1. Exp. 16: Volumetric Analysis: Redox Titration Normality = eq wt of solute L solution Acid/bases: #eq = # H + or OH - ionized Redox reactions – transfer of e - reduction – oxidation reactions Exp. 16 – videoExp. 16 – video(time: 1 hr and 23:08 minutes)

2. Redox reaction Equivalent wt - one equivalent of any oxidizing agent reacts with one equivalent of any reducing agent. This means #eq/mol is equal to the number of e - transferred. MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O (l) (net) MnO 4 - : 5eq same for KMnO 4 mol MnO 4 - Fe 2+ (aq)  Fe 3+ (aq) + 1e - 1eq mol Fe 2+

3. N  M or M  N N (eq) = M (mol) x #eq L L mol Note: N equal to or greater than M 0.1 M KMnO 4  N? Goal: eq KMnO 4 L soln MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O (l) Calc:

4. 4 Solubility Rules for Ionic Compounds (Dissociates 100%) 1.)All compounds containing alkali metal cations and the ammonium ion are soluble. 2.)All compounds containing NO 3 -, ClO 4 -, ClO 3 -, and C 2 H 3 O 2 - anions are soluble. 3.)All chlorides, bromides, and iodides are soluble except those containing Ag +, Pb 2+, or Hg 2 2+. 4.)All sulfates are soluble except those containing Hg 2 2+, Pb 2+, Ba 2+, Sr 2+, or Ca 2+. Ag 2 SO 4 is slightly soluble. 5.)All hydroxides are insoluble except compounds of the alkali metals and Ca 2+, Sr 2+, and Ba 2+ are slightly soluble. 6.)All other compounds containing PO 4 3-, S 2-, CO 3 2-, CrO 4 2-, SO 3 2- and most other anions are insoluble except those that also contain alkali metals or NH 4 +. Generally, compound dissolves > 0.10 M - soluble (aq) < 0.01 M - insoluble (s) in between - slightly soluble (this class we will assume slightly soluble as soluble) Hg 2 Cl 2 (s) insoluble KI(aq) soluble Pb(NO 3 ) 2 (aq) soluble

5. 5 Strong Acids (Ionizes 100%) HCl, HBr, HI, HClO 4, HNO 3, H 2 SO 4 Strong Bases (Dissociates 100%) NaOH, KOH, LiOH, Ba(OH) 2, Ca(OH) 2, Sr(OH) 2

6. 6 A molecular/formula unit equation is one in which the reactants and products are written as if they were molecules/formula units, even though they may actually exist in solution as ions. Calcium hydroxide + sodium carbonate M.E. Ca(OH) 2 Ions in Aqueous Solution Molecular and Ionic Equations + Na 2 CO 3  CaCO 3 + NaOH2 (aq) strong base soluble saltinsoluble salt (aq)(s)(aq) s solid l liquid aq aqueous (acid/bases and soluble salts dissolve in water) g gases

7. 7 An total ionic equation, however, represents strong electrolytes as separate independent ions. This is a more accurate representation of the way electrolytes behave in solution. –A complete ionic equation is a chemical equation in which strong electrolytes (such as soluble ionic compounds, strong acids/bases) are written as separate ions in solution. (note: g, l, insoluble salts (s), weak acid/bases do not break up into ions) M.E. Ca(OH) 2 (aq) + Na 2 CO 3 (aq)  CaCO 3 (s) + 2 NaOH (aq) Total ionic Ions in Aqueous Solution Molecular and Ionic Equations Ca 2+ (aq) + 2OH - (aq ) strong base soluble salt insoluble salt strong base + 2Na + (aq) + CO 3 2- (aq)  CaCO 3 (s) + 2Na + (aq) + 2OH - (aq)

8. 8 Net ionic equations. –A net ionic equation is a chemical equation from which the spectator ions have been removed. –A spectator ion is an ion in an ionic equation that does not take part in the reaction. M.E. Ca(OH) 2 (aq) + Na 2 CO 3 (aq)  CaCO 3 (s) + 2 NaOH (aq) Total Ionic Ca 2+ (aq) + 2OH - (aq) + 2Na + (aq) + CO 3 2- (aq)  CaCO 3 (s) + 2Na + (aq) + 2OH - (aq) Net Ca 2+ (aq) + CO 3 2- (aq)  CaCO 3 (s)

9. 9 Types of Chemical Reactions Oxidation-Reduction Reactions (Redox rxn) –Oxidation-reduction reactions involve the transfer of electrons from one species to another. –Oxidation is defined as the loss of electrons. –Reduction is defined as the gain of electrons. –Oxidation and reduction always occur simultaneously.

10. 10 27.1 Reduction and Oxidation Redox reactions – transfer of e - reduction – oxidation reactions Reduction – gain of e - / gain of H / lost of O Fe 3+ + 1e -  Fe 2+ (lower ox state) note: must balance atoms and charges

11. 11 Oxidation - loss of e - / loss of H / gain of O Fe 2+  Fe 3+ + 1e - (higher ox state) H 2 O + BrO 3 -  BrO 4 - + 2H + + 2e - (Br oxidized: charge 5+  7+) 2H + + 2e -  H 2 (H reduced: charge 1+  0) Oxidizing agent is species that undergoes reduction. Reducing agent is species that undergoes oxidation. Note: need both for reaction to happen; can’t have something being reduced unless something else is being oxidized. Br + 3(-2) = -1 Br = -1 +6 = +5 Br + 4(-2) = -1 Br = -1 +8 = +7

12. 12 27.3 Balancing Redox Reactions - Must know charges (oxidation numbers) of species including polyatomic ions - Must know strong/weak acids and bases - Must know the solubility rules Oxidation Numbers – hypothetical charge assigned to the atom in order to track electrons; determined by rules.

13. 13 Rules to balance redox 1.) Convert to net ionic form if equation is originally in molecular form (eliminate spectator ions). 2.) Write half reactions. 3.) Balance atoms using H + / OH - / H 2 O as needed: –acidic: H + / H 2 O put water on side that needs O or H (comes from solvent) –basic: OH - / H 2 O put water on side that needs H but if there is no H involved then put OH - on the side that needs the O in a 2:1 ratio 2OH - / H 2 O balance O with OH, double OH, add 1/2 water to other side. 4.) Balance charges for half rxn using e -. 5.) Balance transfer/accept number of electron in whole reaction. 6.) Convert equation back to molecular form if necessary (re-apply spectator ions).

14. Zn (s) + AgNO 3(aq)  Zn(NO 3 ) 2(aq) + Ag (s) Total ionic: Net ionic: Zn(s) + Ag + (aq) + NO 3 - (aq)  Zn 2+ (aq) + 2NO 3 - (aq) + Ag(s) Zn(s) + Ag + (aq)  Zn 2+ (aq) + Ag(s) 14

15. Net: Zn (s) + Ag + (aq)  Zn 2+ (aq) + Ag (s) Oxidation: Reduction: Balanced net: Balanced eq: Zn(s)  Zn 2+ (aq)+ 2e - Ag + (aq)  Ag(s) 1e - + Zn(s) + 2 Ag + (aq)  Zn 2+ (aq) + 2 Ag(s) [ ] 2 Zn (s) 15 + 2 AgNO 3(aq)  Zn(NO 3 ) 2(aq) + 2 Ag (s)

16. H + Net: MnO 4 - (aq) + Fe 2+ (aq)  Mn 2+ (aq) + Fe 3+ (aq) Ox: Red: Balanced net: Fe 2+ (aq)  Fe 3+ (aq)+ 1e - [ ] 5 MnO 4 - (aq)  Mn 2+ (aq)+ H 2 O(l)48 H + (aq) +5e - + 8 H + (aq) + MnO 4 - (aq) + 5 Fe 2+ (aq)  Mn 2+ (aq) + 5 Fe 3+ (aq) + 4 H 2 O(l) 16

17. KMnO 4(aq) + NaNO 2(aq) + HCl (aq)  NaNO 3(aq) + MnCl 2(aq) + KCl (aq) + H 2 O (l) Net: Ox: Red: Balanced net: Balanced eq: MnO 4 - (aq)  Mn 2+ (aq)+ NO 2 - (aq) NO 3 - (aq) ++ H + (aq) + H 2 O(l) NO 2 - (aq)  NO 3 - (aq) MnO 4 - (aq)  Mn 2+ (aq)+ 4 H 2 O(l)8 H + (aq) + H 2 O(l) ++ 2 H + (aq) 5 e - + + 2 e - [ ] 5 [ ] 2 2 MnO 4 - (aq) + 5 NO 2 - (aq) + 16 H + (aq) + 5 H 2 O(l)  2Mn 2+ (aq) + 8 H 2 O(l) + 5 NO 3 - (aq) +10 H + (aq) 2 MnO 4 - (aq) + 5 NO 2 - (aq) + 6 H + (aq)  2Mn 2+ (aq) + 3 H 2 O(l) + 5 NO 3 - (aq) 2 KMnO 4 (aq) + KCl2 17 + 5 NaNO 2 (aq)+ 6 HCl(aq)  2MnCl 2 (aq)+ 3 H 2 O(l)+ 5 NaNO 3 (aq)

18. Net: OH - CrI 3 (s) + Cl 2 (g)  CrO 4 2- (aq) + IO 4 - (aq) + Cl - (aq) Ox: Red: Balanced net: CrI 3 (s)  CrO 4 2- (aq) + IO 4 - (aq) Cl 2 (g)  Cl - (aq)2 332 OH - (aq) + + 16 H 2 O(l) 2 e - + + 27 e - [ ] 2 [ ] 27 64 OH - (aq) + 2 CrI 3 (s) + 27 Cl 2 (g)  2 CrO 4 2- (aq) + 6 IO 4 - (aq) + 54 Cl - (aq) + 32 H 2 O(l) 18

19. Exp 16: S 2 O 3 2- (aq) + I 2  S 4 O 6 2- (aq) + I - (aq) thiosulfate ion iodine Ox: Red: Balanced net: Outside exercise II page 199 – posted on my website S 2 O 3 2- (aq)  S 4 O 6 2- (aq) I 2 (aq)  I - (aq) 2 2 + 2 e - 2 e - + 2 S 2 O 3 2- (aq) + I 2 (aq)  S 4 O 6 2- (aq) + 2 I - (aq)

20. S 2 O 3 2- 2eq = 1 eq 2mol S 2 O 3 2- mol S 2 O 3 2- I 2 2eq mol I 2 Exp today First: Standardize thiosulfate against 0.100 N I 2 standard solution. Changes in sample preparation: 10 mL I 2, 30 mL deionized H 2 O, 1 mL starch (20 drops) Starch – indicator (add from beginning) Starch + I 2 gives blue color At end pt (all I 2 consumed), solution will be colorless

21. Since using normality can use N iodine V iodine = N thiosulfate V thiosulfate minimum 3 runs ± 0.005 N (around ± 0.5 mL) report Avg N ± s N thiosulfate ion (S 2 O 3 2- ) Convert average N to M

22. Second: Same exact procedure as standardization except using unknown conc. of I 2. minimum 3 runs ± 0.005 N (around ± 0.5 mL) report Avg N ± s N iodine (I 2 ) unknown Convert average N to M

23. Amount of chemicals to obtain in small beaker per group: Na 2 S 2 O 3. 5H 2 O – 150 mL (source of thiosulfate ions) 0.100 N I 2 standard solution – 50 mL Unknown I 2 solution – 45 mL