1. UNIT V Balancing Redox Reactions and Redox Titrations

2. P REPARING A R EDOX T ABLE Example: Four metals A, B, C, & D were tested with separate solutions of A 2+, B 2+, C 2+ & D 2+. Some of the results are summarized in the following table: List the ions in order from the strongest to weakest oxidizing agent. MetalA 2+ B 2+ C 2+ D 2+ A (1) no reaction (2) reaction B (4) no reaction D (3) reaction

3. P REPARING A R EDOX T ABLE NOTE: For the same element: the more positive species is always the Oxidizing Agent. Ex. A 2+ vs. A Using data: 1. Since B 2+ does not oxidize A, B 2+ must be below A on the table. 2. Since C 2+ reacts with A, C 2+ must be above A. 3. Since A 2+ reacts with D, A 2+ must be above D on the table. But is D 2+ above or below B 2+ ? We don’t know yet. 4. D 2+ does not react with B. Now we know that D 2+ must be below B on the table.

4. P REPARING A R EDOX T ABLE So now we have our complete table: So now we have our answer; The ions in order of strongest to weakest oxidizing agent is: C 2+, A 2+, B 2+, D 2+ The order of reducing agent from strongest to weakest is: D, B, A, C. Oxidizing AgentsReducing Agents C 2+ + 2e - = C A 2+ + 2e - = A B 2+ + 2e - = B D 2+ + 2e - = D

5. P REPARING A R EDOX T ABLE Example: Four non-metal oxidizing agents X 2, Y 2, Z 2 and W 2 are combined with solutions of ions: X -, Y -, Z - and W -. The following results were obtained; X 2 reacts with W - and Y - only. Y - will reduce W 2 List the reducing agents from strongest to weakest

6. P REPARING A R EDOX T ABLE 1. X 2 will be above W - & Y -, but below Z - 2. Since Y - reduces W 2, Y - must be lower on the right of W 2. The reducing agents from strongest to weakest are: Y ‑, W ‑, X ‑, Z ‑

7. P REPARING A R EDOX T ABLE Example: Four solutions A(NO 3 ) 2, B(NO 3 ) 2, C(NO 3 ) 2, and D(NO 3 ) 2 are added to metals, A, B, C, & D. The following information is found: The metal A will not react with any of the solutions C(NO 3 ) 2 reacts spontaneously with B B will not react with D(NO 3 ) 2 Make a small reduction table showing reductions of the metallic ions. (Don’t forget to discard the spectator nitrate ions.)

8. P REPARING A R EDOX T ABLE a) List the oxidizing agents in order of strongest to weakest: _________________________ b) List the reducing agents in order of strongest to weakest: _________________________ c) Would it be safe to store A(NO 3 ) 2 solution in a container made of the metal D? ______________ Hebden Textbook p. 200 Questions #14-18

9. B ALANCING H ALF -R EACTIONS Some half-rxns are on the table, but not all! Given the solution: Is soln acidic or basic? Think: Major Hydroxide (Major  O  H  - charge) (Major atoms  atoms other than O & H)

10. B ALANCING H ALF -R EACTIONS Ex. S 2 O 8 2-  HSO 3 - (acid soln) 1. Balance Major Atoms (S in this case) S 2 O 8 2-  2HSO 3 - 2. Balance “O” atoms by adding H 2 O (to the side with less O’s) S 2 O 8 2-  2HSO 3 - + 2H 2 O 3. Balance “H” atoms by adding H + (to the side with less H’s) S 2 O 8 2- + 6H +  2HSO 3 - + 2H 2 O 4. Balance charge by adding e - ‘s (to the more + side) S 2 O 8 2- + 6H +  2HSO 3 - + 2H 2 O TIC = +4 TIC = -2

11. B ALANCING H ALF -R EACTIONS So the final balanced half-rxn is: S 2 O 8 2- + 6H + + 6 e -  2HSO 3 - + 2H 2 O TIC = -2 TIC = -2

12. B ALANCING H ALF -R EACTIONS Example: MnO 4 -  Mn 2+ (acid soln)

13. B ALANCING H ALF -R EACTIONS In Basic Solution: Do the first steps of the balancing just like an acid. Ex. MnO 2  MnO 4 - (basic solution) 1. Major (Mn already balanced) 2. Oxygen 2H 2 O + MnO 2  MnO 4 - 3. Hydrogen 2H 2 O + MnO 2  MnO 4 - + 4H + 4. Charge 2H 2 O + MnO 2  MnO 4 - + 4H + + 3e -

14. B ALANCING H ALF -R EACTIONS In basic solution, write the reaction: H + + OH -  H 2 O or H 2 O  H + + OH - in whichever way is needed to cancel out the H + ’s and add to the half-rxn. You must write the whole water equation We need 4H + on the left to cancel the 4H + on the right side 2H 2 O + MnO 2  MnO 4 - + 4H + + 3e - 4H + + 4OH -  4H 2 O ________________________________________ MnO 2 + 4OH -  MnO 4 - + 2H 2 O + 3e -

15. B ALANCING H ALF -R EACTIONS Example: Pb  HPbO 2 - (basic soln)

16. B ALANCING O VERALL R EDOX R EACTIONS H ALF -R EACTION M ETHOD 1. Break up rxn into 2 half-rxns. 2. Balance each one (in acidic or basic as given). 3. Multiply each half rxn by whatever is needed to cancel out e - ‘s. (Note: balanced half-rxns show e - ‘s (on left for reduction - on right for oxidation) but balanced redox rxns don’t show e - ‘s). 4. Add the 2 half-rxns canceling e - ‘s and anything else (usually H 2 O’s, H + ’s or OH - ‘s) in order to simplify.

17. B ALANCING O VERALL R EDOX R EACTIONS H ALF -R EACTION M ETHOD Example: U 4+ + MnO 4 -  Mn 2+ + UO 2 2+ (acidic) Balance each half-rxn: U 4+  UO 2 2+ U 4+ + 2H 2 O  UO 2 2+ U 4+ + 2H 2 O  UO 2 2+ + 4H + U 4+ + 2H 2 O  UO 2 2+ + 4H + + 2e - MnO 4 -  Mn 2+ MnO 4 -  Mn 2+ + 4H 2 O MnO 4 - + 8H +  Mn 2+ + 4H 2 O MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O (U 4+ + 2H 2 O  UO 2 2+ 4H + + 2e - ) x 5 (MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O) x 2 5U 4+ + 10H 2 O + 2MnO 4 - + 16H + + 10e -  5U­­O 2 2+ + 20H + + 2Mn 2+ + 8H 2 O + 10e -

18. B ALANCING O VERALL R EDOX R EACTIONS H ALF -R EACTION M ETHOD To simplify:  Take away 10e - from both sides  Take away 16H + ’s from both sides  Take away 8H 2 O’s from both sides 5U 4+ + 2H 2 O + 2MnO 4 -  5UO 2 2+ + 4H + + 2Mn 2+ Quick check by finding TIC’s on both sides. If you have time check all atoms also. If TIC’s are not equal, you messed up!

19. B ALANCING O VERALL R EDOX R EACTIONS H ALF -R EACTION M ETHOD Example: SO 2 + IO 3 -  SO 4 2- + I 2 (basic soln)

20. B ALANCING O VERALL R EDOX R EACTIONS H ALF -R EACTION M ETHOD Quick Notes Some redox equations have just one reactant. Use this as the reactant in both half-rxns. These are called “Self-Oxidation-Reduction” or “Disproportionation Reactions.” Ex. Br 2  Br -- + BrO 3 - (basic soln)

21. B ALANCING O VERALL R EDOX R EACTIONS H ALF -R EACTION M ETHOD Half-rxns are: Br 2  Br - Br 2  BrO 3 - Hebden Textbook p. 207 Question #24

22. B ALANCING O VERALL R EDOX R EACTIONS O XIDATION N UMBER M ETHOD This is optional As long as one method (not guessing!) works for you, that’s fine Read examples on page 208-209 in Hebden Textbook Review Question #25

23. TITRATIONS Find moles of Standard using: Find moles of Sample using: n = C x VC = n/V R EDOX T ITRATIONS Same as in other units (solubility/acids-bases) Coefficient ratios for the “mole bridge” are obtained by the balanced redox equation. MOLE BRIDGE

24. R EDOX T ITRATIONS Example: Acidified hydrogen peroxide (H 2 O 2 ) is used to titrate a solution of MnO 4 - ions of unknown concentration. Two products are O 2 gas and Mn 2+. a) Write the balanced redox equation :

25. R EDOX T ITRATIONS b) It takes 6.50 mL of 0.200 M H 2 O 2 to titrate a 25.0 mL sample of MnO 4 - solution. Calculate the original [MnO 4 - ].

26. F INDING A S UITABLE S TANDARD Use redox table: If sample is on the left (OA)…use something below it on the right (RA) If sample is on the right (RA)…use something above it on the left (OA) Good standards will change colour as they react Acidified MnO 4 - (purple) = Mn 2+ (clear) Acidified Cr 2 O 7 2- (orange) = Cr 3+ (pale green) http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA3/MVHTM/TITRED O/TITR1.HTM http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA3/MVHTM/TITRED O/TITR2.HTM Hebden Textbook p. 213-214 Questions #26, 29