Zumdahl 6 th Ed. Section Z4.10-Z Oxidation-Reduction Reactions 4.11 Balancing Oxidation-Reduction Equations 4.12 Oxidation-Reduction

Transcription

1 Redox Reactions Zumdahl 6 th Ed. Section Z4.10-Z Oxidation-Reduction Reactions 4.11 Balancing Oxidation-Reduction Equations 4.12 Oxidation-Reduction Titrations Revisit the process for making a salt. A metal and a halogen (a nonmetal) react. This is an example of a Redox reaction. Understand Oxidation Numbers, Ox.Nb.s Need Ox.Nb.s to balance redox reactions Balance Redox reactions using half cell method. Problems: Problems to be used on the exam: (See 4.61c) Be sure to identify spectator ions , 4.69, 4.74, 4.76, 4.79, 4.83

2 Figure 4.19: Reaction of solid sodium and gaseous chlorine to form solid sodium chloride: an oxidation reduction reaction.

3 Types of Redox Reactions (Rxns): Burn a metal, a metal oxide (related to a base) Burn a non-metal, a non-metal oxide (an acid)

4 Ox.Nb. Oxidation Numbers Oxidation - A substance gives up electrons to another substance. Mg + ½ O 2 Mg 2+ + O 2- Mg is oxidized (Oxygen is the oxidizing agent; it accepts electrons) Reduction - A substance accept electrons from another substance. Mg + ½ O 2 Mg 2+ + O 2- O is reduced (Notice O s Ox.Nb. (or charge in this case) is reduced; Mg is the reducing agent causing the reduction of Oxygen.) In a chemical reaction, the total number of electrons and the number of charges are conserved. It is convenient to assign fictitious charges to the atoms in a molecule and call them oxidation states or oxidation numbers (Ox.Nb. Or just O.N.). Oxidation numbers are chosen so that (a) charges are conserved, and (b) in ionic compounds the sum of oxidation numbers on the atoms coincides with the charge on the ion.

5 Assigning Ox.Nb.s: Use Periodic Table

6 Rules for giving Ox.Nb.s (Table 4.3 in Text)

7 Ox Nb; Oxidation Numbers Determine the oxidation number (Ox.Nb.) of each element in the following compounds: 1: Iron (III) Chloride 2: Nitrogen Dioxide 3: Sulfuric acid Strategy: We apply the rules in Table 4.3, always making sure that the Ox.Nb. values in a compound add up to zero, and in a polyatomic ion, to the ion s charge. Solution: 1: FeCl 3 This compound is composed of monoatomic ions. The Ox.Nb. of Cl - is -1, for a total of -3. Therefore the ON of Fe is +3, The Roman Numeral, III is the giveaway here.

8 example: NO 2 Examples of Ox.Nb.s The Ox.Nb. of oxygen is -2 for a total of -4. Since the Ox.Nb.s in a compound must add up to zero, the ON of N is +4. example: H 2 SO 4 The Ox.Nb. of H is +1, so the SO 4 2- group must sum to - 2. The ON of each O is -2 for a total of -8. Therefore S has the ON +6. example: OF 2 and O 2 F 2 vs. H 2 O and H 2 O 2

9 Examples (cont.): When hydrogen reacts with a metal, it acts as a nonmetal. BaH 2 (Barium Hydride) H = -1 Ba = +2 NH 4 NO 3 H = +1 O = -2 N = -3 N = +5

10 Oxidation Reduction Summary: An element s oxidation number goes up, it is oxidized (so it is a reducing agent) An element s oxidation number goes down, it is reduced (so it an oxidizing agent) If you think this is confusing try: /watch?v=ls75ntlh3gi

11 More Examples of Ox.Nb.s What is the oxidation number of sulfur in Sodium Sulfate, Na 2 SO 4? What is the oxidation number of iodine in the I 3- ion? What is the oxidation number of oxygen in O 3, ozone? OF 2, and O 2 F 2, H 2 O, H 2 O 2, K 2 O, K 2 O 2, HgS, Hg 2 S.

12 To Balance RedOx Rxns Need Oxidation numbers of species before and after reaction Need the change in OxNb.s to keep the reaction electrically neutral, so there is no charge imbalance at the end of the reaction.

13 Balance Reaction by inspection (hard) ( ) + ( ) ( ) + ( ) Fe s Pt aq Pt s Fe aq What s wrong with setting all coefficients to one? Look at Full reaction: 2+ Fe s + Pt aq + Cl aq ( ) ( ) ( ) ( ) ( ) ( ) 3+ Pt s + Fe aq Cl aq Need to make sure charge is not created (or lost). Reintroduce (fictitious?) spectator ions to balance charge. So that each side is neutral. afe s + bpt aq + 2bCl aq ( ) 2+ ( ) ( ) bpt s + afe aq + 3aCl aq ( ) 3+ ( ) ( ) Chloride balance tells us 2b= 3 a or a= 2

14 Balancing REDOX Equations: The oxidation number method Step 1) Assign oxidation numbers to all elements in the equation. Step 2) From the changes in oxidation numbers, identify the oxidized and reduced species. Step 3) Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. Draw tie-lines between these atoms to show electron changes. Step 4) Choose coefficients for these species to make the electrons lost equal the electrons gained (or total increase in OxNb = total decrease in OxNb) Step 5) Complete the balancing by inspection.

15 Problem: Calculate the mass of metallic Iron that must be added to liters of a solution containing M of Pt 2+ (aq) ions in solution to reclaim all Pt via: Fe(s) + Pt 2+ (aq) Fe 3+ (aq) + Pt(s) Solution: V x M = # moles Pt L x mol Pt 2+ / L = 0.20 mol Pt 2+ Balance the equation to determine how many moles of Fe are needed for every mole of Pt.

16 REDOX Balancing using Ox. No. Method - I The number of electrons must balance (just like the number of atoms must balance). +2 Accepts 2 e - per Pt 0 ( ) + 2+ ( ) 3+ ( ) + ( ) Fe s Pt aq Fe aq Pt s 0 Gives 3 e - per Fe +3 electrons lost must = electrons gained Because the electrons are transferred from one (molecular) species to another Therefore need 3 Pt atoms for every 2 Fe! 2Fe s + 3Pt aq 2Fe aq + 3Pt s ( ) 2+ ( ) 3+ ( ) ( ) Balanced and both sides have same (+6) charge. No change in charge.

17 Problem: Calculate the mass of metallic Iron that must be added to liters of a solution containing M of Pt 2+ (aq) ions in solution to reclaim all Pt via: 2 Fe(s) + 3 Pt 2+ (aq) 2 Fe 3+ (aq) + 3 Pt(s) Solution: moles X = M * V = liters Pt = ( ) 0.20moles Pt liter ( ) ( Fe ) SC Fe = 0.20 * = 0.20 rxn = 2+ ( ) ( 2+ SC Pt 3 Pt ) 3 rxn X moles Pt moles Pt moles Fe X = mol Fe* MW ( Fe) = 0.133mol Fe g = ( ) 7.4g Fe mole

18 Recognizing Oxidizing and Reducing Agents Problem: Identify the oxidizing and reducing agent in each of the Rxn: a) Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g) b) S 8 (s) + 12 O 2 (g) 8 SO 3 (g) c) NiO (s) + CO (g) Ni (s) + CO 2 (g) Plan: First we assign an oxidation number (Ox.Nb.) to each atom (or ion) based on the rules in Table 4.3. The reactant is the reducing agent if it contains an atom that is oxidized (Ox.Nb. increased in the reaction). The reactant is the oxidizing agent if it contains an atom that is reduced ( Ox.Nb. decreased). Solution: a) Assigning oxidation numbers: Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g) HCl is the oxidizing agent, and Zn is the reducing agent!

19 Recognizing Oxidizing and Reducing Agents b) Assigning oxidation numbers: S [0] S[+6] S is Oxidized S 8 (s) + 12 O 2 (g) 8 SO O[0] O[-2] 3 (g) O is Reduced S 8 is oxidized and O 2 is reduced S 8 is the reducing agent and O 2 is the oxidizing agent c) Assigning oxidation numbers: C[+2] C[+4] C is oxidized Ni[+2] Ni[0] Ni is Reduced NiO (s) + CO (g) Ni (s) + CO 2 (g) CO is the reducing agent and NiO is the oxidizing agent No O changes its O.N. but it is the agent for the change.

20 REDOX Balancing using Ox. No. Method - I The number of electrons must balance (just like the number of atoms must balance). 0 Accepts +2 e - per O -2 H 2 (g) + O 2 (g) H 2 O (g) 0 Donates 1 e - per H +1 electrons lost must = electrons gained Because the electrons are transferred from one molecular species to another Therefore need 4 H atoms for the 2 O atoms!

21 REDOX Balancing using Ox. No. Method - I 0 2x +2 e H 2 (g) +_1_ O 2 (g) 2 H 2 O (g) 0 4x -1 e - +1 electrons lost must = electrons gained Therefore need 4 H atoms for the 2 O atoms! The half-cell method (next) makes this electron loss and gain more concrete and is a bit easier to do and more robust.

22 Balance the thermite reaction: Fe 2 O 3 (s) + Al(s) Al 2 O 3 (s) + Fe(s) Assign Ox.Nb. to each element Reactants O: -2 Al:0 Fe:+3 Products O:-2 Al:+3 Fe:0 Identify the elements that are oxidized and reduced Fe 3+ Fe 0 : Iron is reduced (OxNb decreases by 3) Al 0 Al 3+ : Aluminum is oxidized (OxNb increases by 3) Make Ox.Nb. total increase = Ox.Nb. total loss Fe 2 O 3 + 2Al Al 2 O 3 + 2Fe

23 Thermite reaction _ Fe 2 O 3 (s) + Al(s) _ Al 2 O 3 (s) + _ Fe(l) e- per Al +3e- per Fe

24 Thermite reaction 1 Fe 2 O 3 (s) + 2 Al(s) Al 2 O 3 (s) + 2 Fe(l) x(- 3e- per Al) 2x(+3e- per Fe) BALANCED!

25 Half (Cell) Reaction Balancing Method Break a reaction into two half-reactions, one for each element type involved in the redox. Balance each half reaction separately Balance O (with water), H with protons, charge with e When done cross multiply and add to cancel out electrons. This uses the ideas of redox but you don t need to know the oxidation numbers specifically. Just need to identify the species being reduced, and the species being oxidized. These reacitons are not fictitious; they are batteries.

26 Example Half Cell Balance Cu(s) +HNO 3 (aq) Cu 2+ (aq) + NO(g) Identify half-reactions, one is ox, other is red Cu(s) Cu 2+ (aq) Not needed; but nice 0 +2 to know HNO 3 (aq) NO(g) Balance all atoms that are neither H nor O OK as is Balance O by adding H 2 O to side deficient in O Cu(s) Cu 2+ (aq) HNO 3 (aq) NO(g) + 2H 2 O

27 Balance H by adding H + to side deficient in H Cu(s) Cu 2+ (aq) 3H + +HNO 3 (aq) NO(g) + 2H 2 O Balance charge by adding e- to side that has + charge Cu(s) Cu 2+ (aq) + 2e- 3H + + HNO 3 (aq) + 3e- NO(g) + 2H 2 O Multiply each eq by factors so electrons cancel out 3x(Cu(s) Cu 2+ (aq) + 2e-) 2x(3H + + HNO 3 (aq) + 3e- NO(g) + 2H 2 O) Add equations 3Cu(s)+ 6H + + 2HNO 3 (aq) 3Cu 2+ (aq) + 2NO(g) + 4H 2 O

28 The reaction is balanced in an acidic environment because there are protons around. If the problem says, balance in a basic environment, then first balance in acidic solution (as you just did). Now add hydroxides to both sides to cancel out the protons and generate waters. Cancel out excess waters if necessary. You are done. Eg: Add OH- and cancel extra water: Cu s + 6H aq + 2HNO aq 3Cu aq + 2NO g + 4H O aq ( ) ( ) ( ) ( ) ( ) ( ) ( ) + ( ) + ( ) + 3 ( ) 2+ 3Cu ( aq) + 2NO( g ) + 4H 2O( aq) + 6OH ( aq) ( ) + 2 ( ) + 3( ) 2+ 3Cu ( aq) + 2NO( g ) + 4H 2O( aq) + 6OH ( aq) ( ) + 2 ( ) + 3( ) 2+ 3Cu ( aq) + 2NO( g ) + 6OH ( aq) 3Cu s 6H aq 6OH aq 2HNO aq 3Cu s 6H O aq 2HNO aq 3Cu s 2H O aq 2HNO aq

29 The following redox equation is balanced in acidic solution. Balance it for a basic solution. Cr 2 O 7 2- (aq)+2no(g)+6h + (aq) 2Cr 3+ (aq)+2no 3- (aq) +3H 2 O(l) Add OH- equal to number of H + on both sides 6OH - (aq) +Cr 2 O 7 2- (aq)+2no(g)+6h + (aq) 2Cr 3+ (aq)+2no 3- +3H 2 O(l) +6OH - (aq) Combine OH - and H + to form water to max extent possible 6H 2 O(l)+Cr 2 O 7 2- (aq)+2no(g) 2Cr 3+ (aq)+2no 3- +3H 2 O(l) +6OH - (aq) Cancel H 2 O on both sides to max extent possible 3H 2 O(l)+Cr 2 O 7 2- (aq)+2no(g) 2Cr 3+ (aq)+2no 3- +6OH - (aq)

30 RedOx Balancing by Half-Reaction Method ( ) + ( ) ( ) + ( ) Fe aq MnO aq Fe aq Mn aq in acid Identify Oxidation and Reduction Half Reactions Fe aq Fe aq + 1e Oxidation Half Rxn ( ) ( ) MnO aq + 5e Mn aq Reduction Half Rxn ( ) 2 ( ) + 4 Add water as needed to balance O, then add H + to balance H, then e - s H + MnO aq + 5e Mn aq + 4H O ( ) ( ) 4 2 Sum the two half-reactions (multiplied to balance/cancel electrons): ( ) ( ) { } 5x Fe aq Fe aq 1e 2 ( ) ( ) { } x 8H MnO aq 5e Mn aq 4H O ( ) ( ) ( ) ( ) 8H + MnO aq + 5Fe aq 5Fe aq + Mn aq + 4H O

31 Redox Balance Problems Balance these by any of the three methods (Half rxn method is most foolproof). ( ) + ( ) + ( ) ( ) Pb s PbO s H SO aq PbSO s ( ) + ( ) ( ) + ( ) Fe aq MnO aq Fe aq Mn aq in acid ( ) ( ) + ( ) 6 2 ( ) ( ) + ( ) ( ) + ( ) + ( ) Fe CN aq Ce aq Ce OH s Fe OH s CO aq NO aq in acid IO aq + I aq I aq ( ) ( ) ( ) 3 3

32 End of Lecture Many More Worked Examples in EXTRA Redox reaction balancing and determining number of moles of product based on moles of reactants. See RedOx titration notes in EXTRA. Typical stoichiometry problems using a redox reaction to determine the stoichiometry of the reaction.