A current of 0.5 ampere is passed for 30 minutes through a voltameter containing CuSO4 solution.The weight of Cu deposited will be

Question

A current of 0.5 ampere is passed for 30 minutes through a voltameter containing CuSO4 solution.The weight of Cu deposited will be (A) 3.18 g (B) 0.318 g (C) 0.296 g (D) 0.150 g

Tardigrade Admin · Accepted Answer

Q = 1 × t = 0.5 × 30 × 60 = 900 C Cu2+ + 2e- arrow Cu 2 × 96500 C deposit Cu = 63.5 g ∴ 900 C will deposit Cu = (63.5/2× 96500) × 900 = 0.296 g

Q. A current of $0.5$ ampere is passed for $30$ minutes through a voltameter containing $C u S O_{4}$ solution.The weight of $C u$ deposited will be

$3.18 g$

$0.318 g$

$0.296 g$

$0.150 g$

$Q = 1 \times t = 0.5 \times 30 \times 60 = 900 C$
$C u^{2 +} + 2 e^{-} \to C u$
$2 \times 96500 C$ deposit $C u = 63.5 g$
$∴ 900 C$ will deposit $C u$
$= \frac{63.5}{2 \times 96500} \times 900$
$= 0.296 g$

Q. A current of 0.5 ampere is passed for 30 minutes through a voltameter containing CuSO4​ solution.The weight of Cu deposited will be

3.18g

0.318g

0.296g

0.150g