Consider the following unbalanced redox equation:

$\ce{NO_2^- +e^- -> NO3^-}$

This occurs in acidic media.

My professor would note that an negative two oxidation state oxygen "disappears" from left to right.

Oxide anions are strong bases, and since this process occurs in acidic media, and since an oxide anion has "disappeared," it must have been consumed by the strongest acid present - hydronium ion. Specifically, two equivalents of hydronium ion. These two equivalents of hydronium ion react with oxide anion to produce three waters.

Thus we balance the equation as such.

$\ce{NO_2^- +2e^- +2H_3O^+ -> NO3^- + 3H_2O}$

Now, what is the validity of this method, given that redox reactions likely don't involve an oxide anion intermediate? I can find no mention of any $\ce{O^2-}$ intermediates in redox literature.

If that is the case, how can this method still yield a redox equation that is valid in terms of mass and charge? Is there some likely intermediate that is close enough in character to the oxide anion?


1 Answer 1


Your equation is wrong. In acidic medium, water is the one which gives an oxygen to the nitrite. And that should be an oxidation half-reaction (as nitrite oxidizes to nitrate), but you wrote a reduction where two electrons are spent, not released.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.