Thursday

May 26, 2016
Posted by **Nancy** on Monday, April 30, 2007 at 4:03pm.

MnO4(¨C) + H(+) + C2O4(2¨C)--> Mn(2+) + H2O + CO2

Sure but tell us what you don't understand about it. You need to learn how to do them yourself. Tell me what you don't understand, in detail, and I can help you through it.

Well I tried testing it out by doing MnO4¨C + 8H+ + C2O42¨C-> Mn2+ + 4H2O + 2CO2 because that gives them all the same amount of each chemical, but then it was wrong so I don't get if the charges have anything to do with it or if I am just balancing with the wrong numbers because it seems that it should be right

It isn't right.

In order for an equation to balance, it must balance three ways.

a. the atoms must be the same on each side of the equation.

b. the charge must be the same on each side of the equation.

c.

To help, Mn on the left is +7 and it goes to Mn^+2 on the right so that is a change of 5 electrons.

C in C2O4^= on the left is +6 (for both carbon atoms) and it goes to 2CO2 on the right where both C atoms is +8 for a change of 2 electrons.

Let me know if you need further assistance.

I'm sorry, I got kind of confused. How did you get +7 for Mn on the left or +6 for the 2C on the left or 2CO2 on the right?

Wait nm i understand how you got the +7 charge for Mn...but still not the rest....

For MnO4^-, oxygen usually has an oxidation state of -2. We have four of them; therefore, 4*-2 = -8. We must leave a -1 charge on MnO4^-; thus, Mn must be +7. We can check that, and I always do, by +7 for Mn + (-8) for O makes -1 and that is the charge on the ion.

For C on the left. We have C2O4^-2. Oxygen is -2, 4*-2 = -8, we must leave -2 charge on the ion which means both C atoms must be +6. We check that by +6 + (-8) = -2 and that is the charge on the ion. For C on the right, we have 2CO2.

Each O is -2, 2*-2 = -4, for CO2 to be neutral (there is no charge on it), C must be +4 and there are two of them and that makes +8. Voila!

I don't like to call the charge on Mn a +7. The oxidation state is +7 but the charge on the ion is -1 in permanganate. To call the Mn a charge of +7 gets confusing with the actual charge of the ion sometimes.

Here another solution if your algebra is upto speed!

? MnO4– + ?H+ + ?(C2O4)2– --> ?Mn2+ + ?H2O + ?CO2

MnO4– + xH+ + y(C2O4)2– --> Mn2+ + aH2O + bCO2

We have 4 unknowns and can then generate 4 equations.

Charge

x-2y-1=2

x-2y=3---------------------------(1

H atoms

X=2a------------------------------(2

O atoms

4+4y=a+2b---------------------(3

C atoms

2y=b-----------------------------(4

Substitute 2x(4 in (3

4+2b=a+2b

Hence a=4

From (2

x=8

from (1

8-2y=3

y=5/2

and from (4

b=5

substitute the values back into the equation.

MnO4– + 8H+ + 5/2(C2O4)2– --> Mn2+ + 4H2O + 5CO2

Or

2MnO4– + 16H+ + 5(C2O4)2– --> 2Mn2+ + 8H2O + 10CO2