Question

# 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 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)

Part B:

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

Part C:

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

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

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