1. 6.1.3 REDOX REACTIONS

2. Oxidation numbers identify and indicate which element is oxidized and which is reduced. Here's an example - the reaction between sodium metal and chlorine gas: It is often useful to write the oxidation number for every element above the element in the equation. Thus for our reaction we have: balancing coefficients in the equation do not affect the value of the oxidation numbers. 2 Na+Cl 2 →2 NaCl 00 +1 -1 2 Na+Cl 2 →2 NaCl

3. elemen t initial ox no final ox no change in electrons (e - ) oxidized or reduced Na0→+1lost 1 e - oxidized Cl0→gain 1 e reduced

4. EXAMPLE 2 Mg+O2O2 →2 MgO Add in the oxidation numbers which are oxidized and which are reduced? An increase in oxidation number indicates oxidation A decrease in oxidation number indicates reduction

5. Reducing agent the substance that is oxidized. It allows another element to be reduced. Oxidizing agent the substance that is reduced. It allows another element to be oxidized.

6. N2N2 +2H 2 →2 NH 3 EXAMPLE element initial ox no final ox no e-e- oxidized or reduced Agent N0→-3gain 3e - reducedoxidizing agent: N 2 H0→+1lose 1e - oxidizedreducing agent: H 2

7. ASSIGNMENTS Practice 6.1.3 Assignment 6.1.3

8. BALANCING REDOX REACTIONS Many redox reactions cannot easily be balanced just by counting atoms. Consider the following net ionic equation Cu (s) + Ag + (aq) → Cu 2+ (aq) + Ag (s) If you simply count atoms, the equation is balanced but the charges aren’t balanced! Charges represent gain or loss of electrons, and, like atoms, electrons are conserved during chemical reactions.

9. 1. BALANCING EQUATIONS USING OXIDATION NUMBERS Cu (s) + Ag + (aq) → Cu 2+ (aq) + Ag (s 1 Cu (s) + 2Ag + (aq) → 1 Cu 2+ (aq) + 2Ag (s) element initial ox no final ox no change in e - balance for electrons Cu0→+2lost 2× 1=2 Ag+1→0gain 1× 2=2

10. EXAMPLE MnO 4 1- + Fe 2+ + H 1+ → Mn 2+ + Fe 3+ + H 2 O 1 MnO 4 1- + 5 Fe 2+ + H 1+ → 1 Mn 2+ + 5 Fe 3+ + H 2 O H and O are not balanced 1 MnO 4 1- + 5 Fe 2+ + 8 H 1+ → 1 Mn 2+ + 5 Fe 3+ + 4 H 2 O element initial ox no final ox no change in e - balance for electrons Mn+7→+25× 1=5 Fe+2→+31× 5=5

11. EXAMPLE NH 3 + O 2 → NO 2 + H 2 O Before using a multiplier to get the electrons to match, notice the subscript with oxygen - O 2. In our summary chart we base our oxidation number changes on a single atom, but our formula tells us that we must have at least two oxygen. You will save some time and frustration if we take this into account now. So in our summary table we will add some columns to change our minimum number of atoms and electrons involved. Then we complete the chart: element initial ox no final ox no change in e - balance for electrons N-3→+47 O0→-22

12. el e m e nt initial ox no final ox no change in e - No. atom s No. e - balance for electrons N-3→+47x1=7× 4= 2828 O0→-22× 2=4× 7= 2828

13. Since we were counting oxygen atoms in the O 2 molecule on the reactant side of the equation, that's where we'll use the "7". Since nitrogen's oxidation number also changed we will use nitrogen's balancing coefficient there 4 NH 3 + 7 O 2 → 4 NO 2 + H 2 O The last step is to balance for hydrogen atoms (and finishing oxygen), which will mean placing a 6 in front of H 2 O: 4 NH 3 + 7 O 2 → 4 NO 2 + 6 H 2 O

14. K 2 Cr 2 O 7 + NaI + H 2 SO 4 → Cr 2 (SO 4 ) 3 + I 2 + H 2 O + Na 2 SO 4 + K 2 SO 4 Your first concern is to make sure you correctly determine all oxidation numbers. Since the only place you see sulfur in this reaction is in SO 4 2-, sulfur's oxidation number is not going to change. Similarly, hydrogen and oxygen are always in compounds, so their oxidation numbers also won't change during the reaction. That narrows down the list of elements to check.

15. K 2 Cr 2 O 7 + NaI + H 2 SO 4 → Cr 2 (SO 4 ) 3 + I 2 + H 2 O + Na 2 SO 4 + K 2 SO 4 Next, check for any subscripts associated with either of these two elements - we see that Cr always has a subscript of "2" (in both K 2 Cr 2 O 7 and Cr 2 (SO 4 ) 3 ), and I has a subscript in I 2. So we'll add that to our summary chart to get a total number of electrons transferred, and then balance element initial ox no final ox no change in e - No. atoms No. e - balance for electrons Cr+6→+33 I+1→01

16. Our table now tells us to use a balancing coefficient of "1" for Cr on both sides of the equation and "3" for iodine. Since we counted the atoms in I 2 (and not HI), the "3" will go in front of I 2 : 1 K 2 Cr 2 O 7 + NaI + H 2 SO 4 → 1 Cr 2 (SO 4 ) 3 + 3 I 2 + H 2 O + Na 2 SO 4 + K 2 SO 4 With these numbers in place, we now balance for atoms in the remainder of the equation to get our final answer: 1 K 2 Cr 2 O 7 + 6 NaI + 7 H 2 SO 4 → 1 Cr 2 (SO 4 ) 3 + 3 I 2 + 7 H 2 O + 3 Na 2 SO 4 + 1 K 2 SO 4 element initial ox no final ox no change in e - No. atoms No. e - balance for electrons Cr+6→+33×2=6×1=6 I+1→01×2=2×3=6

17. ONE MORE EXAMPLE One more tricky one. Balance Zn + HNO 3 → Zn(NO 3 ) 2 + NO 2 + H 2 O Determine oxidation numbers and create your summary chart: The main thing to notice is that N appears in two separate products - Zn(NO 3 ) 2 and NO 2. Should we consider the subscript for nitrogen from Zn(NO 3 ) 2 ? In this case no, because this compound also contains Zn, the oxidized element. Also, the oxidation number for nitrogen does not change from HNO 3 to Zn(NO 3 ) 2. element initial ox no final ox no change in e - No. atoms No. e - balance for electrons Zn0→+22 N+5→+41

18. We now get our balancing coefficients from our summary table. A "1" will be placed in front of Zn, but which N should we use for the "2"? If you put it in front of both HNO 3 and NO 2 you'll find you cannot balance for nitrogen atoms. Since the oxidation number for nitrogen changed in becoming NO 2, we will try it there first. Some trial-and-error may be required: 1 Zn + HNO 3 → 1 Zn(NO 3 ) 2 + 2 NO 2 + H 2 O With the 2 in place in front of NO 2, we can now balance the rest of the equation for atoms. Doing so gives us the final answer: 1 Zn + 4 HNO 3 → 1 Zn(NO 3 ) 2 + 2 NO 2 + 2 H 2 O

19. PRACTICE PROBLEMS Practice problems 6.1.5- question #1

20. 6.1.5- HALF REACTIONS Another way to balance redox reactions is by the half- reaction method. This technique involves breaking an equation into the oxidation reaction and the reduction reaction. The general technique involves the following: The overall equation is broken down into two half-reactions. If there are any spectator ions, they are removed from the equations. Each half-reaction is balanced separately -atoms and then charge. Electrons are added to one side of the equation to balance charge. Next the two equations are compared to make sure electrons lost equal electrons gained. One will be an oxidation reaction, the other will be a reduction reaction. Finally the two half-reactions are added together, and any spectator ions that were removed are placed back into the equation

21. Mg (s) + Cl 2 (g) → MgCl 2 (s) In this reaction, Mg is oxidized and Cl is reduced (Mg changes from 0 to +2; Cl changes from 0 to -1) Balance the two reactions for atoms. Next balance the equations for charge by adding electrons Mg → Mg +2 Cl 2 → 2 Cl - Mg → Mg +2 + 2 e - Cl 2 + 2 e - → 2 Cl - oxidationreduction

22. Finally, add the two equations together: Mg + Cl 2 → Mg +2 + 2 Cl - and reform any compounds broken apart in the earlier steps: Mg + Cl 2 → MgCl 2 We see that the original equation was already balanced, not just for atoms but for electrons as well

23. Cu (s) + AgNO 3 (aq) → Cu(NO 3 ) 2 (aq) + Ag (s) Identify the elements undergoing oxidation (Cu) and reduction (Ag). The nitrate group (NO 3 ) is a spectator ion which we will not include in our half-reactions. After balancing for atoms and for charge, we see that the two equations do not have the same number of electrons - there are 2 in the copper reaction but only one in the silver reaction. Multiply everything in the silver reaction by 2, then we will add the equations together: Cu → Cu +2 + 2 e - Ag + + 1 e - → Ag oxidationreduction

24. Step 1Step 2Step 3 Write the balanced half-reactions Balance electrons Add half-reactions Cu → Cu +2 + 2 e - Ag + 1 e - → Ag - × 22 Ag + + 2e - → 2 Ag Add equations together Cu + 2 Ag + → Cu +2 + 2 Ag Reform compound/return spectator ions Cu + 2 AgNO 3 → Cu(NO 3 ) 2 + 2 Ag

25. Here is a reaction occurring in an acid solution. MnO 4 - + Fe 2+ + H + → Mn 2+ + Fe 3+ + H 2 O In this example, spectator ions have already been removed. Even though hydrogen and oxygen do not undergo changes in oxidation number they are not spectators and we need to work with them in our half-reactions. We determine that Mn undergoes reduction (+7 to +2) while Fe undergoes oxidation (+2 to +3). The iron half-reaction is straight forward, but the manganese reaction is more complex - we must include hydrogen and oxygen in its half-reaction:

26. To balance the manganese half-reaction - first balance for Mn and O atoms. Next balance the H atoms, and finally add enough electrons to balance the charge on both sides of the equation. Fe 2+ → Fe +3 + 1e - MnO 4 - + 8 H + + 5 e - → Mn 2+ + 4 H 2 O oxidationreduction

27. Step 1Step 2Step 3 Write the balanced half-reactions Add half-reactions Fe 2+ → Fe +3 + 1e - ×5×5 5Fe 2+ → 5Fe +3 + 5e - MnO 4 - + 8H + + 5e - → Mn 2+ + 4H 2 O Add equations together MnO 4 - + 5 Fe 2+ + 8 H + → Mn 2+ + 5 Fe 3+ + 4H 2 O

28. HNO 3 + Cu + H + → NO 2 + Cu 2+ + H 2 O Determine what is oxidized, what is reduced, and write the two balanced half-reactions (Step 1) Balance for electrons lost = electrons gained (Step 2) Add equations together

29. Step 1 Step 2 Step 3 Write the balanced half-reactions Add half-reactions Cu → Cu +2 + 2e - HNO 3 + H + + 1 e - → NO 2 + H 2 O × 22HNO 3 + 2H + + 2e - → 2NO 2 + 2H 2 O Add equations together 2HNO 3 + Cu + 2H + → 2NO 2 + Cu 2+ + 2H 2 O

30. ASSIGNMENT Finish practice problems Assignment 6.1.5- hand-in When balancing redox reactions, either the oxidation number method or the half-reaction method may be used. Often you'll find that one method works best for some equations, while the other method is more suited for other reactions. Or you may find one method just easier to use. The practice exercises and assignments tell you which method to use for a reaction, but as you get get more experience you'll be able to make your own decision as to which method to use. Writing half-reactions, however, is a skill you will need for our final topic in this course - Electrochemistry - so be sure you can write balanced half-reactions.