Question

2.5 amperes of current is passed through copper sulphate solution for 30 minutes. Calculate the number of copper atoms deposited at the cathode (Cu = 63.54).

Answer 1

Given:

Current (I) = 2.5 A

Time (t) = 30 minutes = 30 × 60 = 1800 seconds

Molar mass of Cu = 63.54 g/mol

Valency of Cu in CuSO₄ = 2 (i.e., Cu²⁺)

Faraday’s constant (F) = 96500 C/mol

Avogadro’s number (N_a) = 6.022 × 10²³ atoms/mol

Calculations:

Q = I × t

= 2.5 × 1800

= 4500 C

The half-cell reaction is:

\[\ce{Cu^2+ + 2e- −> Cu}\]

So, 2 Faradays (193000 C) deposit 1 mole (63.54 g) of copper.

Moles of Cu = `(4500 xx 1)/(2 xx 96500)`

= `4500/193000`

= 0.0233 mol

Atoms of Cu = 0.0233 × 6.022 × 10²³

= 1.40 × 10²² atoms