Question

If three Faraday of electricity is passed through the solutions of \[AgN{O_3}\], $CuS{O_4}$ and $AuC{l_3}$, the molar ratio of the metals deposited at the cathode will be-
A.$1:1:1$
B.$1:2:3$
C.$3:2:1$
D.$6:3:2$

Answer 1

Hint: When electricity is passed through the solutions then the compounds will dissolve and break into their respective ions so dissolve the ions and break them into their respective ions. Then calculate the number of electrons needed to make the ions convert into metal. Then we know that the equivalent mass of metals deposited at the cathode will be equal to the number of faradays of electricity passed.
So the molar mass will be equal to the number of faraday of electricity passed divided by the number of electrons needed. Then calculate the ratio of the molar masses of silver, copper and gold.

Complete step by step answer:
Copper ion, silver ions and gold ion are deposited at the cathode when electricity is passed through solutions of\[AgN{O_3}\], $CuS{O_4}$ and$AuC{l_3}$.
When electricity is passed through the solutions the compounds break into their respective ions. The reactions are given as follows-
$AgN{O_3} \to A{g^ + } + NO_3^ - $
$CuS{O_4} \to C{u^{2 + }} + SO_4^{2 - }$
$AuC{l_3} \to A{u^{3 + }} + 3C{l^ - }$
So if three faraday of electricity is passed through the solution then the ratio of the metals deposited at the cathode will be in the ratio-
$ \Rightarrow $ Molar mass of silver: Molar mass of Copper: Molar mass of Gold
$ \Rightarrow \dfrac{3}{1}:\dfrac{3}{2}:\dfrac{3}{3}$
On solving, we get-
$ \Rightarrow 3:\dfrac{3}{2}:1$
On multiplying $2$ on each ratio we get-
$ \Rightarrow 6:3:2$

The correct answer is option D.

Note:
We know that when the same quantity of electricity is passed through different solutions then the masses of different metals deposited at respective electrodes is in proportion to their equivalent weights. So we can write-
$ \Rightarrow \dfrac{W}{E} = $ No. of faraday of electricity passed in electrolytes
This is the second law of Faraday’s law of electrolysis.