I am having trouble with the balancing the oxidation-reduction equations that are in a basic solution.
This is the first one that I'm doing that's in a basic solution, and I can't get it! I get everything right except I get 2H2O instead of one on the left.
MnO4- + NO2- --> MnO2 + NO3-
The method I use is longer than most but it has the advantage of also drilling students on oxidation state. But this method works.
1. Write the skeleton equation.
MnO4- ==> MnO2
2. Idendity the elements that changed oxidation state and place that number just above the element. Mn is +7 on the left and +4 on the right. I can't put those numbers on the ocmputer but you can on a sheet of paper.
3. Add electrons to the appropriate side to balance the change in oxidation state.
MnO4- + 3e==> MnO2
4. Count up the charge on both sides. I see -4 on the left and zero on the right. Now add OH^- to balance the charge.
MnO4- + 3e ==> MnO2 + 4 OH-
5. Now add water to balance the H atoms.
MnO4- + 3e + 2H2O ==> MnO2 + 4 OH-
6. Now check it for three things.
a. do the atoms balance? I see 1 Mn on the left and 1 on the right. There are 6 O on the left and 6 on the right. There are 4 H on the left and 4 on the right.
b. does the charge balance.
I see -4 on the left and -4 on the right.
c. Does the change in oxidatin state balance with the electrons that were added. Mn(+7) + 3e ==> Mn(+4). yes.
If it balances with a,b, and c, we were a success. I hope this helps. By the way, the procedure, except for a slight modification, works very well for acid solutions, too. On step 4, count up the charge and add H+ to balance. Everything else stays the same.
I guess it's obvious that I just balanced half of the equation. You must go through the NO3-==>NO2- part, make the electron change equal, and add the balanced half cells.