0.5 F of electricity is passed through 500 mL of copper sulphate solution. The amount of copper which can be deposited will be

Question

0.5 F of electricity is passed through 500 mL of copper sulphate solution. The amount of copper which can be deposited will be (A) 63.5 g (B) 31.75 g (C) 15.8 g (D) Unpredictable

Tardigrade Admin · Accepted Answer

Cu 2++2 e - longrightarrow Cu 2 F =1 mol of Cu =63.5 g of Cu 0.5 F =(63.5/2) × 0.5=15.8 g

Q. $0.5 F$ of electricity is passed through $500 m L$ of copper sulphate solution. The amount of copper which can be deposited will be

$63.5 g$

$31.75 g$

$15.8 g$

Unpredictable

$C u^{2 +} + 2 e^{-} ⟶ C u$
$2 F = 1 m o l$ of $C u = 63.5 g$ of $C u$
$0.5 F = \frac{63.5}{2} \times 0.5 = 15.8 g$

Q. 0.5F of electricity is passed through 500mL of copper sulphate solution. The amount of copper which can be deposited will be

63.5g

31.75g

15.8g