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Question: Equinormal solutions of two weak acids, HA \(\left( {p{K_a} = 3} \right)\) and HB \(\left( {p{K_a} =...

Equinormal solutions of two weak acids, HA (pKa=3)\left( {p{K_a} = 3} \right) and HB (pKa=5)\left( {p{K_a} = 5} \right) are each placed in contact with equal pressure of hydrogen electrode at25∘C{25^ \circ }C. When a cell is constructed by interconnecting them through a salt bridge, find the emf of the cell.

Explanation

Solution

We know the Hermann Nernst equation is commonly accustomed to calculate the cell potential of a chemical science cell at any given temperature, pressure, and chemical concentration.
Ecell=E0−[RT/nF]lnQ{E_{cell}} = {E_0}-\left[ {RT/nF} \right]lnQ
Where,
Ecell{E_{cell}} = cell potential of the cell
The cell potential below customary conditions is E0{E_0}
R is the universal gas constant
T is temperature
n is amount of electrons transferred within the chemical reaction
Q is the reaction quotient.

Complete step by step solution:
The cell can be represented as,
PtH2(1atm)∣HA2∣∣HA1∣H2(1atm)PtPt{H_2}\left( {1atm} \right)|H{A_2}||H{A_1}|{H_2}\left( {1atm} \right)Pt
At the left hand side,
EH/H+=EOPH/H++0.059log⁡10[H+]2{E_{H/{H^ + }}} = {E_{O{P_{H/{H^ + }}}}} + 0.059{\log _{10}}{\left[ {{H^ + }} \right]_2}
∵−log⁡[H+]=pH\because - \log \left[ {{H^ + }} \right] = pH
EH/H+=EOPH/H++0.059(pH)2{E_{H/{H^ + }}} = {E_{O{P_{H/{H^ + }}}}} + 0.059{\left( {pH} \right)_2}
At the right hand side,
EH/H+= ERPH+/H+0.059log⁡10[H+]1{E_{H/{H^ + }}} = {\text{ }}{E_{R{P_{{H^ + }/H}}}} + 0.059{\log _{10}}{\left[ {{H^ + }} \right]_1}
Thus,EH+/H=ERH+/H−0.059(pH)1{E_{{H^ + }/H}} = {E_{{R_{{H^ + }/H}}}} - 0.059{\left( {pH} \right)_1}
For Acid we know that,
HA1⇔H++A1−H{A_1} \Leftrightarrow {H^ + } + {A_1}^ -
[H+]=KaC\left[ {{H^ + }} \right] = \sqrt {\dfrac{{{K_a}}}{C}}
(pH)1=12pKa1−12log⁡10C{\left( {pH} \right)_1} = \dfrac{1}{2}p{K_{a1}} - \dfrac{1}{2}{\log _{10}}C
(pH)2=12pKa2−12log⁡10C{\left( {pH} \right)_2} = \dfrac{1}{2}p{K_{a2}} - \dfrac{1}{2}{\log _{10}}C
Ecell= EOPH+/H+ERPH+/H {E_{cell}} = {\text{ }}{E_{O{P_{{H^ + }/H}}}} + {E_{R{P_{{H^ + }/H}}}}{\text{ }}
On substituting the values we get,
⇒Ecell=0.059[12pKa2−12pKa1]\Rightarrow {E_{cell}} = {\text{0}}{\text{.059}}\left[ {\dfrac{1}{2}p{K_{a2}} - \dfrac{1}{2}p{K_{a1}}} \right]
⇒Ecell=0.0592[5−3]\Rightarrow {E_{cell}} = \dfrac{{{\text{0}}{\text{.059}}}}{2}\left[ {5 - 3} \right]
⇒Ecell=0.059\Rightarrow {E_{cell}} = {\text{0}}{\text{.059}}

Additional information:
Now we discuss about the uses of Nernst equation as,
The Nernst equation may be accustomed to calculate Single conductor reduction or chemical reaction potential at any conditions.
Standard electrode potentials comparing the relative ability as a subtractive or aerobic agent. Finding the practicability of the mixture of such single electrodes to provide electrical potential. Emf of a chemistry cell Unknown ionic concentrations.
The pHpH scale of answers and solubility of meagerly soluble salts can be measured with the assistance of the Nernst equation.

Note:
Now we discuss about the limitations of Nernst Equation as,
The activity of an ion in an exceedingly} very dilute solution is on the point of eternity and can, therefore, be expressed in terms of the ion concentration. However, for solutions having very high concentrations, the ion concentration isn't up to the particle activity. So as to use the Nernst equation in such cases, experimental livements should be conducted to get actuality activity of the ion. Another defect of this equation is that it can't be accustomed to measure cell potential once there's current flowing through the electrode. This can be as a result of the flow of current that affects the activity of the ions on the surface of the electrode. Also, further factors adore resistive loss and over potential must be thought-about when there is a current flowing through the electrode.