Redox Processes. The basis of electrochemical processes is the transfer of electrons between substances. 11 p 12 n. 9 p 10 n.

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1 1 Electron transfer The basis of electrochemical processes is the transfer of electrons between substances. Na Na+ A e - + A + B + e - B - 11 p 12 n 11 p 12 n 11 p 12 n 9 p 10 n F F- 9 p 10 n 9 p 10 n NaF Oxidation; the reaction with oxygen. 4 Fe(s) + 3O 2 (g) Fe 2 O 3 (s) Why is rust Fe 2 O 3, 2Fe to 3O? 2

2 Electron transfer of iron- Fe Fe3+ + Electron transfer to oxygen 3e- 1/2 O2 + 2e- O Net reaction: 4 Fe(s) + 3O2(g) Fe2O3(s) Fe(+3) O() Fe2O3 : Electrical neutrality 3 4

3 Redox process always occurs together. In redox process, one can t occur without the other. Example: 2 Ca (s) + O 2 2CaO (s) Which is undergoing oxidation? Reduction? Oxidation: Ca Ca +2 Reduction: O 2 O Oxidizing agent; That which is responsible to oxidize another. O 2 ; Oxidizing agent; The agent itself undergoes reduction Reducing agent; That which is responsible to reduce another. Ca; Reducing agent; The agent itself undergoes oxidation 5 Lose electrons: Oxidation Gain electrons: Reduction. 6

4 In all reduction-oxidation (redox) reactions, one species is reduced at the same time as another is oxidized. Oxidizing Agent: The species which causes oxidation is called the oxidizing agent. The substance which is oxidized loses electrons to the other. The oxidizing agent is always reduced Reducing Agent: The species which causes reduction is called the reducing agent. The substance which is reduces gains electrons from the other. The Reducing agent is always oxidized 7 Highest and lowest oxidation numbers of reactive main-group elements. The A group number shows the highest possible oxidation number (Ox.#) for a main-group element. (Two important exception are O, which never has an Ox# of +6 and F, which never has an Ox# of +7.) For nonmetals, (brown) and metalloids (green) the A group number minus 8 gives the lowest possible oxidation number 8

5 Rules for Assigning an Oxidation Number (Ox#) 9 It is common for metal to produce hydrogen gas when they react with acids. For example, the reaction between Mg and HCl: Mg (s) + 2HCI (aq) MgCl 2(aq) + H 2(g). In this rxn, Mg is oxidized and H in HCl is reduced. Note the change in oxidation state for these specie: Mg 0 Mg +2 in MgCl 2 & H + in HCl H 0 in H 2 10

6 It is possible for metals to be oxidized with salt: Fe (s) + Ni(N0 3 ) 2 (aq) Fe(N0 3 ) 2 (aq) + Ni (s). Molecular Equation The net ionic equation shows the redox chemistry well: Fe (s) + Ni +2 (aq) Fe 2+ (aq) + Ni (s) Net ionic Equation In this reaction iron has been oxidized to Fe 2+ while the Ni +2 has been reduced to Ni 0. What determines whether the reaction occurs? 11 Metals can be placed in order of their tendencies for losing electrons. This is called the activity series. 12

7 Consider: Na, Mg, Al, Metallic character decreases left to right. Metal tend to give up electrons. Now consider the reaction: Na + AlCl 3??? (NaCl + Al) To determine if the reaction occurs, the question is to determine which metal has a greater affinity for electrons (or which is willing to lose e- ). Na is more willing to lose e- than Al Al is more willing to accept e- (less metallic) Conclude: The reaction occurs. 3Na + AlCl 3 3NaCl + Al 13 A metal in the activity series can only be oxidized by a metal ion below it. In our example, Na is oxidized by Al. The metals at the top of the activity series are called active metals. 14

8 If we place Cu into a solution of Ag+ ions, will copper plate out of solution? Cu (s) + 2AgNO 3(aq)? [Cu(NO 3 ) 2(aq) + 2Ag (s) ] or Cu (s) + 2Ag+ (aq)? [Cu 2+ (aq) + 2Ag (s) ] Which metal is active? Which is noble? Cu Ag Therefore, Cu 2+ ions is be formed because Cu is above Ag in the activity series. 15 B&L 4.47: Based on the activity series, what is the outcome of the following reaction? b) Ag (s) + PbNO 3 (aq)? c) Cr (s) + NiSO 4 (aq)? e) H 2 (g) + CuCl 2 (aq)? f) Ba (s) + H 2 O (l)? b) Ag vs. Pb, Pb is more active, rxn not occurs c) Cr vs. Ni, Cr is more active, rxn occurs Cr (s) + NiSO 4 (aq) Ni (s) + CrSO 4 (aq) d) H 2 vs. Cu, H 2 is more active, rxn occurs H 2 (g) + CuCl 2 (aq) 2HCl (aq) + Cu (s) e) Ba vs. H 2, Ba is more active, rxn occurs Ba (s) + H 2 O (l) H 2 (g) + Ba(OH) 2 (aq) 16

9 Redox - Oxidation/Reduction reaction Oxidation- Lose electron (LEO) Reduction- Gain electron (GER) Activity Series- Table showing elements relative ease of oxidation. Active M Metal which prefers to lose e- and there fore prefer the oxidized form. Noble M Metal which do not lose e - and therefore prefers the zeroth state Identify oxidized and reduced species Write the half reaction for each. 2. Balance the half rxn separately, the non H and non O elements first followed by H & O s. Balance the non-hydrogen and non-oxygen elements first Balance: Oxygen by H 2 O Balance: Hydrogen by H + Balance: Charge by e - 3. Multiply each half reaction by a coefficient. There should be the same # of e - in both half-rxn. 4. Add the half-rxn together, the electrons should cancel. 18

10 I - + S 2 O 8 I 2 + S 2 O 4 Half Rxn (oxid): I - I 2 Half Rxn (red): S 2 O 8 I 2 + S 2 O 4 Bal. chemical and e- : 2 I - I e - Bal. chemical O and H : 8e - + 8H + + S 2 O 8 S 2 O 4 + 4H 2 O Mult 1st rxn by 4: 8I - 4 I 2 + 8e - Add rxn 1 & 2: 8I - 4 I 2 + 8e - 8e - + 8H + + S 2 O 8 S 2 O 4 + 4H 2 O 8I - + 8H + + S 2 O 8 4 I 2 + S 2 O 4 + 4H 2 O 19 20

11 H 2 O 2 (aq) + Cr 2 O 7 (aq ) Cr 3+ (aq) + O 2 (g) Half Rxn (oxid): 6e H + + Cr 2 O 7 (aq) 2Cr H 2 O Half Rxn (red): ( H 2 O 2 (aq) O 2 + 2H + + 2e - ) x 3 8 H + + 3H 2 O 2 + Cr 2 O 7 2Cr O 2 + 7H 2 O add: 8H 2 O 8 H OH - 8 H + + 3H 2 O 2 + Cr 2 O 7 2Cr O 2 + 7H 2 O 8H 2 O 8 H OH - Net Rxn: 3H 2 O 2 + Cr 2 O H 2 O 2Cr O OH

12 Try these examples: 1. BrO 4 - (aq) + CrO 2 - (aq) BrO 3 - (aq) + CrO 4 (aq) (basic) 2. MnO 4 - (aq) + CrO 4 (aq) Mn 2+ (aq) + Co 2 (aq) (acidic) 3. Fe 2+ (aq) + MnO 4 - (aq) Fe3+ (aq) + Mn2+ (aq) (acidic) 23 24

13 25 Petrucci 7th Ed. p A piece of iron wire weighting g is converted to Fe 2+ (aq) and requires ml of a KMnO 4 (aq) solution for its titration. What is the molarity of the KMNO 4 (aq)? 2. Another substance that may be used to standardized KMNO 4 (aq) is sodium oxalate, Na 2 C 2 O 4. If g of Na 2 C 2 O 4 is dissolved in water and titrated with ml KMnO 4, what is the molarity of the KMnO 4 (aq)? 26