An electrolytic cell is set up using a solution of aqueous Cu(NO3)2, and an external 6V
battery is connected to two inert electrodes which are dipped into the solution. If a current of 113
mA flows through the cell for 30.00 minutes, what mass of solid copper metal will be produced at
the cathode?
This problem is based of Electrolysis,
I = 113 mA = 0.113 A,
t = 30.00 min = 30.00 x 60 = 1800 s
F = Faraday's constant = 96485 C/mol
z = # of electrons transfered = 2
as, (Cu(NO3)2 --------> Cu2+ (aq.) + 2 NO3- (aq.
Cu2+ (aq.) + 2e- ---------> Cu(0)
M =Molar mass of Cu(NO3)2 = 187.56 g/mole.
m =Mass liberated = ?
Formula (by Faradays 1st and 2nd law of electrolysis)
m = ItM /zF
on substituting values,
m = 0.113 x 1800 x 187.56 / (2x96485)
m = 0.198 Kg.
m = 198 g
Mass of solid Cu deposited will be 198 g.
========================XXXXXXXXXXXXXXXXXXXXX======================
Get Answers For Free
Most questions answered within 1 hours.