Posted by humaira on Friday, April 17, 2009 at 4:40am.
i really would love to help but i seriously don't know anything.
1. Pick MnO4^- ==>Mn^+2 in acid solution.
2. Pick what is changing oxidation state. That is Mn.
3. Mn goes from an oxidation state of +7 on the left to +2 on the right. Add electrons to the appropriate side to balance the change in oxidation state.
MnO4^- + 5e ==> Mn^+2
4. Count the charge on the left and right. On the left we have a charge of -6 and on the right +2.
a. If acid solution, add H^+ to balance the charge.
b. If basic solution, add OH^- to balance the charge.
This is an acid solution so we add H^+ to the left side to balance the charge.
MnO4^- + 5e + 8H^+ ==> Mn^+2
Adding the 8H^+ gives a +2 charge on both sides.
5. Now add H2O (usually to the opposite side) to balance the added H^+.
MnO4^- + 5e + 8H^+ ==> Mn^+2 + 4H2O
which usually balances the oxygen (but not always).
6. Check it to make sure it balances for
a. atoms
b. charge
c. oxidation change.
That is the process.
7. You do the C2O4^= to CO2 the same way but its much simpler.
C2O4^= ==> 2CO2 + 2e
8. You want the electrons to cancel. Note that the electrons are on different sides. Since they are to cancel, we multiply equation 1 by 2 (to make 10 e) and we multiply equation 2 by 5 (to make 10 e).
9. Add the multiplied equations, cancel anything that appears on both sides and you have the final equation. It looks like this.
2MnO4^- + 16H^+ + 10e + 5C2O4^= ==> 2Mn^+2 + 8H2O + 10CO2 + 10e.
10. The electrons cancel. Now check it.
I see 2 Mn on left and 2 on right.
I see 28 O on the left and 28 on right.
I see 16 H on the left and 16 on right.
I see 10 C on the left and 10 on right.
I see +4 charge on left and +4 on right.
Check my work.