Acidic Solution. Example #1: ClO3 + SO2 ---> SO4 2 + Cl. Solution: 1) Split into unbalanced half-reactions: ClO3 ---> Cl SO2 ---> SO4 2

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1 Balance the following REDOX reactions in an acidic solution: 1. ClO3 + SO2 ---> SO4 2 + Cl 2. H2S + NO3 ---> S8 + NO 3. MnO4 + H2S ---> Mn 2+ + S8 4. Cu + SO > Cu 2+ + SO2 5. MnO4 + CH3OH ---> HCOOH + Mn 2+ Balance the following REDOX reactions in a basic solution: 1. NH3 + ClO ---> N2H4 + Cl 2. Au + O2 + CN ---> Au(CN)2 + H2O2 3. Br + MnO4 ---> MnO2 + BrO3 4. AlH4 + H2CO ---> Al 3+ + CH3OH 5. Se + Cr(OH)3 ---> Cr + SeO3 2 Acidic Solution Example #1: ClO3 + SO2 ---> SO4 2 + Cl 1) Split into unbalanced half-reactions: ClO3 ---> Cl SO2 ---> SO4 2 2) Balance the half-reactions: 6e + 6H + + ClO3 ---> Cl + 3H2O 2H2O + SO2 ---> SO H + + 2e 3) Make the number of electrons equal: 6e + 6H + + ClO3 ---> Cl + 3H2O 6H2O + 3SO2 ---> 3SO H + + 6e <--- multiplied through by a factor of 3 4) Add the two half-reactions for the final answer: ClO3 + 3H2O + 3SO2 ---> 3SO4 2 + Cl + 6H +

2 Note that items duplicated on each side were cancelled out. The duplicates are 6e, 3H2O, and 6H + Example #2: H2S + NO3 ---> S8 + NO 1) The unbalanced half-reactions: H2S ---> S8 NO3 ---> NO 2) balance each half-reaction: 8H2S ---> S8 + 16H e 3e + 4H + + NO3 ---> NO + 2H2O 3) Make the number of electrons equal: 24H2S ---> 3S8 + 48H e <--- multiplied by a factor of 3 48e + 64H NO3 ---> 16NO + 32H2O <--- multiplied by a factor of 16 Note that 16 and 3 have no common factors except 1, so both 16 and 3 had to be used to obtain the lowest common multiple of 48 for the number of electrons. 4) Add: 24H2S + 16H NO3 ---> 3S8 + 16NO + 32H2O Comment: removing a factor of 8 does look tempting, doesn't it? However, the three in front of the S8 (or the five in the next example) makes it impossible. Also, note that duplicates of 48 electrons and 48 hydrogen ions were removed. Example #3: MnO4 + H2S ---> Mn 2+ + S8 1) Half-reactions:

3 H2S ---> S8 MnO4 ---> Mn 2+ 2) Balance: 8H2S ---> S8 + 16H e 5e + 8H + + MnO4 ---> Mn H2O 3) Make the number of electrons equal (note that there are no common factors between 5 and 16 except 1): 40H2S ---> 5S8 + 80H e <--- factor of 5 80e + 128H MnO4 ---> 16Mn H2O <--- factor of 16 4) The final answer: 40H2S + 48H MnO4 ---> 5S8 + 16Mn H2O Another possibility of removing a factor of 8 destroyed by an odd number, in this case, the 5 in front of the S8. Curses, foiled again! Example #4: Cu + SO > Cu 2+ + SO2 1) The unbalanced half-reactions: Cu ---> Cu 2+ SO > SO2 2) The balanced half-reactions: Cu ---> Cu e 2e + 4H + + SO > SO2 + 2H2O Cu + 4H + + SO > Cu 2+ + SO2 + 2H2O No need to equalize electrons since it turns out that, in the course of balancing the half-reactions, the electrons are equal in amount. Note how easy it was to balance the copper half-reaction. All you needed were the two electrons.

4 Example #5: MnO4 + CH3OH ---> HCOOH + Mn 2+ 1) The balanced half-reactions: 5e + 8H + + MnO4 ---> Mn H2O H2O + CH3OH ---> HCOOH + 4H + + 4e 2) Equalize electrons: 20e + 32H + + 4MnO4 ---> 4Mn H2O <--- factor of 4 5H2O + 5CH3OH ---> 5HCOOH + 20H e <--- factor of 5 12H + + 5CH3OH + 4MnO4 ---> 5HCOOH + 4Mn H2O Basic Solution Example #1: NH3 + ClO ---> N2H4 + Cl 1) The two half-reactions, balanced as if in acidic solution: 2NH3 ---> N2H4 + 2H + + 2e 2e + 2H + + ClO ---> Cl + H2O 2) Electrons already equal, convert to basic solution: 2OH + 2NH3 ---> N2H4 + 2H2O + 2e 2e + 2H2O + ClO ---> Cl + H2O + 2OH Comment: that's 2 OH, not 20 H. Misreading the O in OH as a zero is a common mistake. 2HN3 + ClO ---> N2H4 + Cl + H2O

5 Notice that no hydroxide appears in the final answer. That means this is a basecatalyzed reaction. For the reaction to occur, the solution must be basic and hydroxide IS consumed. It is just regenerated in the exact same amount, so it cancels out in the final answer. Example #2: Au + O2 + CN ---> Au(CN)2 + H2O2 1) the two half-reactions, balanced as if in acidic solution: 2CN + Au ---> Au(CN)2 + e 2e + 2H + + O2 ---> H2O2 2) Make electrons equal, convert to basic solution: 4CN + 2Au ---> 2Au(CN)2 + 2e <--- multiplied by a factor of 2 2e + 2H2O + O2 ---> H2O2 + 2OH 4CN + 2Au + 2H2O + O2 ---> 2Au(CN)2 + H2O2 + 2OH Comment: the CN is neither reduced nor oxidized, but it is necessary for the reaction. For example, you might see this way of writing the problem: Au + O2 ---> Au(CN)2 + H2O2 Notice that CN does not appear on the left side, but does so on the right. Since you MUST balance the equation, that means you are allowed to use CN in your balancing. An important point here is that you know the cyanide polyatomic ion has a negative one charge. Example #3: Br + MnO4 ---> MnO2 + BrO3 1) The two half-reactions, balanced as if in acidic solution:

6 3H2O + Br ---> BrO3 + 6H + + 6e 3e + 4H + + MnO4 ---> MnO2 + 2H2O 2) Make the number of electrons equal: 3H2O + Br ---> BrO3 + 6H + + 6e 6e + 8H + + 2MnO4 ---> 2MnO2 + 4H2O <--- multiplied by a factor of 2 3) Convert to basic solution, by adding 6OH to the first half-reaction and 8OH to the second: 6OH + Br ---> BrO3 + 3H2O + 6e 6e + 4H2O + 2MnO4 ---> 2MnO2 + 8OH 4) The final answer: H2O + 2MnO4 + Br ---> 2MnO2 + BrO3 + 2OH 5) What happens if you add the two half-reactions without converting them to basic? You get this: 2H + + 2MnO4 + Br ---> 2MnO2 + BrO3 + H2O Then, add 2OH to each side: 2H2O + 2MnO4 + Br ---> 2MnO2 + BrO3 + H2O + 2OH Eliminate one water for the final answer: H2O + 2MnO4 + Br ---> 2MnO2 + BrO3 + 2OH The answer to the question? Nothing happens. You get the right answer if convert before adding the half-reactions or after. There will even be cases where balancing one half-reaction using hydroxide can easily be done while the other half-reaction gets balanced in acidic solution before converting. You can add the two half-reactions while one is basic and one is acidic, then convert after the adding (see below for an example of this). Example #4: AlH4 + H2CO ---> Al 3+ + CH3OH

7 1) The two half-reactions, balanced as if in acidic solution: AlH4 ---> Al H + + 8e 2e + 2H + + H2CO ---> CH3OH 2) Converted to basic by addition of hydroxide, second half-reaction multiplied by 4 (note that the hydrogen is oxidized from -1 to +1): 4OH + AlH4 ---> Al H2O + 8e 8e + 8H2O + 4H2CO ---> 4CH3OH + 8OH AlH4 + 4H2O + 4H2CO ---> Al CH3OH + 4OH Example #5: Se + Cr(OH)3 ---> Cr + SeO3 2 1) The unbalanced half-reactions: Se ---> SeO3 2 Cr(OH)3 ---> Cr 2) Note that only the first half-reaction is balanced using the balance-first-in-acid technique, the second is balanced using hydroxide: Se + 3H2O ---> SeO H + + 4e 3e + Cr(OH)3 ---> Cr + 3OH 3) Convert the first half-reaction by adding 6 hydroxide to each side, eliminate duplicate waters, then make the electrons equal (factor of 3 for the first half-reaction and a factor of 4 for the second). The final answer: 6OH + 3Se + 4Cr(OH)3 ---> 4Cr + 3SeO H2O 4) What would happen if we didn't make the first half-reaction basic and just added them? first, make the electrons equal: 3Se + 9H2O ---> 3SeO H e 12e + 4Cr(OH)3 ---> 4Cr + 12OH

8 then, add: 3Se + 4Cr(OH)3 + 9H2O ---> 4Cr + 3SeO H OH combine hydrogen ion and hydroxide ion on the right-hand side: 3Se + 4Cr(OH)3 + 9H2O ---> 4Cr + 3SeO H H2O eliminate water: 3Se + 4Cr(OH)3 ---> 4Cr + 3SeO H + + 3H2O add six hydroxides: 6OH + 3Se + 4Cr(OH)3 ---> 4Cr + 3SeO H2O Note that I combined the H + and the OH to make six waters and then added it to the three waters that were already there.

Directions: Use the rules for Assigning Oxidation numbers to determine the oxidation number assigned to each element in each of the given formulas.

Oxidation Numbers #00 Directions: Use the rules for Assigning Oxidation numbers to determine the oxidation number assigned to each element in each of the given formulas. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.