You are watching: Which is the strongest oxidizing agent?
Many said $ceClF3$ is the most an effective as the oxidises everything, also asbestos, sand, concrete, and can set easily fire to anything i beg your pardon can"t be stopped; it have the right to only it is in stored in Teflon.
And $ceHArF$ could be a very an effective oxidant due to high instability as a compound of argon v fluorine, however was it also used as such?
What link is actually provided as oxidising agent and also was proven to be more powerful then others, by, because that example, conventional reduction potential?
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edited Oct 6 "18 at 17:16
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inquiry Oct 4 "18 in ~ 5:17
Harsh jainHarsh jain
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Ivan"s price is without doubt thought-provoking. Yet let"s have some fun.
IUPAC defines oxidation as:
The complete, network removal that one or an ext electrons native a molecularentity.
My humble questions is therefore - what better way is over there to remove an electron than combining it with a literal anti-electron? Yes, my friends, us shall look for to transcend the difficulty entirely and swat the fly through a thermonuclear bomb. Ns submit as the most powerful entry, the positron.
Since 1932, we"ve well-known that ordinary matter has actually a mirror image, which us now contact antimatter. The antimatter equivalent of the electron ($cee-$) is the positron ($cee+$). To the ideal of our knowledge, they behave specifically alike, other than for their opposite electric charges. I stress that the positron has actually nothing to do with the proton ($cep+$), one more class of bit entirely.
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As you may know, once matter and also antimatter meet, they relax tremendous amounts of energy, thanks to $E=mc^2$. Because that an electron and also positron v no initial energy other 보다 their individual rest masses of $pu511 keV c^-2$ each, the most typical annihilation result is:
$$ cee- + e+ -> 2gamma$$
However, this process is completely reversible in quantum electrodynamics; the is time-symmetric. Opposing reaction is pair production:
$$ ce2gamma -> e- + e+ $$
A reversible reaction? climate there is nothing avoiding us from imagining the adhering to chemical equilibrium:
eginaligncee- + e+ & 2gamma &Delta_r G^circ &= pu-1.022 MeV =pu-98 607 810 kJ mol^-1endalign
The difference between enthalpy and Gibbs totally free energy in such subatomic reaction is totally negligible, together the entropic element is laughably little in comparison, in any kind of reasonable conditions. Ns am just going to brashly take into consideration the over value together the conventional Gibbs free energy readjust of reaction. This massive $Delta_r G^circ$ coincides to an equilibrium continuous $K_mathrmeq = 3 imes 10^17276234$, representing a somewhat product-favoured reaction. Plugging the Nernst equation, the conventional electrode potential because that the "reduction the a positron" is then $mathrmfrac98 607 810 kJ mol^-196 485.33212 C mol^-1 = +1 021 998 V$.
Ivan mentions in his answer using an alpha fragment as one oxidiser. Let"s take the further. Follow to NIST, a turbulent estimate because that the electron affinity the a totally bare darmstadtium cell core ($ceDs^110+$) is $pu-204.4 keV$, so even a stripped superheavy atom can"t enhance the oxidising strength of a positron!