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Write the reduction half-reaction for the reduction of elemental oxygen in the presence of water, producing hydroxide ions

This does not make sense to me - a half reaction (from what I understand) is when you have an "element + some number of electrons -> element charged" or "charged element -> element + some number of electrons"

Reaction:

$\ce{HOH + O -> 2OH}$

Oxidation States: They all seem to be neutral to me - but I may be missing something.

So, how is a half reaction being introduced into this and what would it look like? Also, how can i know which element/compound is reduced/oxidized and is the reducing agent/oxidizing agent?

I am really sorry if this seems simple but I am a chem noob and I was out for a month and my teacher is behind schedule so he is giving us worksheets without teaching and i do not understand a lot. Thank you for your time

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    $\begingroup$ Why the downvote? I don't get how to do the half reaction from the reaction but I put a lot of effort into making this question and trying to figure the solution. $\endgroup$ – John D Apr 23 '18 at 20:10
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    $\begingroup$ Sometimes people downvote for inexplicable reasons. Don't worry too much about it all of us have been there. $\endgroup$ – Avnish Kabaj Apr 24 '18 at 2:52
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O2 + 2HO+ 4e- —> 4OH-
That is the correct answer

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    $\begingroup$ I think you need to reconsider the equation, I think it is O2 + 2H2O + 4e ---> 4OH- $\endgroup$ – Nuclear Chemist May 9 '18 at 20:04

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