Question

When a current of 0.75 ampere is passed through a CuSO₄ solution for 25 minutes, 0.370 g of copper is deposited. Calculate the atomic weight of copper by using the given information.

Answer 1

Given: Current (I) = 0.75 A
Time (t) = 25 minutes 

= 25 × 60 

= 1500 s

Formula: Charge (Q) = I × t

= 0.75 × 1500

= 1125 C

\[\ce{Cu^2+_{(aq)} + 2e- -> Cu_{(s)}}\]

This means 2 moles of electrons (n = 2) are required to deposit one mole of copper atoms.

According to Faraday’s first law of electrolysis, the mass deposited (m) is related to the atomic weight by:

m = `(M xx Q)/(n xx F)`

M = `(m xx n xx F)/Q`

= `(0.370 xx 2 xx 96485)/1125` 

= `71398.9/1125`

= 63.4657 g/mol