Question

a current of 965 ampere is passed for 10 seconds through a solution of copper sulphate then what would be the weight of the copper deposited?​

Answer 1

Answer:

5 amperes is 5 coulombs per second, 5C/s

So the total charge in 965 seconds is Q=5C/s×965s=4825C

Then the number of moles of copper plated out (n) is:

n=

zF

Q

where z is the number of electrons in the half-cell reaction and F is the Faraday constant = 96,485/mol

So, n=

(2×96485)

4825

=0.025mol

And this is 63.546×0.025=1.58g

So 1.58grams are plated deposited at the cathode.

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