# 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. - Chemistry

Sum

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 × 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.

Concept: Electrolytic Cell
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Chapter 5: Electrochemistry - Exercises [Page 118]

#### APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 5 Electrochemistry
Exercises | Q 4.03 | Page 118
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