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
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.
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