Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or reduction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)

Question

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or reduction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)

OH ‾(aq) �¨ O2(g)

I got 2OH^-(aq)-->O_2(g)+2H^+(aq)+4e^-
and its oxidation, but for some reason its wrong.

hint:Your answer contains an improperly or incompletely formatted chemical formula. Your answer contains improper superscript or subscript formatting.
..

Answer 1

I guess your college/school is using an automated response system like the Open Universitiy's iCMA system based on the response. It is difficult to guess what is missing as the formatting is different. My guess would be

2OH^-(aq)--> O_2(g)+2H^+(aq)+4e^-(aq)

Answer 2

I did it correctly though right? I'm trying to figure out what's wrong with the format or maybe i solve the problem wrong. So far, the previous problems, the e^- has no state and they accepted it. I have no idea what's wrong haha.

Answer 3

Yes, your answer balances in terms of charge and atoms.

another (unlikely) possible answer is

2OH^-(aq)--> O_2(g)+H_2(g)+2e^-(aq)

My first choice would have been..

4OH^-(aq)--> O_2(g)+2H2O(l)+4e^-(aq)

Answer 4

First, we need to balance the equation by ensuring that the number of atoms of each element is the same on both sides of the reaction.

The half-reaction given is OH‾(aq) ⟶ O2(g)

To balance the oxygen atoms, we need to add a coefficient of 2 in front of OH‾ on the left side: 2OH‾(aq) ⟶ O2(g)

Next, we balance the hydrogen atoms. Since there are no hydrogen atoms on the left side, we add 2H⁺(aq) on the right side: 2OH‾(aq) ⟶ O2(g) + 2H⁺(aq)

Now, we balance the charge by adding electrons. Since each OH‾ ion has a charge of -1 and there are 2 OH‾ ions on the left side, the total charge is -2. To balance this, we add 4e⁻ on the left side: 2OH‾(aq) + 4e⁻ ⟶ O2(g) + 2H⁺(aq)

Now the equation is balanced.

Regarding whether it is an oxidation or reduction:

In this reaction, OH‾ is being converted to O2. Since oxygen is being gained, this is a reduction reaction.

So, the correct balanced half-reaction is: 2OH‾(aq) + 4e⁻ ⟶ O2(g) + 2H⁺(aq) (reduction)