Question

What is the mass of Cu metal produced at the cathode during the passage of 5 ampere of current through CuSO₄ solution for 100 minutes? [Molar mass of Cu = 63.5 g mol⁻¹].

Answer 1

Given: Current strength = 5 A, Molar mass of Cu = 63.5 g mol^–1
100 minutes = 100 × 60 s = 6000 s

To find: Mass of copper metal produced at cathode

Formulae:

1. Mole ratio = `"Moles of product formed in half-reaction"/"Moles of electrons required in half-reaction"`
2. W = `("I"("A") xx "t" ("s"))/(96500 ("C"//"mol e"^-)) xx "mole ratio" xx "molar mass"`

Calculation:

Stoichiometry for the formation of Cu is

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

Using formula (i),

Mole ratio = `"1 mol"/"2 mol e"^-`

Using formula (ii),

W = `("I"("A") xx "t" ("s"))/(96500 ("C"//"mol e"^-)) xx "mole ratio" xx "molar mass"`

`= (5  "A" xx 6000)/(96500 ("C"//"mol e"^-1)) xx (1  "mol")/("2 mol e"^-) xx 63.5  "g mol"^-1`

= 9.87 g

Mass of copper metal produced at cathode is 9.87 g.