Identify the oxidizing and reducing agents in each of the following redox reactions: (4 points) a) 2 H2O (l) + Al (s) + MnO4 -1(aq) ---→ Al(OH)4 -1(aq) + MnO2 (s)

Answer 1

2 H2O (l) + Al (s) + MnO4 -1(aq) ---→ Al(OH)4 -1(aq) + MnO2 (s)
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
There is a mnemonic for that. "LEO the lion says GER"
LEO (loss electrons oxidation). GER (gain of electrons reduction).
Then you just remember that
The oxidizing agent is the one reduced.
The reducing agent is the one oxidized.
Here is how you do part of it. I'll leave the rest for you.
First you must determine the oxidation state of each of the elements in the equation. Initially that can be time consuming but with a little practice you will see that most of the elements are often the same.
If you don't know how to determine oxidation states or if you just need some practice or clarification, go to google and type in "chem team redox" and read #2 tutorial.
2 H2O (l) + Al (s) + MnO4 -1(aq) ---→ Al(OH)4 -1(aq) + MnO2 (s)
Al is zero on the left and +3 on the right; therefore, it has lost electrons. That makes it oxidized so it must be the reducing agent. Simple, huh?
Look at MnO4^- ==> MnO2 for the rest of it. Post your work if you get stuck or if you want to ask for clarification on an issue.

Answer 2

Al (s) is a Reducig agent
MnO4^- (aq) is an Oxidation agent