a
If the partial pressure of \(NO\) and \(NO _{2}\) gas is taken as \(1\) bar, then Answer is \(4,\) else the question is bonus.

\(NO _{3}^{-}+4 H ^{+}+3 e ^{-} \longrightarrow NO +2 H _{2} O\)

\(\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad E _{ NO _{3} / NO }^{\circ}=0.96 \,V\)

\(NO _{3}^{-}+2 H ^{+}+ e ^{-} \longrightarrow NO _{2}+ H _{2} O\)

\(\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad E _{ NO _{3}^{-} / NO _{2}=0.79}^{\circ}\)

Let \(\left[ HNO _{3}\right]= y \Rightarrow\left[ H ^{+}\right]= y\) and \(\left[ NO _{3}^{-}\right]= y\) for same thermodynamic tendency

\(E _{ NO _{3}^{-} / NO }= E _{ NO _{3}^{-} / NO _{2}}\)

or, \(E _{ NO _{3}^{-} / NO }^{\circ}-\frac{0.059}{3} \log \frac{ P _{ NO }}{ y \times y ^{4}}\)

\(= E _{ NO _{3}^{-} / NO _{2}}^{\circ}-\frac{0.059}{1} \log \frac{ P _{ NO _{2}}}{ y \times y ^{2}}\)

or, \(0.96-\frac{0.059}{3} \log \frac{P_{ NO }}{ y ^{5}}=0.79-\frac{0.059}{1} \log \frac{ P _{ NO _{2}}}{ y ^{3}}\)

or, \(0.17=-\frac{0.059}{1} \log \frac{ P _{ NO _{2}}}{ y ^{3}}+\frac{0.059}{3} \log \frac{ P _{ NO }}{ y ^{5}}\)

\(0.17=-\frac{0.0591}{1} \log \frac{ P _{ NO _{2}}}{ y ^{3}}+\frac{0.0591}{3} \log \frac{ P _{ NO }}{ y ^{5}}\)

\(0.17=-\frac{0.0591}{3} \log \frac{ P _{ NO _{2}}^{3}}{ y ^{9}}+\frac{0.0591}{3} \log \frac{ P _{ NO }}{ y ^{5}}\)

\(0.17=\frac{0.0591}{3}\left[\log \frac{ P _{ NO }}{ y ^{5}}-\log \frac{ P _{ NO _{2}}^{3}}{ y ^{9}}\right]\)

\(0.17=\frac{0.0591}{3}\left[\log \frac{ P _{ NO }}{ y ^{5}} \times \frac{ y ^{9}}{ P _{ NO _{2}}^{3}}\right]\)

Assume \(P _{ No } \simeq P _{ NO _{2}}=1\) \(bar\)

\(\frac{0.17 \times 3}{0.059}=\log y ^{4}=8.644\)

\(\log y =\frac{8.644}{4}\)

\(\log y =2.161\)

\(y =10^{2.16}\)

\(\therefore 2 x =2 \times 2.161=4.322\)