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The equation is $$\ce{2Mn^2+(aq) + 4OH^-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)}$$ I know that the Manganese (II) gets oxidized to Manganese (IV), but I'm not sure about the other half. I know oxygen gets reduced, but using simply $\ce{O2 +2e^-->2O-}$ doesn't work because the equations won't balance out then.

By the way, this chemical equaltion is used in the Winkler titration.

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  • $\begingroup$ No wonder it won't balance. What is the oxidation state of O in the products? $\endgroup$ – Ivan Neretin Mar 23 '17 at 21:43
  • $\begingroup$ The reaction doesn't look right for what occurs in the Winkler titration $\endgroup$ – Tyberius Mar 24 '17 at 1:09
  • $\begingroup$ @Tyberius There are different views for what the first equation is in the winkler method. Wikipedia mentions this. $\endgroup$ – ILoveJesus Mar 24 '17 at 13:26
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The half-equations should look like this:

$$\ce{2Mn^2+(aq) \to 2Mn^4+(aq) + 4e^-}$$

and

$$\ce{4OH^-(aq) + O2(aq) + 4e^- -> 4O- + 2H2O(l)}$$

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Let us consider these half equations: $$\ce{Mn^2+(aq) + 4OH^- \to MnO_2(aq) + 2e^- + 2H2O}$$

and

$$\ce{ 2H2O(l) + O2(aq) + 4e^- -> 4OH- }$$ By multiplying the first equation by two and adding it to the second one, we get the overall reaction.

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  • $\begingroup$ Are these actually half equations though? Is this one correct or is the first answer correct? $\endgroup$ – ILoveJesus Mar 24 '17 at 15:34
  • $\begingroup$ Both answers are correct. However, I prefer mine, because they give directly the overall reaction . The first half equation shows the oxidation of Mn (II) ion into manganese oxide (IV), while the second half equation shows the reduction of two oxygen atoms into ion hydroxide (with the oxidation number -II). Please notice that in the second half equation, the other two atoms of oxygen in the left side keep their oxidation number (-II) as hydroxide ions $\endgroup$ – Yomen Atassi Mar 29 '17 at 17:51

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