Question

Please show how to work this problem?

In some applications, nickel-cadmium batteries (EMF= 1.30 V) have been replaced by nickel-zinc batteries. The overall cell reaction for this new battery is:

2H2O (l) + 2NiO(OH) (s) + Zn (s) --> 2Ni(OH)2 (s) + Zn(OH)2 (s)

a) What is the cathode half reaction?

b) What is the anode half reaction?

c) What voltage would you expect the nickel-zinc battery will produce?

Answer 1

Solution-

The overall cell reaction for this new battery is as follows

2H2O (l) + 2NiO(OH) (s) + Zn (s) --> 2Ni(OH)2 (s) + Zn(OH)2 (s)

1. What is the cathode half reaction?

Cathode half reaction is as follows

2NiO(OH) + 2H2O +2e- --> 2Ni(OH)2 + 2OH-

b) What is the anode half reaction?

Anode half reaction is as follows

Zn + 2OH- --> Zn(OH)2 + 2e

Overall reaction

2NiO(OH) + 2H2O +2e- --> 2Ni(OH)2 + 2OH-

                     Zn + 2OH- --> Zn(OH)2 + 2e-

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2NiO(OH) + 2H2O + Zn --> 2Ni(OH)2 (s) + Zn(OH)2 (s)

1. What voltage would you expect the nickel-zinc battery will produce?

Given

E⁰_cell = 1.3 V

E⁰_Anod = 0.36 V

Zn is the better reducing agent

We know the equation

E⁰_cell = E⁰_cathod - E⁰_Anod

E⁰_cathod = 1.3V + 0.36V

E⁰_cathod = 1.66 V

Answer: voltage expect the nickel-zinc battery will produce = 1.66V