Question

How long would a solution of
Cr_{2}(SO_{4})_{3} have to be electrolyzed
with a current of 4.7 A in order to deposit 11 g of Cr?

What is the average current passing through a solution of NiSO4 if 5.91 g of Ni were deposited in 4.27 hours?

Answer #2

Ans 1.

The oxidation of Cr in Cr2(SO4)3 = +3

So each mole of Cr^{+3} utilises 3 moles of electrons to
get reduced to Cr.

11 g of Cr = 11 / 52 = 0.212 moles

0.212 moles would require 3 times the electrons , that is 0.636 moles.

Each mole of electrons uses 1 faraday or 96485 Coloumb ,

so 0.636 moles will use 0.636 x 96485 = 61634.46 C

Q = I. T

where Q is charge in Coloumbs , I is current in amperes and T is time in seconds

So putting all the values , we get ,

61634.46 = 4.7 x T

T = 13113.71 seconds

or 3.64 hours

answered by: anonymous

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