Question

A current of 3.56 A is passed through a Cu(NO3)2 solution. How long (in hours) would this current have to be applied to plate out 8.60 g of copper?

Can someone list out the steps to solving this problem, would help a lot.

Answer 1

Electrode reduction potential for Cu2+|Cu half cell,

Cu2+ +2e Cu(s) Eo(reduction)=0.3419V

Using equation,

Current,I=Q(charge passed) per unit time,t

I=Q/t

2 mol electrons are required to be passed to plate out 1 mol Cu(s)

Let's find out mol of Cu to be plated out=mass/molar mass of Cu=8.60 g/63.546 g/mol=0.135 mol

So mol of electrons required to be passed=2*0.135mol=0.271mol electrons

Also ,Q=F*n ,F=faraday's constant=96485C/mol electrons ,n=mol of electrons passed

I=Q/t

t=Q/I=F*n/I=96485 C/mol*0.271mol/3.56 A [A=C/s]

t=7344.785s=7344.785 s*(1h/3600s)=2.04 hours

t=2.04 hours