What I have been told is that the acidity of the four oxyacids of Chlorine increase in the order $\ce{HOCl} < \ce{HClO2} < \ce{HClO3} < \ce{HClO4}$.

I have also been told that the oxidising strength goes in the reverse order. In my knowledge acidic strength should increase with oxidising strength. Where am I going wrong?

For your reference, the Ka values of the acids are as follows:

  1. $\ce{HOCl} = 2.9 \times 10^{-8}$
  2. $\ce{HClO2} = 1.1 \times 10^{-2}$
  3. $\ce{HClO3} = 5.0 \times 10^2$
  4. $\ce{HClO4} = 1 \times 10^3$
  • 4
    $\begingroup$ Don't you mean the oxyacids $\ce{HClO}$, $\ce{HClO2}$, $\ce{HClO3}$, and $\ce{HClO4}$? $\endgroup$ – Philipp Aug 29 '14 at 12:31

For many acids, the proton $\ce{H+}$ is the oxidizing agent. For example, in the oxidation of zinc with $\ce{HCl}$, the oxidizing agent is $\ce{H+}$.

$$\ce{Zn + 2H+ -> Zn^{2+} + H2}$$

However, for some acids, the conjugate base (anion) is also an oxidant (and usually a more powerful oxidant than $\ce{H+}$. The oxyacids of chlorine are all oxidizing agents through their anions, as is nitric acid and some others:

$$\ce{Zn + 2H+ + 2OCl- -> Zn^{2+} + H2O + 2Cl-}$$

In the above example, hypochlorite is the oxidizing agent. For the oxyacids of chlorine and others, the strength of the oxidizing agent is independent of the strength of the acid.

  • 1
    $\begingroup$ Why exactly would the proton often times be a weak oxidizer and why exactly would the conjugate bases sometimes be good oxidizers? Isn't the formation of H2 entropically favorable? $\endgroup$ – Dissenter Aug 29 '14 at 16:07
  • 1
    $\begingroup$ Formation of $\ce{H2}$ is entropically favorable, but so is conversion of $\ce{2HOCl}$ to $\ce{H2O + 2Cl-}$ (three particles from 2). Additionally, you cannot ignore the contribution to spontaneity from enthalpy. $\endgroup$ – Ben Norris Aug 30 '14 at 13:25

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