Question

How much copper is deposited on the cathode if a current of $3A$ is passed through aqueous $CuS{O}_4$ solution for $15$ minutes?

Answer 1

3 amperes is 5 coulombs per second, 3 C/s

So the total charge in 30 minutes is $Q=3C/s\times 15min\times 60s/min=2700C$

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

$n=\frac{Q}{zF}$

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, $n=\frac{2700}{(2\times 96485)}=0.0139mol$

And this is $63.546\times 0.0139=4.54g$

So 4.54 grams are plated deposited at the cathode.