Question

During electrolysis of copper(ii) tetraoxosulphate(vi) solution, a steady current of 4.0*10^2A flowing for one hour liberated 0.48 of copper. Calculate the mass of copper liberated by one coulomb of charge.

Answer 1

Answer:

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

So the total charge in 30 minutes is Q = 5C/s x 30min x 60s/min = 9000C

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

n = Q/zF where z is the number of electrons in the half-cell reaction (in this case, 2) and F is the Faraday constant = 96,485/mol

So, n = 9000/(2x96485) = 0.0466mol

And this is 63.546 x 0.0466 = 2.96g

So 2.96 grams are plated deposited at the cathode.