Question

A current of 2 Amperes is passed through a litre of 0.100 molar solution of copper sulphate for 4825 seconds. The molarity of copper sulphate at the end of experiment is

• A. 0.075
• B. 0.050
• C. 0.025
• D. 0

Answer 1

Answer: (B)

0.050

Answer 2

$Cu^{2 +} + 2e^{-} \rightarrow Cu$

$\mathbf{Q = 2}\mathbf{\times}\mathbf{4825 = 9650}$ Coulombs

$\therefore$ 96500 Coulombs of electricity deposits 0.5 mole copper

$\therefore$ 9650 C $= \frac{0.5 \times 9650}{96500} = 0.05$mole of copper

$\therefore$ Molarity at the end of the experiment

$= 0.1 - 0.05 = 0.050$