Chemistry - Why is hypochlorite a stronger oxidizing agent than other oxychlorides?


Solution 1:

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: (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 (

Solution 2:

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, 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.