The 10 Steps* to Balance:

The Mother of All REDOX:

Total Molecular Equation:

(Cr(N2H4CO)6)4(Cr(CN)6)3 + KMnO4 + H2SO4 à

K2Cr2O7 +  MnSO4 +  CO2  +  KNO3 +  K2SO4 +  H2O

*The Ion Electron or Half Equation Method

Step#1: Net Ionic Equation:

(Cr(N2H4CO)6)4(Cr(CN)6)3 +  MnO41- + H1+  à

Cr2O72- +    Mn2+ +    CO2 +   NO31- +    H2O

Step #2: Acid Media:

Helpers:     H1+  and   H2O

Step #3: Write Half Equations:

Half Equation:

(Cr(N2H4CO)6)4(Cr(CN)6)3 à Cr2O72- +   CO2 +   NO31-

Half Equation:

MnO41-  à Mn2+

Step #4: Balance the Element(s) Changing:

1st Half Equation:

(Cr(N2H4CO)6)4(Cr(CN)6)3 à Cr2O72- +   CO2 +   NO31-

Seven Cr on Left but just Two on Right (LCM 14)

2(Cr(N2H4CO)6)4(Cr(CN)6)3à7Cr2O72-+  CO2 +   NO31-

Now 14 Cr on Left and 14 Cr on Right; Now Balance the Carbons:

84 C’s on Left and One on right. (Add 84 coefficient to the CO2)

2(Cr(N2H4CO)6)4(Cr(CN)6)3à7Cr2O72-+ 84 CO2 + NO31-

Now Balance the Nitrogens:  132 on the Left and Just one on the right:

2(Cr(N2H4CO)6)4(Cr(CN)6)3à7Cr2O72-+ 84 CO2 +132 NO31-

2nd  Half Equation:

MnO41-  à Mn2+

The Manganese is balanced!

Step #5: Balance the Oxygen by adding H2O helper:

1st Half Equation:

2(Cr(N2H4CO)6)4(Cr(CN)6)3à7Cr2O72-+84CO2+132NO31-

Here are 48 Oxygen on left and 613 Oxygen on the Right,

add (613-48) = 565 Water to the left of the half equation:

2(Cr(N2H4CO)6)4(Cr(CN)6)3 + 565 H2O à7Cr2O72-+84CO2+132NO31-

2nd  Half Equation::

MnO41-  à Mn2+

There are 4 Oxygen on left and no Oxygen on the Right, add 4 H2O molecules to the right:

MnO41-  à Mn2+ + 4 H2O

Step #6: Balance the hydrogen by adding H1+ ions:

1st Half Equation:

2(Cr(N2H4CO)6)4(Cr(CN)6)3 + 565 H2O à7Cr2O72-+84CO2+132NO31-

There are 1322 hydrogen atoms on the left, but none of the right,
1322 hydrogen ions to the right:

2(Cr(N2H4CO)6)4(Cr(CN)6)3 + 565 H2O à7Cr2O72-+84CO2+132NO31- + 1322 H1+

2nd  Half Equation:

MnO41-  à Mn2+ + 4 H2O

No Hydrogen atoms on the left and 8 on the right, add 8 H1+ ions to the left:

8 H1+ + MnO41-  à Mn2+ + 4 H2O

Step #7: Charge balance the half equations by adding e1- :

1st Half Equation:

2(Cr(N2H4CO)6)4(Cr(CN)6)3 + 565 H2O à7Cr2O72-+84CO2+132NO31- + 1322 H1+

There is no charge on the left , but +1176 on the right

(0 + 0) = 7x(-2) + 0 +132x(-1) + 1322x(+1)
0 = +1176  (add 1176 electrons to the right) (
This is the oxidation half equation)

2(Cr(N2H4CO)6)4(Cr(CN)6)3 + 565 H2O à7Cr2O72-+84CO2+132NO31- +1322H1+ +1176e1-

2nd  Half Equation:

8 H1+ + MnO41-  à Mn2+ + 4 H2O

The charge on the left is +8 – 1 or +7 = +2 + 0 on the right; add 5 e1- to the left:

5 e1- + 8 H1+ + MnO41-  à Mn2+ + 4 H2O

The second half equation is the reduction step and the permanganate ion is the oxidizer(or oxidizing agent)!

Step #8: Make e1- Gain (Reduction) = e1- Loss (Oxidation):

The Least Common Multiple (LCM) of 1176 and 5 is 5880

Therefore 5 oxidations must occur for every 1176 reductions:

1st Half Equation (Oxidation) x 5:

5 [2(Cr(N2H4CO)6)4(Cr(CN)6)3+565H2O à7Cr2O72-+84CO2+132NO31- +1322H1+ +1176e1-]

10(Cr(N2H4CO)6)4(Cr(CN)6)3+2825H2Oà35Cr2O72-+420CO2+660NO31-+6610H1++5880e1-

2nd  Half Equation(Reduction) x 1176:

1176 [5 e1- + 8 H1+ + MnO41-  à Mn2+ + 4 H2O]

5880e1-+ 9408 H1++ 1176 MnO41-  à1176 Mn2++ 4704 H2O

Step #9: Add Half Equations for Balanced Net Ionic:

1st Half Equation (Oxidation):

10(Cr(N2H4CO)6)4(Cr(CN)6)3+2825H2Oà35Cr2O72-+420CO2+660NO31-+6610H1++5880e1-

2nd  Half Equation(Reduction):

5880 e1- + 9408 H1+ + 1176 MnO41-  à1176 Mn2+ + 4704 H2O

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

1.   Cancel 5880 electron on both sides

2.   Subtract 6610 Hydrogen ion on Right from 9408 on the left = Net 2798 Hydrogen Ions on Left

3.   Subtract 2285 Water Molecules on Left from 4704 on the Right= Net 1879 Water Molecules on Right

Balanced Net Ionic Equation:

10 (Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176 MnO41- + 2798 H1+  à

35 Cr2O72- + 1176 Mn2+420 CO2 + 660 NO31- + 1879 H2O

Step #10: Substitute Balanced Net Ionic into Total Molecular Equation and Balance the Spectator Ions by Inspection:

Balanced Net Ionic Equation:

10 (Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176 MnO41- + 2798 H1+  à

35 Cr2O72- + 1176 Mn2+420 CO2 + 660 NO31- + 1879 H2O

Total Balanced Molecular Equation:

10 (Cr(N2H4CO)6)4(Cr(CN)6)3+ 1176 KMnO4+ 1399 H2SO4 à

35 K2Cr2O7+1176 MnSO4+420 CO2660KNO3+ ? K2SO4+ 1879 H2O

1.   Balance Potassium atoms: 1176 on Left; 70+660 + ? on Right; ? = 446 divided by 2 = 223 K2SO4

2.   Check Sulfate Ions: 1399 on Left; 1176 + 223 on Right or 1399=1399

3.   ?  is  223 K2SO4  Formula Units