Question

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?

Answer 1

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.

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