What does oxidizing strength mean?

Does it mean the strength of a oxidation agent that is needed to oxidize the reduction agent, or does it just mean the tendency of a substance to lose electrons (oxidate)?

  • $\begingroup$ If X has a higher oxidising strength than Y, then X would probably be able to oxidise Y (even if X/Y are both oxidizing agents). $\endgroup$
    – t.c
    Nov 6 '14 at 12:34
  • $\begingroup$ See also Comparing Strengths of Oxidants and Reductants. $\endgroup$
    – Sparkler
    Jun 10 '15 at 3:01

The strengths of oxidizing and reducing agents are indicated by their standard electrode potentials. Let's consider standard potentials of some redox couples:

$$ \begin{align} E^\circ(\ce{Li^+/Li}) &= \pu{-3.04 V} \\ E^\circ(\ce{K^+/K}) &= \pu{-2.92 V} \\ E^\circ(\ce{Cu^2+/Cu}) &= \pu{+0.34 V} \\ E^\circ(\ce{F_2/F^-}) &= \pu{+2.87 V} \end{align} $$

The values above are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.


Oxidizing strength is a relative term used to denote how certain elements oxidize other elements with respect to each other.

As t.c said in the comments, if $\text{X}$ has a higher oxidizing strength than $\text{Y}$, then $\text{X}$ can a higher likelihood oxidize $\text{Y}$ faster than $\text{Y}$ has to oxidize $\text{X}$. The same definition could apply for reducing strength, and you could say that if $\text{X}$ has a higher oxidizing strength than $\text{Y}$, then $\text{Y}$ has a higher reducing strength than $\text{X}$.

  • $\begingroup$ I think that you are confusing kinetics here with rates (thermodynamics). $\endgroup$
    – Dissenter
    Jan 6 '15 at 6:26
  • 1
    $\begingroup$ -1. This is incorrect - if X is a stronger oxidising agent than Y, that means that X will oxidise the reduced form of Y. $\ce{Cl2}$ is a stronger oxidising agent than $\ce{SO4^2-}$ but that doesn't mean that $\ce{Cl2}$ will oxidise $\ce{SO4^2-}$. If you're still not convinced, think about this. $E^\circ(\ce{MnO4-}/\ce{Mn^2+}) = +1.51~\text{V}$ and $E^\circ(\ce{H2O2}/\ce{H2O}) = +1.78~\text{V}$ but $\ce{MnO4-}$ actually oxidises $\ce{H2O2}$. $\endgroup$
    – orthocresol
    Apr 5 '16 at 22:15

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