[kookaburra1701]

I'm trying to get started on this problem, but none of the methods we were taught in class (or that I can find on the internet) seem to work for this reaction! Here is the method my professor taught us for balancing redox reactions:


1. Assign oxidation numbers to all species
2. write the half-reactions separately
3. balance the half-reactions

• balance the species being oxidized or reduced
• balance by inspection any other elements except H2O
• put in elections (gained as reactants/lost as products)
• balance the charge by adding H+ (in acidic solution) or OH- (in basic solution)
• balance the oxygen by adding H2O
4. repeat step 3 for other half-reaction
5. multiply half-reactions as needed to give equal numbers of electrons lost and gained
6. add half-reactions together, reduce coefficients if necessary and cancel out excess H+ or OH- and H2O[/li]

The problem I am stuck on is this one.
In basic solution:

Mn²⁺(aq) + O₂(g) → MnO2(s)

I wrote the first half reaction as:

Mn²⁺ +2H₂O →MnO₂+4H⁺+2e^-

But I am completely stumped about how to construct the second half-reaction. This is all I can come up with:

O₂ → MnO₂

Both of those are neutral, charge balanced atoms. There's nothing to balance. None of the tutorials I can find online talk about this situation.

Edited to make the nested lists legible.

[sjb]

But I am completely stumped about how to construct the second half-reaction. This is all I can come up with:

O₂ → MnO₂

Both of those are neutral, charge balanced atoms. There's nothing to balance. None of the tutorials I can find online talk about this situation.

Edited to make the nested lists legible.

What is the oxidation number of O in O₂2, and in MnO₂? (edit for format, sjb)

[kookaburra1701]

In O₂, it is zero, because that is a naturally occurring state for that element, in MnO₂, it is -2, which gives both oxygens a combined total of -4, and Mn +4 to make a neutral compound. So two e^- go from the O's to the Mn.

Did I get that right?

[sjb]

In O₂, it is zero, because that is a naturally occurring state for that element, in MnO₂, it is -2, which gives both oxygens a combined total of -4, and Mn +4 to make a neutral compound. So two e^- go from the O's to the Mn.

Did I get that right?

In O₂ it is zero, but not because it's "a naturally occurring state for that element", or at least not really - I think what you meant was the element in its standard state (may be a language thing)

So this half reaction is O₂ 2O^2-. Can you balance that?

[kookaburra1701]

In O2 it is zero, but not because it's "a naturally occurring state for that element", or at least not really - I think what you meant was the element in its standard state (may be a language thing)

That's just what it says on the oxidation number handout my professor gave us - I thought it might be equivalent to "standard state" but we just started learning about those and enthalpy and I didn't want to use it incorrectly.

So this half reaction is O2 2O2-. Can you balance that?

The half reaction would be 2e^- + O₂ 2O^2-

So does this mean I would set it up like this?

2e^- + O₂ 2O^2-
Mn²⁺ + 2H₂O MnO₂ + 4H⁺ +2e^-
-----------------------------------------------
Mn²⁺ + O₂ + 2H₂O MnO₂ + 4H⁺

Is this right? I think I ended up with 2 less electrons on the products side.

Edit: Wait, to balance the elements alone I think it should be 2Mn²⁺ + O₂ + 2H₂O 2MnO₂ + 4H⁺

The elements and charges balance in this, but it's like it's for an acidic solution instead of a basic solution.

Edit II Electric Boogaloo: WAIT I should add 4 OH- to the reactants side right? Then it would be 4OH^- + 2Mn²⁺ + O₂  2MnO₂ + 2H₂O

Does that balance? I think that balances!

[kookaburra1701]

When we went over the worksheet problems in lab today, it was right! Thank you so, so much for taking the time to walk me through the problem. I still need lots of practice before the exam on Wednesday, but I feel much more confident now.