Question: 83. 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 \mathrm{Cu}: 63 \mathrm{~g} \mathrm{~mol}^{-1}, 1 \mathrm{~F}=96487 \mathrm{C} )
(1) 3.15 g
(2) 0.315 g
(3) 31.5 g
(4) 0.0315 g
Answer: Option (2)
Explanation:
The reaction occurring during electrolysis of copper sulphate solution is
\mathrm{Cu}^{2+} + 2e^{-} \rightarrow \mathrm{Cu}This shows that 2 moles of electrons are required to deposit 1 mole of copper.
Total charge passed is given by
Q = I \times t Q = 9.6487 \times 100 = 964.87 \ \mathrm{C}Charge required to deposit 1 mole of copper is
2 \times 96487 = 192974 \ \mathrm{C}Number of moles of copper deposited is
\text{Moles of Cu} = \dfrac{964.87}{192974} = 0.005Mass of copper deposited is
\text{Mass} = 0.005 \times 63 = 0.315 \ \mathrm{g}Therefore, the mass of copper deposited is 0.315 \ \mathrm{g}.