1. Balancing Acidic Redox Reactions

2. Step 1: Assign oxidation numbers to all elements in the reaction. MnO 4  1 + SO 2  Mn +2 + SO 4  2 22 22 22 +2+4+7+6

3. Step 2: List the changes in oxidation numbers. MnO 4  1 + SO 2  Mn +2 + SO 4  2 22 22 22 +2+4+7 +6 Mn+7  +2 S change 55 +4  +6+2

4. Step 3: Label the species being oxidized and reduced. Mn +7  +2  5 change S +4  +6 +2 oxidized reduced

5. Step 4: Label the oxidizing and reducing agents. Mn +7  +2  5 change S +4  +6 +2 oxidized reduced MnO 4  1 + SO 2  Mn +2 + SO 4  2 oxidizing agent reducing agent

6. Step 5: Balance the change. This is done by multiplying each change by a value to attain the least common multiple of the two numbers. reduced Mn +7  +2  5 change oxidized S +4  +6 +2 22 55 = =  10 +10 0

7. Step 6: Using the values chosen to balance the change, add coefficients to the particular species. Note: take into account any subscripts present. MnO 4  1 + SO 2  Mn +2 + SO 4  2 5522 It is necessary to multiply the manganeses by 2 and the sulfurs by 5.

8. Step 7: Make sure that all elements (with the exception of hydrogen and oxygen) are balanced. Add coefficients as necessary to balance extra elements. Note: If hydrogen or oxygen is the species being oxidized or reduced, it must be balanced at this step. 2 MnO 4  1 + 5 SO 2  2 Mn +2 + 5 SO 4  2

9. Step 8: Balance the charge. Part a: Multiply the coefficient by the charge on the ion or molecule. 2 MnO 4  1 + 5 SO 2  2 Mn +2 + 5 SO 4  2 (2)(  1) + (5)(0)  (2)(+2) + (5)(  2) (  2) + (0)  (+4) + (  10)  2   6

10. Step 8: Balance the charge. Part b: Add hydrogen ions to account for the extra charges. 2 MnO 4  1 + 5 SO 2  2 Mn +2 + 5 SO 4  2  2   6 + 4 H +1  2   2 2 MnO 4  1 + 5 SO 2  2 Mn +2 + 5 SO 4  2 + 4 H +1

11. Step 9: Count the hydrogens and oxygens on each side of the equation. 2 MnO 4  1 + 5 SO 2  2 Mn +2 + 5 SO 4  2 + 4 H +1 H  H O  O 04 18 20

12. Step 10: Balance the hydrogens and oxygens by adding water molecules. 2 MnO 4  1 + 5 SO 2  2 Mn +2 + 5 SO 4  2 + 4 H +1 0 H  4 H 18 O  20 O 2 H 2 O +

13. Step 11: Re-write the equation and box the entire balanced reaction. 2 MnO 4  1 + 5 SO 2 + 2 H 2 O  2 Mn +2 + 5 SO 4  2 + 4 H +1

14. It will be preferable for you to use the following method – called the half-reaction method of balancing redox equations. Given your prior knowledge and understanding, this will be much easier!

15. Step 1: Given the reaction to balance, separate the two half-reactions. MnO 4  1  Mn +2 SO 2  SO 4  2 MnO 4  1 + SO 2  Mn +2 + SO 4  2

16. Step 2: Balance all of the atoms except H and O. For an acidic solution, next add H 2 O to balance the O atoms and H +1 to balance the H atoms. In a basic solution, we would use OH -1 and H 2 O to balance the O and H. Be careful: On this example the atoms except H and O are already balanced. Most of the time, they won’t already be balanced. WATCH OUT. Do that first!! MnO 4  1  Mn +2 SO 2  SO 4  2 + 4 H 2 O8 H +1 + 2 H 2 O ++ 4 H +1

17. Step 3: Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is accomplished by adding electrons to the reactions: 8 H +1 + MnO 4  1  Mn +2 + 4 H 2 O (+8) + (  1)  (+2) (+7)  (+2) 2 H 2 O + SO 2  SO 4  2 + 4 H +1 (0)  (  2) + (+4) (0)  (+2) 5 e  1 + + 2 e  1

18. Step 4: Now multiply the oxidations numbers so that the two half-reactions will have the same number of electrons and can cancel each other out: (Remember the LCM?? Multiply the first reaction by 2 and the second reaction by 5.) 8 H +1 + MnO 4  1  Mn +2 + 4 H 2 O (+8) + (  1)  (+2) (+7)  (+2) 2 H 2 O + SO 2  SO 4  2 + 4 H +1 (0)  (  2) + (+4) (0)  (+2) 5 e  1 + + 2 e  1 10 16 22 8 10 5520 10

19. Step 5: Add the two half-reactions. 10 e  1 + 16 H +1 + 2 MnO 4  1  2 Mn +2 + 8 H 2 O 10 H 2 O + 5 SO 2  5 SO 4  2 + 20 H +1 + 10 e  1 16 H +1 + 2 MnO 4  1 +10 H 2 O + 5 SO 2  2 Mn +2 + 8 H 2 O + 5 SO 4  2 + 20 H +1

20. Step 6: Get the overall equation by canceling out the electrons and H 2 O, H +1, and OH -1 that may appear on both sides of the equation: 16 H +1 + 2 MnO 4  1 +10 H 2 O + 5 SO 2  2 Mn +2 + 8 H 2 O + 5 SO 4  2 + 20 H +1 becomes 2 MnO 4  1 +2 H 2 O + 5 SO 2  2 Mn +2 + 5 SO 4  2 + 4 H +1

21. Balance the following using the half-reaction method: a. Br  1 + MnO 4  1  Br 2 + Mn +2 b. As 2 O 3 + NO 3  1  H 3 AsO 4 + NO 16 H +1 + 10 Br  1 + 2 MnO 4  1  5 Br 2 + 2 Mn +2 + 8 H 2 O 4 H +1 + 7 H 2 O + 4 NO 3  1 + 3 As 2 O 3  6 H 3 AsO 4 + 4 NO