Question

The e.m.f of the cell pt,${H_2}$ ( $1$ atm)|$HA$ ( $0.01$ )||$HCL$ ( $1.0$ M)|${H_2}$ ( $1$ atm) Pt is $0.295$ V.Thee dissocial constant of acid $HA$ is
1) $1.0 \times {10^{ - 4}}$
2) $1.0 \times {10^{ - 6}}$
3) $1.0 \times {10^{ - 8}}$
4) $1.0 \times {10^{ - 5}}$

Answer 1

Hint: First of all we will write Nernst equation. We will write the reaction of anode and cathode separately. Then we will add them both to get the complete cell equation. Then we will try to find the value of concentration from the cell equation and then we will find the value of ${K_a}$ by using the formula $[{H^ + }] = {H^ + } + {A^ - }$

Complete Step by step answer: First lets write the anodic and cathodic reaction separately
Reaction at anode:
${H_2} \to 2{H^ + } + 2{e^ - }$
Now we will write equation of cathode
$2{H^ + } + 2{e^ - } \to {H_2}$
Now let’s sum this reaction to get the cell reaction
${H_2} \to 2{H^ + } + 2{e^ - }$
$2{H^ + } + 2{e^ - } \to {H_2}$
---------------
${H^ + } \to {H^ + }$
${E_{cell}} = {E^0}_{cell} - \dfrac{{0.059}}{n}\log (\dfrac{{anode}}{{cathode}})$
Now using this formula we have $[C]anode = [C]cathode$
$[C]cathode = 1$ M
Here n=1 electron
${E_{cell}} = 0.295$ v
 So by formula we have,
$0.295 = 0 - \dfrac{{0.059}}{1}\log \dfrac{{[C]}}{1}$
$ \Rightarrow \dfrac{{0.295}}{{0.059}} = \log [C]$
We have $[C] = {10^{ - 5}}$
Now we got the concentration as ${10^{ - 5}}$
We have $HA \to {H^ + } + {A^ - }$
So $[{H^ + }] = \sqrt {c{k_a}} $
 So ${10^{ - 5}} = \sqrt {0.01{k_a}} $
$ \Rightarrow \dfrac{{{{10}^{ - 10}}}}{{{{10}^{ - 2}}}} = {k_a} = {10^{ - 8}}$

Hence, answer is option 3.

Additional information: Electrolysis: This is a process of decomposing ionic compounds into their elements by passing a direct current through the compound in a fluid form.
Anode: In electrolysis an anode is an electrode through which the conventional current enters into the electrical device.
Cathode: It is an electrode through which the conventional current leaves into the electrical device.

Note: while calculating the Nernst equation we should take care of sign and we should calculate the value carefully because there are chances of getting the wrong answer. The log table can be used wherever necessary.