Question

Three mole of electrons are passed through three solutions in succession containing $AgN{O}_3,CuS{O}_4$ and $AuC{l}_3$ respectively. The ratio of amounts of cations reduced at cathode will be___________.

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

Answer 1

Answer: (D)

$6:3:2$

Faraday's first law:-

The amount of substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.

i) $A{g}^{+}+1e^{-}\to Ag$-----------[reduction at cathode]

1 mole of electron reduces 1 mole of $Ag$

$\to$ 1 mole of $e^{-}\to$ 1 mole of $Ag$.

$\therefore$ 3 moles of $e^{-}\to ?$

$\therefore$ Moles of $Ag$ reduced$=3$.

ii) $C{u}^{2+}+2e\to Cu$.

$\to 2mole{e}^{-}\to 1moleCu$

$\therefore$ 3 moles of $e^{-}\to ?$

$\therefore$ Moles of $Cu$ reduced$=\frac{3}{2}$.

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

$\to$ 3 mole of$e^{-}\to 1moleAu$

$\therefore$ moles of $Au$ reduced$=1$.

$\therefore Ratio=3:\frac{3}{2}:1\to$Converting into whole number$=6:3:2$