Manganese redox reactions

  1. Hello,

    I've been working on some redox reactions for the oxidation of manganese.


    Write the complete formulae for the redox reactions below:

    i.) Manganese reacts with nitric acid (V) to form Mn2+ ions.
    ii.) The Mn2+ ions reacts with potassium iodate, KIO4 to form MnO4- (VII)

    My Solution:


    Oxidation: 3Mn (s) --> 3Mn2+ (aq) + 6e- (aq)
    Reduction: 6HN03 (aq) + 6e- (aq) --> 6N03- (aq) + 6H+ (aq)
    Overall reaction: 3Mn + 6HN03 -> 6NO3- + 3Mn2+ + 6H+

    Is this correct? If so, does the single protons regenerate the acid and/or become attached to substances like water?


    Oxidation: Mn2+ + X -> MnO4- + ?H+
    Reduction: IO4- + ?H+ -> I? + 4H2O
    Overall reaction: Mn2+ + IO4- -> MnO4- + I2-

    I'm having trouble with this part of the question. I does not have -II as one of its common oxidation number.

    Any help or hints would be greatly appreciated :smile:

    Best Regards,
  2. jcsd
  3. Gokul43201

    Gokul43201 11,046
    Staff Emeritus
    Science Advisor
    Gold Member

    Take a step back and ask yourself,

    metal + acid --> salt + (?)

    You do not get protons. Look at the reduction reaction you've written. The charge is not balanced. Fix that one mistake using above hint and you're good.

    On the RHS, you have Mn, H and O. On the left, you have Mn. Ergo, X is simple H2O.

    This is a little tricky. The reaction actually depends on the conditions. Mostly, you just have to know what happens to IO4- under different conditions. Typically, it will first get reduced to IO3-. Unless someone else suggests otherwise, you could go with that.
    Last edited: May 31, 2006
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