We were taught that nitric acid can oxidize $\ce{Cl-}$ to $\ce{Cl2}$, but sulfuric acid cannot. Is this due to its smaller size, or structure, or what?


2 Answers 2


The nitrate ion is a strong oxidant whereas the sulfate ion is a milder oxidant that only shows oxidizing properties in high concentrations. The oxidizing atom in question is the oxygen atom in the acid, not the hydrogen ion. This is because hydrogen ion is not very electrophilic, which is why non-oxidizing acids cannot dissolve copper: even copper has a higher affinity to electrons than hydrogen.

In a nitrate ion, the molecular structure is inherently not very stable, as the nitrogen atom has to bear a positive charge to maintain the molecular structure. Besides, nitrogen is in itself a very electronegative element, so that makes the nitrate ion even more unstable, hence its oxygen atom readily reacts with other atoms or molecules that oxygen can oxidize, including chloride ions:
$\ce{Cl- + O -> OCl-}$
$\ce{H+ + OCl- <=> HOCl}$
$\ce{H+ + Cl- <=> HCl}$
$\ce{HCl + HOCl <=> H2O + Cl2 }$

  • $\begingroup$ Oxygen in these acids is in lowest ox. state so it can oxidate anything. $\endgroup$
    – Mithoron
    Commented Mar 15, 2015 at 14:32
  • $\begingroup$ @Mithron isn't the formal oxidation state of oxygen here -2? Otherwise are you talking about partial charge? $\endgroup$
    – busukxuan
    Commented Mar 15, 2015 at 15:06
  • 1
    $\begingroup$ Sorry I meant it cannot because it's -2. $\endgroup$
    – Mithoron
    Commented Mar 15, 2015 at 15:08
  • $\begingroup$ @Mithoron not entirely. Formal oxidation state serves as no more than a reference for determining stability of compounds, while the reality isn't too accurately reflected by this. It is indeed true that the nitrogen atom takes a much greater oxidizing role than oxygen here, the reaction still takes place through the oxygen atom as if oxygen is the oxidant here, and the stability of whatever product is formed is determined by oxygen's oxidative power and not nitrogen's. $\endgroup$
    – busukxuan
    Commented Mar 15, 2015 at 16:04

The acids $\ce{HNO3,H2SO4}$ , and the ions $\ce{NO3-,SO4^2-}$ are involved in redox reactions . standard reduction potentials of these ions are

$\ce{(1) SO4^2- +4H+ +2e- -> H2SO3 + H2O (E^0= -0.93V)}$ and

$\ce{(2) NO3- +2H+ +e- ->NO2 + H2O (E^0= +0.83V)}$; Reduction potentials reveal that $\ce{NO3-}$ is more prone to be reduced i.e. nitrate ion is stronger oxidizing agent; The possible reasons for this behavior are:-

(1) $\ce{SO4^2-}$ is tetrahedral and sulfur is comparatively blocked to be attacked moreover, charged oxygens in sulfate repel the incoming $\ce{Cl-}$/ electrons, while, $\ce{NO3-}$ is planar & nitrogen is more open to be attacked& less repelled.

(2) N of $\ce{HNO3}$ is more electronegative than S of $\ce{H2SO4}$ for the reasons that N stays positively charged (formal charge in the lewis structure) & $\mathrm{sp^2}$ hybridized while S stays with no formal charge & $\mathrm{sp^3}$ hybridized. These factors can account for nitric acid's enhanced oxidizing nature. Hope this answer is helpful.


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