What does oxidizing strength mean?

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)?

• 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).
– t.c
Nov 6 '14 at 12:34
• Jun 10 '15 at 3:01

\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}
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 $\text{X}$.
• -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 $\ce{H2O2}$. Apr 5 '16 at 22:15