Question

Given,$N{O_3} \to N{O_2}$(acid medium), $E = 0.790V$
$N{O_3} \to N{H_3}O{H^ - }$(acid medium), $E = 0.731V$
At what $pH$, the above two will have the same $E$ value?
Assume the concentration of all other species $N{H_3}OH$ except $\left[ {{H^ - }} \right]$ to be unity.

Answer 1

Hint: Nernst Equation is used to calculate Electrode potential or EMF of a cell under non-standard conditions. It is a very important equation in Electrochemistry. This equation was given by German physicist Walther Nernst. This equation also helps us in calculating the accurate value of equilibrium constant as well as relating cell potential and reaction quotient.

Complete answer:
Let us consider an electrochemical reaction-
$aA + bB \rightleftharpoons cC + dD - - - - (1)$
The Nernst equation is given as follows-
${E_{cell}} = E{^\circ _{cell}} - \dfrac{{RT}}{{nF}}\ln \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}} - - - - (2)$
Equation (2) can also be written as-
${E_{cell}} = E{^\circ _{cell}} - \dfrac{{RT}}{{nF}}\ln Q - - - - (3)$
Where ${E_{cell}}$ is Electrode potential of cell
$E{^\circ _{cell}}$ is Standard electrode potential that is calculated as $E{^\circ _{cell}} = E{^\circ _{Cathode}} - E{^\circ _{Anode}}$
$Q$ is the reaction quotient of the cell given as $\dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$
$T$ is temperature in Kelvin
$R$ is a gas constant whose value is $8.314J/moleK$.
$n$ is the number of electrons transferred in the balanced reaction
$F$ is Faraday’s constant $(1F = 96500C/mole)$
Writing Equation (3) in the form of ${\log _{10}}$we have-
${E_{cell}} = E{^\circ _{cell}} - \dfrac{{2.303RT}}{{nF}}{\log _{10}}Q - - - - (4)$
At standard conditions of temperature $T = 298K$we can rewrite equation as-
${E_{cell}} = E{^\circ _{cell}} - \dfrac{{0.0591}}{n}{\log _{10}}Q - - - - (5)$
In order to solve the question first we need to write balanced chemical Equations in Acidic medium-
$\begin{gathered}
  N{O_3}^ - + 2{H^ + } + {e^ - } \to N{O_2} + 2{H_2}O - - - - (6) \\
  N{O_3}^ - + 8{H^ + } + 6{e^ - } \to N{H_3}O{H^ + } + 2{H_2}O - - - - (7) \\
\end{gathered} $
In the question it is specified to take all the species concentration except $\left[ {{H^ + }} \right]$ to be unity.
Now we use Equation (5) to solve the question-
For the Equation (6) we can write Nernst equation as under-
$E = 0.790 - \dfrac{{0.0591}}{1}\log \dfrac{1}{{{{\left[ {{H^ + }} \right]}^2}}} - - - - (8)$ (here value of $n = 1$)$$
For the Equation (7) we can write Nernst equation as follows-
$E = 0.731 - \dfrac{{0.0591}}{6}\log \dfrac{1}{{{{\left[ {{H^ + }} \right]}^8}}} - - - - (9)$(here value of $n = 6$)
We are given that both the half cells have the same EMF values so we will equate Equation (8) and (9).
$0.790 - \dfrac{{0.0591}}{1}\log \dfrac{1}{{{{\left[ {{H^ + }} \right]}^2}}} = 0.731 - \dfrac{{0.0591}}{6}\log \dfrac{1}{{{{\left[ {{H^ + }} \right]}^8}}}$
$ - \log \left[ {{H^ + }} \right] = 1.5$
We know $pH = - \log \left[ {{H^ + }} \right]$
Therefore the value of $pH = 1.5$

Note:
Nernst equation is very useful for calculation of not only the EMF but also the pH, Solubility product, Equilibrium constant, Thermodynamic properties like Enthalpy and also in Potentiometric titrations. While solving questions on the Nernst equation it is important to write the balanced equation first and find the reaction quotient and the number of electrons required in that reaction.