Why chlorate(V) is one of the primary products of disproportionation reaction between chlorine gas and hot concentrated NaOH?

Question

Cold diluted $\ce{NaOH}$ reacts with $\ce{Cl2}$ producing hypochlorite:

$$\ce{2 NaOH(aq, dil, cold) + Cl2(g) -> NaClO(aq) + NaCl(aq) + H2O(l)},\tag{R1}$$

whereas hot concentrated $\ce{NaOH}$ yields chlorate(V):

$$\ce{6 NaOH(aq, conc, hot) + 3 Cl2(g) -> NaClO3(aq) + 5 NaCl(aq) + 3 H2O(l)}\tag{R2}$$

The explanation I gathered is that the reduction potential of $\ce{NaOH}$ increases with temperature, making it a stronger oxidizing agent causing further oxidation of $\ce{Cl}$.

If that is the case, why $\ce{Na\overset{+5}{Cl}O3}$ and not chlorite $\ce{Na\overset{+3}{Cl}O2}$ or perchlorate $\ce{Na\overset{+7}{Cl}O4}$? What is so special about chlorate(V)?

Answer 1

The explanation you have gathered is not correct. $\ce{NaOH}$ is not an oxidizing agent. The conversion of hypochlorite to chlorate is a very complicated reaction so there is no direct formation of chlorate as your direct equation shows. Instead, this is a third order reaction (at least in an electrochemical cell) and if you are seriously interested in the mechanism, start with Wanngård and Wildlock [1] and go to the earlier references therein. I am sure the heat accelerates the collisions of hypochlorite ions to form chlorate ions.

The other part of the question as to why there is no intermediate chlorite is relatively easy, at least from an electrochemical point of view. If you are familiar with Frost diagrams, then it clearly shows chlorite is very unstable with respect to disproportionation. Look at the red curve (for low pH) and blue curve for alkaline pH. On the $x$-axis we have the oxidation number and on the $y$-axis we have the free energy. Lower the value, more stable the species is—thermodynamically. Chlorite ion is very slightly raised in that red curve as compared to its adjacent neighbors, it means that it is thermodynamically unstable and chlorine would like to stay either as hypochlorite and chlorate but not as chlorite. Similarly, the highest oxidation state species is on the extreme right, which means that is also less stable than chlorate.

It is then obvious that there are relatively two choices for chlorine atoms—stay as chloride ions which it does all the time (sea water for millions of years is still chloride) but under oxidizing conditions, the next stable choice is hypochlorite, and then under even more harsh oxidizing conditions, chlorate but not chlorite.

Figure 5.18 A Frost diagram for chlorine. The red line refers to acid conditions $(\mathrm{pH} = 0)$ and the blue line to $\mathrm{pH} = 14.$ Note that because $\ce{HClO3}$ and $\ce{HClO4}$ are strong acids, even at $\mathrm{pH} = 0$ they are present as their conjugate bases.

Image source: [2, p. 185].

Bonus trivia: Mars' soil has huge amounts of perchlorate salts and this was discovered by an electrochemical sensor's mistake sent to Mars. It was sent to detect nitrates but nobody checked that the same sensor also responds to perchlorate. So natural formation of even perchlorate is not out of the blue. It is just that Earth's atmosphere is not too oxidizing... otherwise if there were no atmosphere, all sea salts would have met the same fate as Mars. Solar radiation would have fried seasalt to perchlorate.

References

1. Wanngård, J.; Wildlock, M. The Catalyzing Effect of Chromate in the Chlorate Formation Reaction. Chem Eng Res Des 2017, 121, 438–447. DOI: 10.1016/j.cherd.2017.03.021.
2. Shriver, D.; Weller, M.; Overton, T.; Armstrong, F.; Rourke, J. Inorganic Chemistry, 6th ed.; W. H. Freeman: Oxford, 2014. ISBN 978-0-19-875717-7.

Answer 2

The following is an addon to the existing answer. Frost diagram is indeed a simple and intuitive visual aid helping to identify which oxidation states of an element are most stable in solution. However, the textbook Frost diagrams, like the majority of thermodynamic data, are usually reported for 25 °C. Since OP is asking about boiling hot solution, we need to compile the Frost diagram for that ourselves.

The standard reduction potentials alongside with their temperature coefficients $\mathrm dE^\circ/\mathrm dT$ have been collected and reported by Bratsch [1]. The temperature dependence of reduction potentials is approximately linear between $\pu{0 °C}$ and $\pu{100 °C}:$

$$E^\circ_T = E^\circ_{298} + (T - 298.15)\left(\frac{\mathrm dE^\circ}{\mathrm dT}\right)_{298}$$

For the further reference I made a tableau with calculated $(\pu{60 °C}$ to $\pu{100 °C})$ potentials for the range of elevated temperatures for the half-reactions of interest:

\begin{array}{llrrrrr} \hline & \left(\frac{\mathrm dE^\circ}{\mathrm dT}\right)_{298} & & & & & \\ \text{Half-reaction} & /\pu{mV K^-1} & E^\circ_{298}/\pu{V} & E^\circ_{333}/\pu{V} & E^\circ_{343}/\pu{V} & E^\circ_{353}/\pu{V} & E^\circ_{363}/\pu{V} & E^\circ_{373}/\pu{V} \\ \hline \ce{Cl2(g)}/\ce{Cl-} & -1.248 & 1.360 & 1.404 & 1.417 & 1.429 & 1.442 & 1.454 \\ \ce{ClO^-}/\ce{Cl2(g),OH-} & -0.90 & 0.420 & 0.389 & 0.380 & 0.371 & 0.362 & 0.353 \\ \ce{ClO2^-}/\ce{Cl^-,OH-} & -1.267 & 0.786 & 0.742 & 0.729 & 0.716 & 0.704 & 0.691 \\ \ce{ClO3-}/\ce{Cl2(g),OH-} & -1.35 & 0.465 & 0.418 & 0.404 & 0.391 & 0.377 & 0.364 \\ \ce{ClO4-}/\ce{Cl2(g),OH-} & -1.322 & 0.446 & 0.399 & 0.385 & 0.372 & 0.358 & 0.345 \\ \hline \end{array}

Using this, one can make a Latimer diagram for the oxidation numbers (O.N.s) of interest, which I'm going to make in a tabular form for $\pu{25 °C}$ and $\pu{100 °C}:$

\begin{array}{rrr} \hline \text{O.N.} & E^\circ_{298}/\pu{V} & E^\circ_{373}/\pu{V} \\ \hline -1 & -1.360 & -1.267 \\ 0 & 0.000 & 0.000 \\ 1 & 0.420 & 0.353 \\ 3 & 2.146 & 1.958 \\ 5 & 2.325 & 1.819 \\ 7 & 3.122 & 2.413 \\ \hline \end{array}

Plotting this data gives a Frost diagram for chlorine at room temperature and boiling hot solution. Note a more pronounced local minimum for O.N. 5 for the hot solution, which implies tendency of both chlorite and perchlorate to comproportionate forming chlorate $\ce{ClO3-}$ as the more thermodynamically stable oxyanion. Chlorite always lies above the line connecting two adjacent species and would tend to undergo disproportionation at room temperature and much faster (negative slope) in hot solution.

Reference

1. Bratsch, S. G. Standard Electrode Potentials and Temperature Coefficients in Water at 298.15 K. J Phys Chem Ref Data 1989, 18 (1), 1–21. DOI: 10.1063/1.555839.

Answer 3

Latimer Diagrams may also helpful to find unstable species with respect to disproportionation. I'd list Latimer Diagrams for chlorine in acidic and basic solutions given in here:

$$\text{In acid: } \ce{ClO4- ->[\pu{1.19 V}] ClO3- ->[\pu{1.21 V}] HClO2 ->[\pu{1.65 V}] HClO ->[\pu{1.63 V}] Cl2 ->[\pu{1.36 V}] Cl-}$$ $$\text{In base: } \ce{ClO4- ->[\pu{0.36 V}] ClO3- ->[\pu{0.35 V}] ClO2- ->[\pu{0.65 V}] ClO- ->[\pu{0.40 V}] Cl2 ->[\pu{1.36 V}] Cl-}$$

Accordingly, Latimer diagrams provide a very easy way to determine if disproportionation is spontaneous: If the voltage to the right of the species in question is greater than the voltage to the left of the species, the species is unstable with respect to disproportionation.

Thus, $\ce{ClO2-}$ is not stable with respect to disproportionation in basic medium as $\pu{0.65 V} \ \gt \ \pu{0.35 V}$.