Question

Three moles of electrons are passed through three solutions containing $CuS{O}_4$, $AgN{O}_3$ and $AuC{l}_3$ respectively. The ratio of moles of cations reduced at cathode will be :

• A. 6 : 3 : 2
• B. 2 : 1 : 3
• C. 3 : 6 : 2
• D. 3 : 1 : 2

Answer 1

Answer: (C)

3 : 6 : 2

$C{u}^{2+}+2e^{-}\to Cu$

3 moles of electrons $⟹$$\frac{3}{2}$ moles Cu.
$A{g}^{+}+e^{-}\to Ag$

3 moles of electrons $⟹$ 3 moles Ag.

$A{u}^{3+}+3e^{-}\to Au$

3 moles of electrons $⟹$ 1 mole Au.

Ratio of moles of Cu, Ag and Au $=3/2:3:1=3:6:2$