We have metallic iron, reacting with oxygen gas and water
Fe + O_2 + H_2O
What would happen with a bar of iron, left outside (where the air is humid)? Rust, of course. Rust is the name of a bunch of stuff, the main ones being ferric oxide Fe_2O_3 and ferric hydroxide Fe(OH)_3, however as it turns out, the former is just what happens when you dehydrate the second, so we can say the reaction goes like
Fe + O_2 + H_2O rarr Fe(OH)_3
Which isn't quite balanced because of that pesky water, now thinking mathematically, we must have the same number of Hs, Os and Fes on both sides.
If we have x moles of water and y moles of O_2 and z moles of Fe we have
2x H on LHS, x + 2y O on LHS and z Fe on LHS, to
3z H on RHS, 3z O on RHS and z Fe on RHS.
If we pick z = 1, we'll have
2x = 3 rarr x = 3/2
3/2 + 2y = 3 rarr 2y = 3 - 3/2 rarr y = 3/4
Fe + 3/4O_2 + 3/2H_2O rarr Fe(OH)_3
Multiplying by 4,
4Fe + 3O_2 + 6H_2O rarr 4Fe(OH)_3
Now, as the name implies oxygen oxidizes the other, because it likes to go from 0 charge to -2 charge, while it'll make the iron go from 0 charge to +3 charge, which means Oxygen reduces itself and is the oxydizing agent, whereas Iron oxidizes itself and is the reducing agent.
This, like all redox equations can be accelerated by heat.
