)+2e?	?Mg(s)	-2.37
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
Cu2+(aq)+2e?	?Cu(s)	0.16

Part A:

Pb2+(aq)+Mg(s)?Pb(s)+Mg2+(aq)

Express your answer using three significant figures.

Part B:

Br2(l)+2Cl?(aq)?2Br?(aq)+Cl2(g)

Express your answer using two significant figures.

Part C:

MnO2(s)+4H+(aq)+Cu(s)?Mn2+(aq)+2H2O(l)+Cu2+(aq)

Express your answer using two significant figures.

Answer 1

Calculation of equilibrium constant

Part A)

Eo = Ecathode - Eanode

     = -0.13 - (-2.37) = 2.24 V

Using,

nFEo = RTlnKc

Kc = equilibrium constant

R = gas constant

T = 25 oC + 273 = 298 K

F = Faraday's constant

n = number of electrons

So,

2 x 96485 x 2.24 = 8.314 x 298 lnKc

Kc = 5.9 x 10^75

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Part B)

Part A)

Eo = Ecathode - Eanode

     = 1.09 - 1.36 = -0.27 V

Using,

nFEo = RTlnKc

2 x 96485 x -0.27 = 8.314 x 298 lnKc

Kc = 7.36 x 10^-10

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Part C)

Eo = Ecathode - Eanode

     = 1.21 - 0.16 = 1.05 V

Using,

nFEo = RTlnKc

2 x 96485 x 1.05 = 8.314 x 298 lnKc

Kc = 3.30 x 10^35