Answer to Question #145806 in Physical Chemistry for deepak

Question

Electrorefining of copper using acidic copper sulphate solution is carried out at a current of
15000 A and 0.4 V. If the output per day (uninterrupted cell operation) is 410 kg on a total cathode
area of 10 m2

, calculate: a) current efficiency, b) current density and c) thickness of deposit. {Data:
Relative atomic mass of copper is 64; Faraday constant = 96500 Coulomb/gm eqv wt; density of
copper is 8930 kg m-3}.

Answer 1

Question (a) current efficiency

n=2 [Cu²⁺+2e⁻→Cu]

Atomic mass=64 g

t=60×60×24sec

t=86400sec

I=15000A

Mass deposited=410kg=410000g

m= [Atomic Mass/nF] ​It

m=64/ [2×96500] ×15000×86400

m=426404g, m=426.404kg

Current efficiency is the ratio of the actual mass deposited to the theoretical mass liberated according to Faraday's law.

Current Efficiency= [Theoretical Mass /Actual Mass​] ×100

[410/426.404]100%=96.15%

The current efficiency is 96.15%

Question (b) current density

J = I / A

I= 15000A        =15000/10

A= 10m²           =1500A/m²

Question (c) thickness of deposit

Volume of the deposited mass= mass/density

=410/8930= 0.04591m³

Thickness = Volume/Area

=0.04591/10= 0.004591m