Question

Do reduction and oxidation occur together?

Answer 1

Well, yes.....we conceive of `"oxidation"` as the loss of electrons.....

Explanation:

And we conceive of `"reduction"` as the gain of electrons.

Since charge, as well as mass, is CONSERVED in every chemical reaction, for every oxidation there must be a corresponding reduction, and we can assign `"oxidation numbers"` as arbitrary numbers with which we can assess `"conceptual electron transfer....."`

Now ammonia can oxidized to nitrate.......a formal oxidation of `N(-III)` to `N(+V)`, an EIGHT electron transfer.....

And thus for oxidation we write.....

`NH_3 +3H_2O rarrNO_3^(-)+9H^(+) +8e^(-)` `(i)`

Mass is balanced, and charge is balanced....so this is a reasonable representation of chemical change. Now the electrons are presumed to GO somewhere on oxidation; i.e. they cause a CORRESPONDING reduction of some other reagent (which of course is the OXIDIZING agent). A typical oxidizing agent is permanganate ion, `MnO_4^(-)`, which is REDUCED from `Mn(VII+)` to colourless `Mn^(2+)`, a 5 electron reduction.....

`MnO_4^(-) +8H^+ + 5e^(-)rarr Mn^(2+) + 4H_2O` `(ii)`

Charge and mass are balanced as required. Are they?

To complete the entire redox reaction, we cross multiply to remove the electrons.....`5xx(i) + 8xx(ii)`:

`8MnO_4^(-) +64H^+ +5NH_3 +15H_2O+40e^(-)rarr 8Mn^(2+) + 32H_2O +5NO_3^(-)+45H^(+) +40e^(-)`

And then we cancel out the common reagents......

`8MnO_4^(-) +cancel(64)19H^+ +5NH_3 +cancel(15H_2O)+cancel(40e^(-))rarr 8Mn^(2+) + cancel(32)17H_2O +5NO_3^(-)+cancel(45H^(+)) +cancel(40e^(-))`

To give finally.......

`8MnO_4^(-) +19H^+ +5NH_3 rarr 8Mn^(2+) +5NO_3^(-)+ 17H_2O`

Which, despite the whack coefficients, is balanced with respect to mass and charge. And we would see the deep purple colour of permanganate dissipate to give COLOURLESS `Mn^(2+)`.