Question

During electrolysis of a solution of$AgN{{O}_{3}}$, 9650 coulomb of charge pass through the electrolytic cell; the mass of silver deposited on the cathode will be:
(A).21.6 g
(B).108 g
(C).10.8 g
(D).1.08 g

Answer 1

For finding the deposited mass of silver on the cathode we have to write the cathode reaction and anode reaction. And we have to find out the charge transfer.

Complete step by step solution:
We know that electrolysis, the process by which electric current is passed through a substance to effect a chemical change. This process is carried out in an electrolytic cell, and apparatus consisting of positive and negative electrodes held apart and dipped into a solution containing positively and negatively charged ions.
Here, the half cell reaction of silver (Ag) is:
$A{{g}^{+}}+1{{e}^{-}}\to A{{g}_{(s)}}$
Mole ratio = (moles of Ag formed) / (moles of electrons transferred)
= (1 / 1) =1
We know that moles of Ag = $\dfrac{Q(c)}{96500Cmo{{l}^{-1}}{{e}^{-}}}\times moleratio$
Q(c) = charge pass through the electrolytic cell
Given, Q(c ) = 9650 C
Now we will put all the data given and calculated in the above equation:
Moles of Ag = $\dfrac{Q(c)}{96500Cmo{{l}^{-1}}{{e}^{-}}}\times moleratio$
= $\dfrac{9650C}{96500Cmo{{l}^{-1}}{{e}^{-}}}\times 1$
$\dfrac{Weight}{molecularWeight}$= $\dfrac{9650C}{96500Cmo{{l}^{-1}}{{e}^{-}}}\times 1$
And, molecular weight of Ag = 108$gmo{{l}^{-}}$
By rearranging the equation:
Weight = $\dfrac{9650\times 108}{96500}$
= 10.8 g
So, the mass of silver deposited = 10.8 g
Hence the correct answer is option “C”.

Note:
Here there is no need to write the complete cell equation, because we have to find only the weight of the silver (Ag). Number of electron transfers is equal to the change in the oxidation state.