Question

Passage of three faradays of charge through an aqueous solution of $AgN{O}_3,CuS{O}_4,Al(N{O}_3{)}_3\text{and}NaCl$ will deposit metals at the cathode in the molar ratio of :

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

Answer 1

The correct option is B 6 : 3 : 2 : 6

Electrolysis of $aq.AgN{O}_3$
$AgN{O}_3\left(aq\right)+e^{-}\to Ag\left(s\right)+N{O}_3^{-}\left(aq\right)$
Thus,
1 mole of$AgN{O}_3$ requires 1 mole of $e^{-}$ or 1 Faraday of electricity to produce one mole of $Ag$.
Hence, $3F$ is used to produce 3 mole of $Ag$.

Electrolysis of $aq.CuS{O}_4$
$CuS{O}_4\left(aq\right)+2e^{-}\to Cu\left(s\right)+S{O}_4^{2-}\left(aq\right)$
1 mole of $CuS{O}_4$ requires 2 Faraday of Charge to produce 1 mole of $Cu$
Hence, $3F$ is used to produce $\frac{3}{2}$ mole of $Cu$.

Electrolysis of $aq.Al(N{O}_3{)}_3$
$AlN{O}_3\left(aq\right)+3e^{-}\to Al\left(s\right)+3N{O}_3^{-}\left(aq\right)$
1 mole of$Al(N{O}_3{)}_3$ requires 3 mole of $e^{-}$ or 3 Faraday of electricity to produce one mole of $Al$.
Hence, $3F$ is used to produce 1 mole of $Al$.

Electrolysis of $aq.NaCl$
$NaCl\left(aq\right)+e^{-}\to Na\left(s\right)+C{l}^{-}\left(aq\right)$
1 mole of $NaCl$ requires 1 mole of $e^{-}$ or 1 Faraday of electricity to produce one mole of $Na$.
Hence, $3F$ is used to produce 3 mole of $Na$.

The molar ratio will be :
$3:\frac{3}{2}:1:3$
or
$6:3:2:6$