Question

If three faradays of electricity is passed through the solution of $AgN{O}_3,CuS{O}_4$ and $AuC{l}_3$, the molar ratio of the cations deposited at the cathodes will be:

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

Answer 1

The correct option is D $6:3:2$

$AgN{O}_3+e^{-}\to Ag+N{O}_3^{-}$
$CuS{O}_4+2e^{-}\to Cu+S{O}_4^{2-}$
$AuC{l}_3+3e^{-}\to Au+3C{l}^{-}$
If three faradays of electricity is passed through the solution of
$AgN{O}_3,CuS{O}_4$ and $AuC{l}_3$
$3$ moles, $\frac{3}{2}$ moles and $1$ mole of the cations will be deposited at the cathodes .
The molar ratio of the cations deposited at the cathodes will be $3:\frac{3}{2}:1$ or
$6:3:2$