# Disproportionation with aqueous NaOH

Which of the following undergoes disproportionation with $\ce{NaOH}$?

1. $\ce{SiO2}$ (silicon dioxide)
2. $\ce{CO2}$ (carbon dioxide)
3. $\ce{SO2}$ (sulfur dioxide)
4. $\ce{SO3}$ (sulfur trioxide)
5. $\ce{NO2}$ (nitrogen dioxide)

From the above we can conclude that $\ce{SiO2}$, $\ce{CO2}$ and $\ce{SO3}$ do not undergo disproportionation as they are all in the maximum oxidation states. $\ce{NO2}$ is in the +4 oxidation states so it disproportionates in to +3 and +5 upon addition of aqueous $\ce{NaOH}$. But what about $\ce{SO2}$? It can also disproportionates because it is in +4 oxidation no. But the answer states only $\ce{NO2}$. Can anyone please explain the reason?

Disproportionation is a redox reaction. Since $$\Delta G = -zF\Delta E$$ you may compare the redox potentials for both half-reactions in order to determine whether disproportionation occurs spontaneously ($\Delta G < 0$).

Note that nitrogen dioxide ($\ce{NO2}$) exists in equilibrium with dinitrogen tetroxide ($\ce{N2O4}$) $$\ce{2 NO2 <=> N2O4}$$ and literature values for redox potentials are often given for $\ce{N2O4}$ (values at $\mathrm{pH}=14$):

\begin{alignat}{2} \ce{N2O4 + 2 e- \;&<=> 2 NO2- }\quad &&E_1^\circ = +0.867\ \mathrm{V}\\ \ce{2 NO3- + 2 H2O + 2 e- \;&<=> N2O4 + 4 OH-}\quad &&E_2^\circ = -0.86\ \mathrm{V} \end{alignat}

The potential $E_1^\circ = +0.867\ \mathrm{V}$ for the reduction of $\ce{N2O4}$ is larger than the potential $E_2^\circ = -0.86\ \mathrm{V}$ for the oxidation of $\ce{N2O4}$; therefore, disproportionation should occur.

In equlibrium:

\begin{align} E_1^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},1}}{a_{\text{red},1}} &= E_2^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}}\\ E_1^\circ - E_2^\circ &= \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}} - \frac{0.059\ \mathrm{V}}{z}\log \frac{a_{\text{ox},1}}{a_{\text{red},1}} \\ \left(E_1^\circ - E_2^\circ\right) \cdot \frac{z}{0.059\ \mathrm{V}} &= \log \frac{a_{\text{ox},2} \cdot a_{\text{red},1}}{a_{\text{red},2} \cdot a_{\text{ox},1}} \\ \left(0.867\ \mathrm{V} + 0.86\ \mathrm{V}\right) \cdot \frac{2}{0.059\ \mathrm{V}} &= \log\frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{\left[\ce{N2O4}\right] \cdot \left[\ce{N2O4}\right]} \\ {10^{58.5}} &= \frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{{\left[\ce{N2O4}\right]}^2} \\ {10^{29}} &= \frac{\left[\ce{NO3-}\right] \cdot \left[\ce{NO2-}\right]}{{\left[\ce{N2O4}\right]}} \end{align}

Therefore, $\ce{N2O4}$ is unstable with respect to disproportionation: $$\ce{N2O4 + 2 OH- -> NO2- + NO3- + H2O}$$