Question

If three Faraday of electricity is passed through the solutions of $Ag{NO}_3,Cu{SO}_4$ and $Au{Cl}_3$, the molar ratio of the metals deposited at the cathode will be:

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

Answer 1

Answer: (B)

$Cu$ metal deposited as ${Cu}^{2+}$ ion

$Ag$ metal deposited as ${Ag}^{+}$ ion

$Au$ metal deposited as ${Au}^{3+}$ ion

$\therefore$ to form 1 atom of $Cu$, $Ag$ and $Au$

$\downarrow$ $\downarrow$ $\downarrow$

${2e}^{-}$ ${1e}^{-}$ ${3e}^{-}$

if 3 mol Faraday of electricity is passed,

then

1 mol of $Au$, 3 mol of $Ag$ and 3/2 mol of $Cu$

i.e. $Ag,Cu,Au$

$3:\frac{3}{2}:1$

$6:3:2$