22sec3
- 1. 22.3 Balancing Redox Reactions
- 2. Identifying Redox Equations In general, all chemical reactions can be assigned to one of two classes: oxidation-reduction, in which electrons are transferred: Single-replacement, combination, decomposition, and combustion this second class has no electron transfer, and includes all others: Double-replacement and acid-base reactions
- 3. Identifying Redox Equations In an electrical storm, oxygen and nitrogen react to form nitrogen monoxide: N 2(g) + O 2(g) -> 2NO (g) Is this a redox reaction? If the oxidation number of an element in a reacting species changes, then that element has undergone either oxidation or reduction; therefore, the reaction as a whole must be a redox. YES!
- 4. Problem Use the change in oxidation number to identify whether a reaction is redox. Identify the element reduced and oxidized. Cl 2(g) 2NaBr (aq) -------2NaCl (aq) + Br 2(g) Cl goes from 0 to -1. so it is reduced Br goes from -1 to 0, so it is oxidized It is a redox reaction
- 5. Balancing Redox Equations It is essential to write a correctly balanced equation that represents what happens in a chemical reaction Fortunately, two systematic methods are available, and are based on the fact that the total electrons gained in reduction equals the total lost in oxidation. The two methods: Use oxidation-number changes Use half-reactions
- 6. Using Oxidation-Number Changes Sort of like chemical bookkeeping, you compare the increases and decreases in oxidation numbers. start with the skeleton equation Step 1 : assign oxidation numbers to all atoms; write above their symbols Step 2 : identify which oxidized/reduced Step 3 : use bracket lines to connect Step 4: use coefficients to equalize Step 5: make sure they are balanced for both atoms and charge
- 7. Problem Balance, using oxidation numbers, the redox reaction below: K 2 Cr 2 O 7(aq) + H 2 O (l) + S (s) --------- KOH (aq) + Cr 2 O 3(s) + SO 2(g) Step 1: +1 +6 -2 +1 -2 0 K 2 Cr 2 O 7(aq) + H 2 O (l) + S (s) --------- +1-2+1 +3 -2 +4-2 KOH (aq) + Cr 2 O 3(s) + SO 2(g)
- 8. Problem con’td Step 2: Cr is reduced S is oxidized Step 3: -3 K 2 Cr 2 O 7(aq) + H 2 O (l) + S (s) --- KOH (aq) + Cr 2 O 3(s) + SO 2(g) +4
- 9. Step 4: -3(4)=-12 2 K 2 Cr 2 O 7(aq) + H 2 O (l) + 3 S (s) --- KOH (aq) + 2 Cr 2 O 3(s) + 3 SO 2(g) +4(3)=+12 Step 5: 2 K 2 Cr 2 O 7(aq) + 2 H 2 O (l) + 3 S (s) --- 4 KOH (aq) + 2 Cr 2 O 3(s) + 3 SO 2(g)
- 10. Using half-reactions A half-reaction is an equation showing just the oxidation or just the reduction that takes place they are then balanced separately, and finally combined Step 1: write unbalanced equation in ionic form Step 2: write separate half-reaction equations for oxidation and reduction Step 3: balance the atoms in the half-reactions
- 11. Using half-reactions continued Step 4: add enough electrons to one side of each half-reaction to balance the charges Step 5: multiply each half-reaction by a number to make the electrons equal in both Step 6: add the balanced half-reactions to show an overall equation Step 7: add the spectator ions and balance the equation Rules shown on page 665 – bottom Sample Problem 22.7 pg 667
- 12. Problem Write the balanced ionic equation for the following equations which occur in acid solution. Use the half-reaction method. Sn 2+ (aq) + Cr 2 O 7 2- (aq) ----Sn 4+ (aq) + Cr 3+ (aq)
- 13. Problem Step 2: Sn 2+ ----Sn 4+ Cr 2 O 7 2- ----Cr 3+ Step 3 Sn 2+ ----Sn 4+ Cr 2 O 7 2- + 14H + ----2Cr 3+ + 7H 2 O Step 4 Sn 2+ ----Sn 4+ + 2e - Cr 2 O 7 2- + 14H + + 12e - ---2Cr 3+ + 7H 2 O
- 14. Problem Step 5 3Sn 2+ ----3Sn 4+ + 12e - Cr 2 O 7 2- + 14H + + 12e - ---2Cr 3+ + 7H 2 O Step 6 3Sn 2+ + Cr 2 O 7 2- + 14H + + ------ 2Cr 3+ + 7H 2 O + 3Sn 4+ Step 7 3Sn 2+ + Cr 2 O 7 2- + 14H + + ------ 2Cr 3+ + 7H 2 O + 3Sn 4+
- 15. Choosing a Balancing Method The oxidation number change method works well if the oxidized and reduced species appear only once on each side of the equation, and there are no acids or bases. The half-reaction method works best for reactions taking place in acidic or alkaline solution.