Question

The amount of metals deposited when $965{\text{ C}}$ of electricity is passed through $NaCl$ and $AgN{O_3}$ solutions are-
A.$23{\text{ g}}$ and $108{\text{ g}}$
B.$11.5{\text{ g}}$ and $54{\text{ g}}$
C.$0.23{\text{ g}}$ and $1.08{\text{ g}}$
D.$2.3{\text{ g}}$ and $10.8{\text{ g}}$

Answer 1

Hint: Use the faraday law of electrolysis according to which the amount of metal that gets deposited is directly proportional to the quantity of electricity passed through the solution. According to Faraday law of electrolysis-
$ \Rightarrow W = \dfrac{{MQ}}{{nF}}$ Where Q= charge, M =molar mass, n is the number of electrons and F is the faraday constant$\left( { = 96500} \right)$. Put the given values in the equation and solve for both sodium and silver to get the answer.

Complete step by step answer:
Given charge passed through solutions Q=$965{\text{ C}}$
Given, solutions are$NaCl$ and $AgN{O_3}$ solutions.
We have to find the amount of metal deposited.
When electricity is passed through the $NaCl$solution, sodium chloride breaks into sodium ion and chloride ion. The sodium ion will gain one electron and get deposited.
$ \Rightarrow N{a^ + } + {e^ - } \to Na$
And its molar mass is given as-
$ \Rightarrow {M_{Na}} = 23$
When electricity is passed through the $AgN{O_3}$ solution, silver nitrate breaks into silver ions and nitrate ions. The silver ion will gain one electron and get deposited.
$ \Rightarrow A{g^ + } + {e^ - } \to Ag$
 And its molar mass is given as-
$ \Rightarrow {M_{Ag}} = 108$
Now we know that from Faraday’s first law of electrolysis the amount of metal that gets deposited is directly proportional to the quantity of electricity passed through the solution. Its formula is given as-
$ \Rightarrow W = \dfrac{{MQ}}{{nF}}$ --- (i)
Where Q= charge, M =molar mass, n is the number of electrons and F is the faraday constant$\left( { = 96500} \right)$
For sodium, M=$23$, Q=$965$ , n=$1$
Then on using the faraday formula and putting given values we get,
$ \Rightarrow {W_{Na}} = \dfrac{{23 \times 965}}{{1 \times 96500}}$
On solving we get,
$ \Rightarrow {W_{Na}} = \dfrac{{23}}{{100}} = 0.23$ g -- (ii)
For silver, M=$108$, Q=$965$ , n=$1$
Then on putting the given values in eq. (i) we get,
$ \Rightarrow {W_{Ag}} = \dfrac{{108 \times 965}}{{1 \times 96500}}$
On solving we get,
$ \Rightarrow {W_{Ag}} = \dfrac{{108}}{{100}} = 1.08$g -- (iii)
So from eq. (ii) and (iii) it is clear that-

The correct answer is C.

Note:
Here the student may mistake the $965{\text{ C}}$to be the current and may try to use the formula-
$ \Rightarrow W = \dfrac{{MIt}}{{nF}}$ Where I is current and t is time {Q=It} but here time is not provided in the given question so you may not be able to solve it.