0. 5F of electricity is passed through 500mL of copper sulphate...

${Cu}^{2 +} + 2e^{-} ⟶ Cu... \left(i\right)$

According to Faraday's first law, when an electric current is passed through an electrolyte, amount of substance deposited is directly proportional to the quantity of electric charge passed through the electrolyte.

$∵ 1mol$ of $63. 5g$ of $Cu$ is obtained by passing $2F$ of electricity.

Mass of $Cu$ obtained by passing $1F$ of electricity $= \dfrac{63. 5}{2}$ g$</p><p>$∴$Mass of$Cu$obtained by passing 0. 5F$ of electricity
$= \dfrac{63. 5}{2} \times0. 5 = 15. 80g$