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Balancing Acidic Redox Reactions

Balancing Acidic Redox Reactions. Step 1: Assign oxidation numbers to all elements in the reaction. MnO 4 1 + SO 2  Mn +2 + SO 4 2. +7. 2. +4. 2. +2. +6. 2. Step 2: List the changes in oxidation numbers. MnO 4 1 + SO 2  Mn +2 + SO 4 2. +7. 2. +4. 2.

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Balancing Acidic Redox Reactions

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  1. Balancing Acidic Redox Reactions

  2. Step 1: Assign oxidation numbers to all elements in the reaction. MnO41 + SO2  Mn+2 + SO42 +7 2 +4 2 +2 +6 2

  3. Step 2: List the changes in oxidation numbers. MnO41 + SO2  Mn+2 + SO42 +7 2 +4 2 +2 +6 2 change Mn +7  +2 5 S +4  +6 +2

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

  5. Step 4: Label the oxidizing and reducing agents. MnO41 + SO2  Mn+2 + SO42 oxidizing agent reducing agent change reduced Mn +7  +2 5 oxidized S +4  +6 +2

  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. change reduced Mn +7  +2 5 2 = 10 oxidized S +4  +6 +2 5 = +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. It is necessary to multiply the manganeses by 2 and the sulfurs by 5. 2 5 2 5 MnO41 + SO2  Mn+2 + SO42

  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 MnO41 + 5 SO2  2 Mn+2 + 5 SO42

  9. Step 8: Balance the charge. Part a: Multiply the coefficient by the charge on the ion or molecule. 2 MnO41 + 5 SO2  2 Mn+2 + 5 SO4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 MnO41 + 5 SO2  2 Mn+2 + 5 SO42 2  6 + 4 H+1 2  2 2 MnO41 + 5 SO2  2 Mn+2 + 5 SO42 + 4 H+1

  11. Step 9: Count the hydrogens and oxygens on each side of the equation. 2 MnO41 + 5 SO2  2 Mn+2 + 5 SO42 + 4 H+1 H  HO  O 0 4 18 20

  12. Step 10: Balance the hydrogens and oxygens by adding water molecules. 2 MnO41 + 5 SO2  2 Mn+2 + 5 SO42 + 4 H+1 0 H  4 H18 O  20 O 2 H2O +

  13. Step 11: Re-write the equation and box the entire balanced reaction. 2 MnO41 + 5 SO2 + 2 H2O  2 Mn+2 + 5 SO4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. MnO41 + SO2  Mn+2 + SO42 MnO41  Mn+2 SO2  SO42

  16. Step 2: Balance all of the atoms except H and O. For an acidic solution, next add H2O to balance the O atoms and H+1 to balance the H atoms. In a basic solution, we would use OH-1 and H2O 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!! MnO41  Mn+2 SO2  SO42 8 H+1 + + 4 H2O 2 H2O + + 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 + MnO41  Mn+2 + 4 H2O (+8) + (1)  (+2) (+7)  (+2)2 H2O + SO2  SO4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 + MnO41  Mn+2 + 4 H2O (+8) + (1)  (+2) (+7)  (+2)2 H2O + SO2  SO42 + 4 H+1 (0)  (2) + (+4) (0)  (+2) 10 5 e1 + 2 2 16 8 5 5 20 10 + 2 e1 10

  19. Step 5: Add the two half-reactions. 10 e 1+ 16 H+1 + 2 MnO41  2 Mn+2 + 8 H2O10 H2O + 5 SO2  5 SO42 + 20 H+1 +10 e 1 16 H+1 + 2 MnO41 +10 H2O + 5 SO2  2 Mn+2 + 8 H2O + 5 SO42 + 20 H+1

  20. Step 6: Get the overall equation by canceling out the electrons and H2O, H+1, and OH-1 that may appear on both sides of the equation: 16 H+1 + 2 MnO41 +10 H2O + 5 SO2  2 Mn+2 + 8 H2O + 5 SO42 + 20 H+1 becomes 2 MnO41 +2 H2O + 5 SO2  2 Mn+2 + 5 SO42 + 4 H+1

  21. Balance the following using the half-reaction method: a. Br 1 + MnO41  Br2 + Mn+2 16 H+1 + 10 Br 1 + 2 MnO41  5 Br2 + 2 Mn+2 + 8 H2O b. As2O3 + NO31  H3AsO4 + NO 4 H+1 + 7 H2O + 4 NO31 + 3 As2O3  6 H3AsO4 + 4 NO

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