Question

Please explain: How to identify the reducing agent in a reaction of magnesium with oxygen? @aaronq Thanks <3

Answer 1

Sure. First of all, are you good with assigning oxidation states/numbers?

Answer 2

not really ;/

Answer 3

It's kind of a lengthy process to learn, but it's relatively easy once you get the hang of it.
You'll need a periodic table for this, so pull it out.

You start off with a basic set of rules and you figure the rest out with simple arithmetic.

Rules (most often true, except in rare occasions):
Group 1 always +1
Group 2 always +2
Oxygen always -2
Group 7 always -1
Noble gases, elemental or homonuclear (the same element) molecules are zero

The rest you have to figure out by subtracting or adding to achieve whatever charge the compound bears.

eg. H2O
H is group 1, so H = +1, there are 2 of them
O must balance the charge so, O=-2

HCl
H=+1, Cl=-1

MgSO4,
you have to know that SO4^2- is a polyatomic ion, treat that as it's own separate thing:
O=-2
Total contribution of O=-2(4) = -8
^4 oxygen atoms

Charge on ion=-2, so S has to fulfil the rest which is 6, so S=+6

back to MgSO4, so we treat SO4 as entity and it bears a -2 charge,
Mg therefore has to be +2 to balance the charges.

Answer 4

Thank you so much!

Answer 5

we'll thats not really the answer.

that's what you need to know before you can understand how to get the answer.

When an atom is reduced is decreases in oxidation number.
For example, +3 to -2, or 0 to -1.

An oxidation is the opposite, the atom increases in oxidation number.
For example -2 to -1, or 0 +1 to +3.

This makes sense because what's being lost/gained are electrons which are negatively charged.