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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 18.54

Write the equilibrium constant expression for each of the following reactions, and determine the numerical value of Keq at 25 oC.

(a) Ag+(aq) + Fe2+(aq)Fe3+(aq) + Ag(s)

(b) MnO2(s) + 4 H+(aq) + 2 Cl(aq) Mn2+(aq) + 2 H2O(l) + Cl2(g)

(c) 2 OCl(aq) 2 Cl(aq) + O2(g) (basic solution)




Answer:
(a) Ag+(aq) + Fe2+(aq)Fe3+(aq) + Ag(s)

The reaction is already balanced so the mass action expression for the equilibirum constant is easily found by inspection. Since a thermodynamic equilibrium constant is required, aqueous components are given as concentrations and gases are given as partial pressures (pure liquids and pure solids are ignored):

To find the value of this equilibrium constant, use the Table of Standard Reduction Potentials to find the cell potential and then use this value to find the equilibrium constant.

Ag+(aq) + eAg(s)

Eored = 0.800 V

Fe2+(aq) Fe3+(aq) + e

Eoox = –0.771 V

Eocell = 0.800 – 0.771 = 0.029 V

 

Keq = exp(nFEocell/RT)

n = 1
F = 96485 coul/mol
R = 8.314 J/mol•K
T = 298 K
 

Keq = exp[(1)(96485)(0.029)/(8.314)(298)] = 3.1

(b) MnO2(s) + 4 H+(aq) + 2 Cl(aq) Mn2+(aq) + 2 H2O(l) + Cl2(g)

The reaction is already balanced so the mass action expression for the equilibirum constant is easily found by inspection. Since a thermodynamic equilibrium constant is required, aqueous components are given as concentrations and gases are given as partial pressures (pure liquids and pure solids are ignored):

To find the value of this equilibrium constant, use the Table of Standard Reduction Potentials to find the cell potential and then use this value to find the equilibrium constant.

MnO2(s) + 4 H+(aq) + 2 eMn2+(aq) + 2 H2O(l)

Eored = 1.23 V

2 Cl(aq) Cl2(g) + 2 e

Eoox = –1.358 V

Eocell = 1.23 –1.358 = –0.13 V

 

Keq = exp(nFEocell/RT)

n = 2
F = 96485 coul/mol
R = 8.314 J/mol•K
T = 298 K
 

Keq = exp[(2)(96485)(–0.13)/(8.314)(298)] = 4×10–5

(c) 2 OCl(aq) 2 Cl(aq) + O2(g) (basic solution)

The reaction is already balanced so the mass action expression for the equilibirum constant is easily found by inspection. Since a thermodynamic equilibrium constant is required, aqueous components are given as concentrations and gases are given as partial pressures (pure liquids and pure solids are ignored):

To find the value of this equilibrium constant, use the Table of Standard Reduction Potentials to find the cell potential and then use this value to find the equilibrium constant.

OCl(aq) + H2O(l) + 2 eCl(aq) + 2 OH(aq)

Eored = 0.890 V

4 OH(aq) O2(g) + 2 H2O(l) + 4 e

Eoox = –0.401 V

Eocell = 0.890 –0.401 = 0.489 V

 

Keq = exp(nFEocell/RT)

n = 4
F = 96485 coul/mol
R = 8.314 J/mol•K
T = 298 K
 

Keq = exp[(4)(96485)(0.489)/(8.314)(298)] = 1×1033