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.
Solution
Given:
Mass of Cu = 2.4 g,
Molar mass of Cu = 63.5 g mol–1
1 hours = 1 × 60 × 60 s = 3600 s
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.
