Question

Answer the following:

What current strength in amperes will be required to produce 2.4 g of Cu from CuSO₄ solution in 1 hour? Molar mass of Cu = 63.5 g mol^–1.

Answer 1

Given:

Mass of Cu = 2.4 g,
Molar mass of Cu = 63.5 g mol^–1
1 hours = 1 × 60 × 60s = 3600s

To find: Current strength (in amperes)

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:

1) Stoichiometry for the formation of Cu is

`"Cu"_("s")^(2+) + 2"e"^(-) -> "Cu"_(("s"))`

Using formula (i),

Mole ratio = `(1  "mole")/("2 mole")`

2) Using formula (ii),

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

`2.4 "g" = ("I"("A") xx "t"("s"))/(96500 ("C"//"mol"  "e"^(-))) xx (1  "mole")/("2 mole"  "e"^-1) xx 63.5 "g mol"^-1`

I(A) = `(2.4 xx 96500 xx 2)/(63.5 xx 3600)` = 2.03 A

Current strength in amperes required to produce 2.4 g of Cu from CuSO₄ is 2.03 A.