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In my chemistry book they write $\ce{MnO4^- -> Mn^{2+}}$ is a reduction but is this correct? Electrons are lost aren't they because it becomes less negative so is it not an oxidation?

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2 Answers 2

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You have to write out the complete half-cell reaction, and then it is obvious that $\ce{MnO4^−}$ is reduced.

$\ce{MnO4^− + 8 H^+ + 5 e^− <=>[reduction][oxidation] Mn^{2+} + 4H2O\quad\quad\quad\rm{E}^0 = +1.51~\rm{V}}$

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A simple analogy to see whether oxidation or reduction has occur is as following. If oxidation number of element decreases,there will be reduction and if oxidation number increases there will be oxidation.In your case,Mn had +7 oxidation state initially and finally it has a oxidation state of +2 so it is a reduction.

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  • $\begingroup$ So how does your technique work with hydrogen peroxide? $\endgroup$
    – MaxW
    Commented Oct 12, 2016 at 23:10
  • $\begingroup$ in peroxides,oxygen atom always have -1 oxidation state. $\endgroup$ Commented Oct 13, 2016 at 1:29

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