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-
Thank you!!
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.
To balance the oxidation-reduction equation in a basic solution for the reaction MnO4- + NO2- --> MnO2 + NO3-, you can follow these steps:
1. Write the skeleton equation:
MnO4- --> MnO2
2. Identify the elements that changed their oxidation state and place that number just above the element. In this case, Mn changes from +7 on the left side to +4 on the right side.
3. Add electrons (e-) to the appropriate side to balance the change in oxidation state. In this case, Mn changes by 3 electrons:
MnO4- + 3e- --> MnO2
4. Count up the charge on both sides. On the left side, there is a -1 charge from MnO4-. On the right side, there is no charge. To balance the charge, add OH- ions to the side needing more electrons. In this case, add 4 OH- ions to the right side:
MnO4- + 3e- + 4 OH- --> MnO2
5. Now, add water (H2O) to balance the hydrogen atoms on both sides. In this case, add 2 water molecules to the left side:
MnO4- + 3e- + 4 OH- + 2 H2O --> MnO2
6. Finally, check if the equation is balanced. Verify if the number of each element is the same on both sides, if the charges balance, and if the change in oxidation state balances with the number of electrons added. In this case, there is 1 Mn on both sides, 6 O on both sides, and 4 H on both sides. The charges also balance (-1 on the left, 0 on the right). The change in oxidation state is balanced by the 3 electrons added.
Therefore, the balanced equation in a basic solution for the reaction MnO4- + NO2- --> MnO2 + NO3- is:
MnO4- + NO2- + 3 OH- --> MnO2 + NO3- + 2 H2O
Remember that this method can also be applied to balancing oxidation-reduction equations in acidic solutions. The only difference is that instead of adding OH- ions, you would add H+ ions to balance the charges.