Alkaline batteries operate in a potassium hydroxide liquid electrolyte which all

Question

Alkaline batteries operate in a potassium hydroxide liquid electrolyte which allows current to flow at room temperature. The two half-cell reactions, written as reduction reactions, are given below along with their standard potentials in the alkaline solution. Zn(OH)_2 + 2e^- rightarrow + 2OH^- v degree = -1.25 v V degree = 0.30 V a) What material is the anode in an AA battery? b) What is the valence of the cathode cation after reduction? c) Write the overall reaction for the battery. d) Calculate the value of Delta G degree (in kJ/mol) for the overall reaction. e) Calculate the value of the equilibrium constant at room temperature. f) Calculate Delta G degree for MnO(OH) formation at room temperature.

Explanation / Answer

a) The more negative the standard potential of a couple, the more difficult it is to reduce the element. Consequently, it is easier to oxidise the element. Zn(OH)2/Zn2+ exhibits a more negative standard potential and hence this will constitute the anode of the AA battery.

The anode reaction is

Zn + 2 OH- ------> Zn(OH)2 + 2 e-; V0 = +1.25 V (reversing the reaction reverses the sign of V0).

b) MnO2/MnO(OH) couple will comprise the cathode. MnO2 is reduced to MnO(OH). Let the oxidation number of the metal (Mn) at the cathode be x.

Since the compound is electrically neutral and the oxidation states of O and OH are -2 and -1 respectively, we must have

x – 2 – 1 = 0

===> x – 3 = 0

===> x = 3

The valence of the cathode cation after the oxidation is +3 (ans).

c) The two half reactions are given:

2 MnO2 + 2 H2O + 2 e- -------> 2 MnO(OH) + 2 OH-; V0 = 0.30 V (cathode)

Zn + 2 OH- -----> Zn(OH)2 + 2 e-; V0 = +1.25 V (anode)

Since both the half reactions contain equal number of moles of electrons, simply add:

2 MnO2 + 2 H2O + Zn + 2 OH- ------> 2 MnO(OH) + Zn(OH)2; V0cell = V0cathode + V0anode = (0.30 V) + (1.25 V) = 1.55 V (ans).

d) The free energy change for the reaction is given by

G0 = -n*F*V0cell where n = number of moles of electrons transferred, F = 1 Faraday = 96485 J V-1 mol-1 and V0 = 1.55V. Plug in the values

G0 = -(2)*(96485 J V-1 mol-1)*(1.55 V) = 299103.5 J/mol = (299103.5 J/mol)*(1 kJ/1000 J) = -299.1035 kJ/mol (ans).

e) The equilibrium constant can be found out from the relation

G0 = R*T*ln K

Plug in values ( T= 298 K),

-299.1035 kJ/mol = -(8.314 J/mol.K)*(298 K)*ln K

===> -299.1035 kJ = -(2477.572 J)*ln K

===> ln K = (299.1035*1000 J)/(2477.572 J) = 120.724

===> K = e^(120.724) = 2.69*1052

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