Question

${NO}_3^{-}\to {NO}_2$ (acidic medium): $E^o=0.79V$ and ${NO}_3^{-}\to {NH}_{2}OH$ (Acidic medium). $E^o=0.731V$. At what pH the above two half reactions will have same $E$ values. Assume the concentration of all the species to be unity.

Answer 1

For the first equation

$N{O}_3^{-}\to N{O}_2$

Balancing O atoms of above equation

$N{O}_3^{-}\to N{O}_2+H_{2}O$

Balancing H atoms

$N{O}_3^{-}+2H^{+}\to N{O}_2+H_{2}O$

Balancing charges

$N{O}_3^{-}+2H^{+}+e^{-}\to N{O}_2+H_{2}O$

For the second equation

$N{O}_3^{-}\to N{H}_{3}O{H}^{+}$

$N{O}_3^{-}+8H^{+}+6e^{-}\to N{H}_{3}O{H}^{+}+2H_{2}O$

Using Nernst equation

$E=E_0-\frac{0.059}{n}lnQ$

where n is the no. of electrons

For first equation

$Q-\frac{\left[N{O}_2\right]\left[H_{2}O\right]}{\left[N{O}_3^{-}\right][H^{+}{]}^2}$ (CONCENTRATION OF ALL THE SPECIES IS UNITY)

$Q=\frac{1\times 1}{1\times [H^{+}{]}^2}$

$Q=\frac{1}{[H^{+}{]}^2}$

$E=E_0-0.06log\frac{1}{[H^{+}{]}^2}$

$E=0.79-0.06log\frac{1}{[H^{+}{]}^2}$ ( EQ 1)

Similarly for second eqation

$E=E_0-\frac{0.06}{6}log\frac{1}{[H^{+}{]}^8}$

$E=0.73-\frac{0.06}{6}log\frac{1}{[H^{+}{]}^8}$ (EQ 2)

As E should be same $(EQ1=EQ2)$

$0.79-0.06log\frac{1}{[H^{+}{]}^2}$$=$$0.73-\frac{0.06}{6}log\frac{1}{[H^{+}{]}^8}$

$0.79-0.73=0.06log\frac{1}{[H^{+}{]}^2}-\frac{0.06}{6}log\frac{1}{[H^{+}{]}^8}$

$0.06=-0.06\times 2log\left[H^{+}\right]+\frac{0.06\times 8}{6}log\left[H^{+}\right]$

$1=(\frac{8}{6}-2)log\left[H^{+}\right]$

$\frac{3}{2}=-log\left[H^{+}\right]$

$pH=-log\left[H^{+}\right]$

$pH=\frac{3}{2}=1.5$