Question

When the nickel-zinc battery, used in digital cameras, is recharged, the following cell reaction occurs:

2Ni(OH)2(s)+Zn(OH)2(s)→ 2Ni(OH)3(s)+Zn(s)

How many grams of zinc are formed when 3.39×10−2 g of Ni(OH)2 are consumed?

How many minutes are required to fully recharge a dead battery that contains 6.13×10−2 g of Zn with a constant current of 0.100 A?

Thank You!!

Answer 1

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Given,

mass of Ni(OH)2 =3.39 X 10^-2

molecular mass of Ni(OH)2 = 92

moles of Ni(OH)2 = 3.39 X 10^-2 / 92

                           = 3.6 X 10^-4

Thus ;

2Ni(OH)2(s)+Zn(OH)2(s)→ 2Ni(OH)3(s)+Zn(s)

in the above reaction we can see that;

2 moles of Ni(OH)2 gives 1 mole of Zn

1 moles of Ni(OH)2 gives 1/2 mole of Zn

so 3.6 X 10^-4 moles will give = 3.6 X 10-4 / 2 moles

                                              = 1.8 X 10-4 moles

Hence mass of Zinc produced = 65 X 1.8 X 10-4

                                               = 0.0117 gm of Zinc.

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Given

Mass of Zinc = 6.13×10−2 g

Current = 0.100 A

Atomic mass of Zn is given as 65.

Valency = 2, since 2 moles of electrons can be used to combine with another atom.

Equivalent weight = Atomic weight of Zinc/Valency of Zinc = 65/2 =32.5 g

Hence by using faraday law of electrolysis

m = E x I x t /96,500

6.13 X 10-2 = 32.5 X 0.100 X t / 96500

t = 1820 seconds.

hence time required = 1820 sec = 30 minutes (appox)