Question

How do you write oxidation reduction half reactions?

Answer 1

See here and links.

Explanation:

In redox reactions we invoke electrons as elementary particles, whose LOSS corresponds to oxidation, and whose gain corresponds to reduction. Rather than rabbit on about stuff that may appear in the links, let us consider an actual redox reaction, i.e. the oxidation of ammonia to nitrate ion by metallic zinc.

Now ammonia is OXIDIZED ... from `N(-III)` to `N(V+)`:

`NH_3(aq) +3H_2O(l) rarr NO_3^(-)+9H^+ + 8e^(-)` `(i)`

And as always the difference in oxidation numbers is accounted for by `"formal electron transfer"`, the which is conceived to operate in the reaction. Again as always, because CHARGE as well as mass ARE CONSERVED in ALL chemical reactions, the electrons are conceived to GO somewhere...i.e. to cause a corresponding reduction reaction. Now certainly a manganese salt, say `MnO_2` could be reduced down to `Mn^(2+)`:

`MnO_2(s) +4H^+ + 2e^(-) rarr Mn^(2+) + 2H_2O(l)` `(ii)`

And so `"manganese(IV) dioxide"`, even though it is as soluble as a brick, has acted as an `"oxidizing agent"` that is formally reduced in the redox reaction to give colourless (and soluble) `Mn(+II)` salt.

The overall redox reaction eliminates the electrons....i.e. we takes `(i)+4xx(ii):`

`4MnO_2(s)+NH_3(aq) +3H_2O(l) +16H^+ + 8e^(-) rarr 4Mn^(2+) + NO_3^(-)+9H^+ + 8e^(-)+8H_2O(l)`

And after cancellation...

`4MnO_2(s)+NH_3(aq) +7H^+ rarr 4Mn^(2+) + NO_3^(-) +5H_2O(l)`

And again, I stress that while the assignment of oxidation numbers and introduction of electrons as elementary particles may seem a bit fanciful, and abstract, such redox reactions as written do REPRESENT actual chemical transformations, which is of course why we learn how to assign them, and why we utilize them when we write chemical reactions.