1 / 21

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.

kameko-best
Download Presentation

Balancing Acidic Redox Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related