Question

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol–1 , 1 F = 96487 C)

• A. 3.15 g
• B. 0.315 g
• C. 31.5 g
• D. 0.0315 g

Answer 1

Answer: (B)

0.315 g

Answer 2

Step 1 : Faraday's second law of electrolysis:

Mass of Cu deposited = $\frac{I \times t \times M}{n \times F}$

$M$ = Molar mass of Cu, $n$ = Number of electrons, $F$ = Faraday's constant

Step 2 : Substituting values:

Mass = $\frac{9.6487 \times 100 \times 63}{2 \times 96487} = 0.315 \, g$