Question

An electrolytic cell contains a solution of $A{g_2}S{O_4}$ and has platinum electrodes. A current is passed until $1.6g$ of ${O_2}$ has been liberated at anode. The amount of silver deposited at cathode will be
A. $108g$
B. $1.6g$
C.$0.8g$
D.$21.60g$

Answer 1

Hint: An electrolytic cell, which is an electrochemical device that uses electrical energy, aids a non-spontaneous redox reaction. Electrolytic cells are electrochemical cells that can electrolyze a wide range of materials.

Formula used:
\[ \text{Number of Equivalent} = \dfrac{\text{Given Mass}}{\text{Equivalent mass}} \]

Complete answer:
We know that the number equivalent of ${O_2}$ liberated is equal to the number equivalent of $Ag$ deposited.
So, number of equivalent of ${O_2}$can be calculated using the above formula,
Substituting the values in above formula we get ,
Number of equivalent of ${O_2}$$ = \dfrac{{1.6}}{8} = 0.2$
Number of equivalent of $Ag$$ = \dfrac{{mass}}{{108}} = 0.2$
Hence , the mass of silver can be calculated by ,
Mass of $Ag = 108 \times 0.2 = 21.6g$
Hence, the amount of silver deposited at cathode will be D.$21.60g$

Hence, the correct option is D.$21.60g$.

Additional Information:
Electrolytic cells can be used to generate oxygen and hydrogen gas from water by electrolyzing it. These devices can also extract chlorine gas and metallic sodium from sodium chloride aqueous solutions (common salt). Electroplating is another prominent application of electrolytic cells.

Note:
A cation or neutral molecule is reduced at the cathode and an anion or neutral molecule is oxidised at the anode when an electrical current passes through a solution (typically of electrolytes).