The reduction potentials of perchlorate, chlorate, chlorous acid and chlorine dioxide in acidic and basic solutions are listed in the table below: $$ \begin{array}{lcc} \hline \text{Oxychloride} & E^\circ_\text{acidic}/\pu{V} & E^\circ_\text{basic}/\pu{V} \\ \hline \ce{ClO4-} & 1.19 & 0.56 \\ \ce{ClO3-} & 1.21 & 0.63 \\ \ce{HClO2} & 1.65 & 0.78 \\ \ce{ClO2} & 1.63 & 0.89 \\ \hline \ce{} \end{array} $$

Why is $\ce{ClO-}$ such a strong oxidizing agent compared to other chloride oxyanions, despite its low oxidation number? Is oxidation number representative of oxidizing strength?

  • $\begingroup$ The TLDR here is that you provided data out of context, but the context matters. Reduction potentials apply in the context of a half reaction, not a single oxidant. In the series you have provided, there's no way to tell if this series has any kind of homology because in essence, you can't tell how much each species is reduced. $\endgroup$
    – Zhe
    Jul 7, 2019 at 19:19

2 Answers 2


First, hypochlorite ($\ce{ClO−}$) is not a strong oxidizing agent compared to other chloride oxyanions, at least in acidic medium. Under standard conditions in acidic medium, chlorous acid ($\ce{HClO2}$) is the best oxidizing agent among them due to the largest positive number for standard electrode potential:

$$\begin{align} \ce{HClO2 + 2H+ + 2 e- &<=> HClO + H2O} &\quad E^\circ &= \pu{1.645 V} \label{rxn:1}\tag{1}\\ \ce{HClO2 + 3 H+ + 3 e- &<=> 1/2 Cl2 + 2 H2O} &\quad E^\circ &= \pu{1.628 V} \label{rxn:2}\tag{2}\\ \ce{HClO2 + 3 H+ + 4 e- &<=> Cl– + 2 H2O} &\quad E^\circ &= \pu{1.570 V} \label{rxn:3}\tag{3}\\ \end{align}$$

Keep in mind that in the reduction half-reaction \eqref{rxn:1}, $\ce{HClO2}$ reduced to $\ce{HClO}$ with its largest positive standard electrode potential $(E^\circ = \pu{1.645 V})$ among chlorine oxoacids. For comparison, we should look at half-reactions of which each of these two reagents reducing to the same product. Thus, look at half-reactions \eqref{rxn:2} versus \eqref{rxn:4}, and \eqref{rxn:3} versus \eqref{rxn:5} where $\ce{HClO2}$ shows its superiority over $\ce{HClO}$:

$$\begin{align} \ce{HClO + H+ + e- &<=> 1/2 Cl2 + H2O} &\quad E^\circ &= \pu{1.611 V} \label{rxn:4}\tag{4}\\ \ce{HClO + H+ + 2 e- &<=> Cl– + H2O} &\quad E^\circ &= \pu{1.482 V} \label{rxn:5}\tag{5}\\ \end{align}$$

Perchlorate ($\ce{ClO4-}$) and chlorate ($\ce{ClO3-}$) ions, on the other hand, are also good oxidizing agents, but not as strong as $\ce{HClO2}$ or $\ce{HClO}$, yet closed. This fact is clear when compared to the half-reactions where they directly reduce to $\ce{Cl2}$ or $\ce{Cl-}$ similar to $\ce{HClO2}$ and $\ce{HClO}$:

$$\begin{align} \ce{ClO4- + 8H+ + 7e- &<=> 1/2 Cl2 + 4H2O} &\quad E^\circ &= \pu{1.39 V} \label{rxn:6}\tag{6}\\ \ce{ClO4- + 8H+ + 8e- &<=> Cl– + 4H2O} &\quad E^\circ &= \pu{1.389 V} \label{rxn:7}\tag{7}\\ \ce{ClO3- + 6H+ + 5e- &<=> 1/2 Cl2 + 3H2O} &\quad E^\circ &= \pu{1.47 V} \label{rxn:8}\tag{8}\\ \ce{ClO3- + 6H+ + 6e- &<=> Cl– + 3H2O} &\quad E^\circ &= \pu{1.451 V} \label{rxn:9}\tag{9}\\ \end{align}$$

However, the standard reduction potentials OP is mentioning is for different reduction half-reactions $\ce{ClO3-}$ and $\ce{ClO4-}$ involved (\eqref{rxn:12} and \eqref{rxn:10}, respectively):

$$\begin{align} \ce{ClO4- + 2H+ + 2e- &<=> HClO3 + H2O} &\quad E^\circ &= \pu{1.189 V} \label{rxn:10}\tag{10}\\ \ce{ClO3- + 2H+ + e- &<=> ClO2 + H2O} &\quad E^\circ &= \pu{1.152 V} \label{rxn:11}\tag{11}\\ \ce{ClO3- + 3H+ + 2e- &<=> HClO2 + H2O} &\quad E^\circ &= \pu{1.214 V} \label{rxn:12}\tag{12}\\ \end{align}$$

Accordingly, when consider Latimer diagrams $(\ce{ClO4- -> ClO3- -> HClO2 -> HClO -> Cl2 -> Cl-}),$ you'd see, $\ce{HClO2}$ cannot oxidizes $\ce{ClO3-},$ although it is the strongest oxidizing agent. However, $\ce{HClO2}$ can oxidize another species, which $\ce{ClO3-}$ cannot oxidize.

Note that all reduction potential values are from: http://depa.fquim.unam.mx/amyd/archivero/TablasdepotencialesREDOX_26700.pdf (Petr Vanẏsek), which cited the following references:

  1. G. Milazzo, S. Caroli, V. K. Sharma, Tables of Standard Electrode Potentials; Wiley, Chichester, 1978.
  2. A. J. Bard, R. Parsons, and J. Jordan, Standard Potentials in Aqueous Solutions; Marcel Dekker, New York, 1985 (Series: Monographs in electroanalytical chemistry and electrochemistry).
  3. S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” J. Phys. Chem. Ref. Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).

Besides the the thermodynamic aspects discussed by Matthew, consider the kinetic aspects. Oxidation by a chlorine oxyanion involves displacement of oxygen from its bond with the chlorine. Such a displacement, in a protic solvent such as water, must involve protonation of the oxygen:

  • Oxygen without the proton would have to be displaced as oxide ion, which would be a high-energy intermediate in any protic solvent.

  • In this article from the University of Utah[1], molecular orbitals for some phosphorous-oxygen species are discussed. These show that oxygen can back-donate its $p$ electrons to phosphorous, strengthening the bond. A similar back-donation to chlorine would make the oxygen harder to displace in a redox reaction, but protonation would cut down on this back-donation.

Given these considerations, oxidation by a chlorine oxyanion is kinetically favored if the oxygen can act as a base and thus form a species that is more easily displaced than a bare oxide ion. Hypochlorite ion does that relatively well, whereas in higher oxyanions the needed basic character weakens rapidly unless we force the issue with a strongly acidic solution.


1. Benjamin Gamoke, Diane Neff, and Jack Simons, "Nature of PO Bonds in Phosphates", J. Phys. Chem. A 2009, 113, 5677–5684. http://simons.hec.utah.edu/papers/317.pdf

  • 3
    $\begingroup$ In practice, kinetics trumps thermodynamics. All the tables of half-reactions are useless if you can't isolate a certain half-reaction; and you can't. The tabulated values for chlorine oxides are calculated from heats of reaction, and not from actual voltages. $\endgroup$ Jul 7, 2019 at 14:29

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