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Redox in basic solution

  1. Nov 27, 2004 #1
    Balance in basic (check my work)

    NO + MnO4- ------ NO3- + MnO2

    4OH- + 4H+ + NO ---- NO3- + 2H2O + 4OH-
    4OH- + 4H+ + MnO4- --- MnO2 + 2H2O + 4OH-
    Balanced with 5 e- on each side

    2H2O + NO + MnO4- ----- NO3- + MnO2 + 8OH-

  2. jcsd
  3. Nov 27, 2004 #2


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    You need help, my friend :smile:

    First, prepare atomic-based electron balance:

    [tex]N^{2+} \longrightarrow N^{5+} + 3e^-[/tex]
    [tex]Mn^{7+}+5e^- \longrightarrow Mn^{2+}[/tex]

    Then balance them.

    [tex]5N^{2+}+3Mn^{7+}\longrightarrow 5N^{5+}+3Mn^{2+}[/tex]

    You'll see that both electron counts and atom counts are balanced now.

    When you write the "real" ions, I mean, NO and MnO4-, you'll have to add OH- and H2O to the side with less oxygen and less hydrogen, etc.
  4. Nov 28, 2004 #3
    Cool, this is what I got

    5NO + 3MNO4 = 5NO3 + 3MnO2

    It appears that all are balanced. Does this look right.
  5. Nov 28, 2004 #4
    Another class mate said that the original reaction was already balanced. He said he worked through it and kept coming up with the original equation.

    Could this be right??? Or does mine look fine???

  6. Nov 29, 2004 #5


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    Well, it seems that atom counts are okay, but you've forgotten ionic counts. This is why redox reactions are hard to study.

    You said that this redox should be in basic solution; so just put OH- ions to the left side. However, there is a serious error; in basic solutions, permanganate is only reduced to manganese dioxide, as you correctly wrote. So the coefficients cancel to reflect this:

    [tex]N^{2+}+Mn^{7+}\longrightarrow N^{5+}+Mn^{4+}[/tex]

    So if we write its real forms, we'll get this one:

    [tex]NO + KMnO_4 \longrightarrow KNO_3 + MnO_2 [/tex]
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