Can someone balance this equation for me please. Its a Redox?

(Cr(N2H4CO)6)4(Cr(CN)6)3 + KMnO4 + H2SO4 -> K2Cr2O7 + MnSO4 + CO2 + KNO3 + K2SO4 + H2O

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  • 1 decade ago
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    This is quite the reaction - I found it balanced online (without half reactions), but thought I'd try it by half-reactions to see if I could do it.

    [Cr(N2H4CO)6]4[Cr(CN)6]3 + KMnO4 + H2SO4 ---> K2Cr2O7 + MnSO4 + CO2 + KNO3 + K2SO4 + H2O

    assign oxidation numbers

    Cr in the complex has +3 and +2 (first and second part)

    N in the complex has -3 in both parts

    C in the urea ligand has +4, in the CN ligand +2 (this is the only C that changes)

    H is always +1

    C in CO2 is +4

    Cr in K2CrO7 is +6

    N in KNO3 is +5

    set up the half reactions

    oxidations: there are several

    Cr+3 ---> Cr+6 + 3e- (there are 4 Cr+3 in the complex)

    4Cr+3 ---> 4Cr+6 + 12e-

    Cr+2 ---> Cr+6 + 4e- (there are 3 Cr+2 in the complex)

    3Cr+2 ---> 3Cr+6 + 12e-

    N-3 ---> N+5 + 8e- (there are a total of 66 N in the complex)

    66N-3 ---> 66N+5 + 528e-

    C+2 ---> C+4 + 2e- (there are a total of 18 C+2)

    18C+2 ---> 18C+4 + 36e-

    add oxidation half-reactions together

    4Cr+3 + 3Cr+2 + 66N-3 + 18C+2 ---> 7Cr+6 + 66N+5 + 18C+4 + 588e-

    there are 2 Cr in K2Cr2O7, so we have to have an even number of Cr on the right side - multiply all by 2

    8Cr+3 + 6Cr+2 + 132N-3 + 36C+2 ---> 14Cr+6 + 132N+5 + 36C+4 + 1176e-

    reduction: only one half-reaction

    Mn(+7) + 5e- ---> Mn(+2)

    balance electrons - mult oxidation by 5, reduction by 1176

    40Cr+3 + 30Cr+2 + 660N-3 + 180C+2 ---> 70Cr+6 + 660N+5 + 180C+4 + 5880e-

    1176Mn(+7) + 5880e- ---> 1176Mn(+2)

    add together

    40Cr+3 + 30Cr+2 + 660N-3 + 180C+2 + 1176Mn(+7) ---> 70Cr+6 + 660N+5 + 180C+4 + 1176Mn(+2)

    now, put back into original equation

    10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 180CO2 + 660KNO3 + K2SO4 + H2O

    need to add in the C+4 from the urea that also goes to CO2 (no reduction or oxidation here) - there are 240 of them

    10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + K2SO4 + H2O(not yet balanced - balance K)

    1176 K on left side

    (35 x 2) + 660 = 730 on right

    1176 - 730 = 446

    446 / 2 = 223 K2SO4 needed

    10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + H2O (now balance SO4)

    1176 + 223 = 1399 on right side

    1399 H2SO4 needed

    10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + 1399H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + H2O (now balance H and O)

    (4 x 6 x 4 x 10) + (1399 x 2) = 3758 H on left side

    3758 / 2 = 1879 H2O needed

    10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + 1399H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + 1879H2O

    (and this does match what I found online)

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