Question

How many grams of Cu can be produced from 1.0 L of 0.50 M aqueous Cu(NO3)2 when a current of 0.50 A is run through the solution for 2 hours?

Answer 1

1.186g

Explanation

Cu²⁺(aq) + 2e -------> Cu(s)

1mole of Cu is deposited by 2 moles of electrons

1A = 1coulombs per second

0.5A = 0.5 coulombs per second

cuurent duration = 2hours = 7200s

Total coulombs passed = 0.5C/s × 7200s = 3600C

Faraday constant = 96485Coulombs per mole of electrons

No of moles of electrons = 3600C/96485C/mol = 0.03731mol

No of moles of Cu deposited = 0.03731mol/2 = 0.01866mol

Molar mass of Cu = 63.55g/mol

mass of Cu deposited = 0.01866mol × 63.55g/mol = 1.186g