3
$\begingroup$

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

Improve this question
$\endgroup$
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ Sometimes people downvote for inexplicable reasons. Don't worry too much about it all of us have been there. $\endgroup$ – Avyansh Katiyar Apr 24 '18 at 2:52
-1
$\begingroup$

O2 + 2HO+ 4e- —> 4OH-
That is the correct answer

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.