Question

Current of i ampere was passed for t sec, through three cells P,Q and T connected in series. These contain respectively silver nitrate, mercuric nitrate, and mercurous nitrate. At the cathode of the cell P 0.216g of Ag was deposited. The weights of mercury deposited in the cathode of $Q$ and $R$ respectively are: ( at wt. of Hg: 200.59g)
A. 0.4012 and 0.8024g
B. 0.4012 and 0.2006g
C. 0.2006 and 0.4012g
D. 0.1003 and 0.2006g

Answer 1

We know that electrolytic cells use electrical energy to get a non-spontaneous redox reaction. An electrolytic cell is a type of electrochemical cell. It is mentioned that the cells are connected in series. This gives us the information that the same amount of current is passing through all the electrolytic cells in the given situation. This would be helpful while solving.

Formula used: Faraday’s second Law
$\dfrac{W_1}{W_2} = \dfrac{E_1}{E_2}$
Where, ${E_1}$ and ${E_2}$ are the equivalent weights of the two electrolytes
And ${W_1}$ and ${W_2}$ are the mass of the substances deposited at the electrolytes
We also know that $E = \dfrac {\text {atomic weight}} {\text {valency of the atom (its combining power)}}$.

Complete step by step answer:
The amount of silver deposited at $P$ electrode is already given to us, we will use this to calculate the amount of mercury deposited at electrode $Q$ and $R$ .
At electrode $Q$ For Mercuric nitrate $(H{g^{2 + }})$
$\dfrac{W_1}{W_2}(Ag) = \dfrac{E_1}{E_2}(Ag)$
On rearranging, $W_1 = \dfrac{E_1}{E_2}( Ag) \times W_2(Ag)$
$\Rightarrow W_1 = \dfrac{200.59}{108\times 2} \times 0.26$
On solving we get the answer ${W_1}{{ }} = {{ }}0.2006g$
At electrode $R$ For mercurous nitrate $(H{g^ + })\;$
$\dfrac{W_1}{W_2}(Ag) = \dfrac{E_1}{E_2}(Ag)$
we know that E for Ag = 108
On rearranging, $W_1 = \dfrac{E_1}{E_2}( Ag) \times W_2(Ag)$
$\Rightarrow W_1 = \dfrac{200.59}{108} \times 0.26$
On solving we get the answer ${W_1}{{ }} = {{ }}0.4012g$

So, Option C is correct.

Additional Information:
Faraday’s second law tells us that the amount of substances that will get deposited by the passing of the same amount of electric current will be directly proportional to their equivalent weights.
This law is particularly useful while calculating the amount of reaction or mass involved for different substances for the passage of same quantity of current

Note: One need to keep in mind the valence while calculating equivalent weight, for which we should know the valency of the atom in the compound, for example here in this question the valency of mercury was 1and 2 in mercurous nitrate and mercuric nitrate, respectively.