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Problem 18.54

Write the equilibrium constant expression for each of the following reactions, and determine the numerical value of K_{eq} at 25 ^{o}C.

(a)Ag^{+}(aq) + Fe^{2+}(aq)Fe^{3+}(aq) + Ag(s)(b)MnO_{2}(s) + 4 H^{+}(aq) + 2 Cl^{–}(aq) Mn^{2+}(aq) + 2 H_{2}O(l) + Cl_{2}(g)(c)2 OCl^{–}(aq) 2 Cl^{–}(aq) + O_{2}(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.

Ag

^{+}(aq) + e^{–}Ag(s)E ^{o}_{red}= 0.800 VFe

^{2+}(aq) Fe^{3+}(aq) + e^{–}E ^{o}_{ox}= –0.771 VE

^{o}_{cell}= 0.800 – 0.771 = 0.029 V

K

_{eq}= exp(nFE^{o}_{cell}/RT)n = 1

F = 96485 coul/mol

R = 8.314 J/mol•K

T = 298 K

K

_{eq}= exp[(1)(96485)(0.029)/(8.314)(298)] = 3.1

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.

MnO

_{2}(s) + 4 H^{+}(aq) + 2 e^{–}Mn^{2+}(aq) + 2 H_{2}O(l)E ^{o}_{red}= 1.23 V2 Cl

^{–}(aq) Cl_{2}(g) + 2 e^{–}E ^{o}_{ox}= –1.358 VE

^{o}_{cell}= 1.23 –1.358 = –0.13 V

K

_{eq}= exp(nFE^{o}_{cell}/RT)n = 2

F = 96485 coul/mol

R = 8.314 J/mol•K

T = 298 K

K

_{eq}= exp[(2)(96485)(–0.13)/(8.314)(298)] = 4×10^{–5}

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) + H_{2}O(l) + 2 e^{–}Cl^{–}(aq) + 2 OH^{–}(aq)E ^{o}_{red}= 0.890 V4 OH

^{–}(aq) O_{2}(g) + 2 H_{2}O(l) + 4 e^{–}E ^{o}_{ox}= –0.401 VE

^{o}_{cell}= 0.890 –0.401 = 0.489 V

K

_{eq}= exp(nFE^{o}_{cell}/RT)n = 4

F = 96485 coul/mol

R = 8.314 J/mol•K

T = 298 K

K

_{eq}= exp[(4)(96485)(0.489)/(8.314)(298)] = 1×10^{33}