(Cr(N2H4CO)6)4(Cr(CN)6)3 + KMnO4 + H2SO4 ---> K2Cr2O7 + MnSO4 + CO2 + KNO3 + K2SO4 + H2O
Solution:
1) Assign oxidation numbers (ignore Mn for the moment):
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 second C 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
2) Set up the oxidation half-reactions:
(a) Cr3+ ---> Cr6+ + 3e¯Because there are 4 Cr3+ in the complex, the following will be used:4Cr3+ ---> 4Cr6+ + 12e¯
(b) Cr2+ ---> Cr6+ + 4e¯
Because there are 3 Cr2+ in the complex, the following will be used:3Cr2+ ---> 3Cr6+ + 12e¯
(c) N3¯ ---> N5+ + 8e¯
Because there are 66 N in the complex, the following will be used:66N3¯ ---> 66N5+ + 528e¯
(d) C2+ ---> C4+ + 2e¯
Because there are 18 C2+, the following will be used18C2+ ---> 18C4+ + 36e¯
3) Add oxidation half-reactions together
4Cr3+ + 3Cr2+ + 66N3¯ + 18C2+ ---> 7Cr6+ + 66N5+ + 18C4+ + 588e¯
4) There are 2 Cr in K2Cr2O7, so we have to have an even number of Cr on the right side. Multiply by 2:
8Cr3+ + 6Cr2+ + 132N3¯ + 36C2+ ---> 14Cr6+ + 132N5+ + 36C4+ + 1176e¯
5) There is only one reduction half-reaction:
Mn7+ + 5e¯ ---> Mn2+
6) Balance electrons: multiply oxidation by 5 and reduction by 1176.
40Cr3+ + 30Cr2+ + 660N3¯ + 180C2+ ---> 70Cr6+ + 660N5+ + 180C4+ + 5880e¯
1176Mn7+ + 5880e¯ ---> 1176Mn2+
7) Add together:
40Cr3+ + 30Cr2+ + 660N3¯ + 180C2+ + 1176Mn7+ ---> 70Cr6+ + 660N5+ + 180C4+ + 1176Mn2+
8) Put back into original equation:
10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 180CO2 + 660KNO3 + K2SO4 + H2O
9) Add in the 240 C4+ from the urea that go to CO2 (no reduction or oxidation here).
10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + K2SO4 + H2O
10) Balance K:
1176 K on left side(35 x 2) + 660 = 730 already present 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
11) Balance SO4:
1176 + 223 = 1399 on right side1399 H2SO4 needed to provide 1399 SO4
10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + 1399H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + H2O
12) Balance H and O:
(4 x 6 x 4 x 10) + (1399 x 2) = 3758 H on left side3758 / 2 = 1879 H2O needed
10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + 1399H2SO4 ---> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + 1879H2O
This balancing website gives the above answer.