Question

(a) How many electrons flow through a metallic wire if a current of 0.5 ampere is passed for 2 hrs? (Given 1F = 96500 C/mol)
(b) Give the products obtained at cathode and anode during electrolysis of $AgN{O}_3$ solution using $Ag$ electrodes.

Answer 1

Solution:-

(a) As we know that,

$q=i\times t$

Given:-

$i=0.5A$

$t=2hrs=2\times 60\times 60=7200sec$

$\therefore q=0.5\times 7200=3600C$

Now,

No. of electrons in $96500C=6.023\times {10}^{23}$

Therefore,

No. of electrons in $3600C=\frac{6.023\times {10}^{23}}{96500}\times 3600=2.25\times {10}^{22}{e}^{-}s$

Hencee the no. of electrons flow through a metallic wire is $2.25\times {10}^{22}$.

(b) At cathode: The following reduction reactions compete to take place at the cathode.

${Ag}^{+}+e^{-}\stackrel{A}{\to }g$

$H^{+}+e^{-}⟶\frac{1}{2}{H}_2$

The reaction with a higher value of $E^{\circ }$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode: The $Ag$ anode is attacked by ${N{O}_3}^{-}$ ions. Therefore, the silver electrode at the anode dissolves in the solution to form ${Ag}^{+}$.

$Ag⟶{Ag}^{+}+e^{-}$