1 Chapter 17 Oxidation-Reduction Cu (s) + 2AgNO 3(aq) 2Ag (s) + Cu(NO 3 ) 2(aq) copper wire colorless solution silver crystals pale blue solution Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison and Susan Arena
2 Chapter Outline 17.1 Oxidation Number 17.2 Oxidation-Reduction 17.3 Balancing Oxidation- Reducing Equations 17.4 Balancing Ionic Redox Equations 17.5 Activity Series of Metals 17.6 Electrolytic and Voltaic Cells
3 Oxidation Number The oxidation number (oxidation state) is an integer assigned to each element in a particle that allows us to keep track of electrons associated with each atom. An oxidation number of 0 means the atom has the same number of electrons assigned to it as there are in the free neutral atom. (Elements are 0.) A positive oxidation number means the atom has fewer electrons assigned to it than in the neutral atom. A negative oxidation number means the atoms has more electrons assigned to it than in the neutral atom.
4 Molecular Substances Elements and molecules whose electrons are equally shared have zero oxidation numbers: Polar bonds are made of unequally shared electron pairs. The electrons are assigned to the more electronegative element Assigned to Cl.
5 Rules for Assigning Oxidation Numbers 1. Elements in the free state are H is +1 except in metal hydrides where it is O is 2 except in peroxide where it is 1 and in OF 2 where it is In covalent compounds the negative oxidation number is assigned to the most electronegative atom. 5. The sum of the oxidation numbers in a compound is zero. 6. The sum of the oxidation numbers in a polyatomic ion is the charge of the ion.
6 Finding the Oxidation Number 1. Write the oxidation number of each known atom below the atom in the formula. 2. Multiply each oxidation number by the number of atoms of that element in the compound. 3. Write an expression indicating the sum of all the oxidation numbers in the formula. a. Sum = 0 for a compound b. Sum = charge for a polyatomic ion.
7 Nitrogen Oxides Oxygen is -2 since it is the more electronegative element. N 2 O -2 2N +(-2)=O N = +1 N 2 O 3-2 2N +3(-2)=O N = +3 NO 2-2 N +2(-2)=O N = +4 N 2 N 2 O NO N 2 O 3 NO 2 N 2 O 5 N oxidation number
8 Oxidation Number of Ions Cr 2 O 7 2-2Cr + 7(-2) = -2 Cr = +6 HSO - 4 S + 1(+1) + 4(-2) = -1 S = +6 CO 3 2- C + 3(-2) = -2 C = +4
9 Your Turn! What is the oxidation number of manganese in MnO 2? a. 0 b. +2 c. +4 d. -2 e. -4
10 Your Turn! What is the oxidation number of sulfur in H 2 SO 3? a. +2 b. +4 c. +6 d. -4 e. -6
11 Your Turn! What is the oxidation number of carbon in C 2 O 2-4? a. 0 b. +1 c. +2 d. +3 e. +4
12 Oxidation-Reduction (Redox) Redox reactions are chemical processes in which the oxidation numbers of an element are changed. Oxidation occurs whenever the oxidation number increases from loss of electrons. Reduction occurs whenever the oxidation number decreases from gain of electrons.
13 Redox Easy ways to remember which is which: OIL RIG: Oxidation Is Loss, Reduction Is Gain LEO the lion goes GER: Lose Electrons Oxidation, Gain Electrons Reduction Oxidizing Agent the substance that causes an increase in the oxidation state of another substance by gaining electrons. It is reduced in the process. Reducing Agent - the substance that causes an decrease in the oxidation state of another substance by losing electrons. It is oxidized in the process..
14 Zinc and Hydrochloric Acid In the reaction between Zn and HCl, we see vigorous bubbles of H 2. Zn (s) + 2HCl (aq) ZnCl 2(aq) + H 2(g) net ionic equation: Zn (s) + 2H + (aq) Zn 2+ (aq) + H 2(g) Oxidation: Zn 0 Zn 2+ +2e - Reduction: 2H + +2e - H 2 0
15 Your Turn! Which reactant was the reducing agent in the reaction between zinc and hydrochloric acid? Zn (s) + 2HCl (aq) ZnCl 2(aq) + H 2(g) a. Zn b. HCl c. ZnCl 2 d. H 2
16 Your Turn! Which reactant is reduced in the following equation? a. Ca 2 Ca (s) + O 2(g) 2 CaO (s) b. O 2 c. CaO
17 Balancing Redox Equations Half-Reaction Method: unbalanced equation: Al + Cl 2 AlCl 3 Oxidation half-reaction: 2Al 3+ (aq) + 3e - (s) 2Al 3+ (aq) + 6e - Reduction half-reaction: 3Cl 2e 2Cl 2 + 6e - 6Cl - 2Al +3 Cl 2 2(Al Cl - ) 2AlCl 3 Make The balanced sure number equation of electrons is the sum lost of = the number two half of electrons reactions. gained.
18 Balancing Redox Equations Change-in-oxidation-number strategy 1. Assign oxidation numbers to every element. 2. Write 2 half-reactions using only the 2 elements that changed. One half-reaction must produce electrons and the other must use electrons. 3. Multiply the half-reactions by the smallest whole number to make sure numbers of electrons lost are equal to numbers of electrons gained.
19 Balancing Redox Equations Change-in-oxidation-number strategy (continued) 4. Transfer the coefficient in front of each substance in the balanced half-reaction to the substance in the original equation. 5. Balance the remaining elements that are not oxidized or reduced. 6. Check to make sure both sides of the equation have the same number of atoms of each element.
20 Example 1 Sn (s) + HNO 3(aq) SnO 2(s) + NO 2(g) + H 2 O (l) Step 1: Assign oxidation numbers Step 2: Write the two half reactions
21 Example 1 (continued) Sn (s) + HNO 3(aq) SnO 2(s) + NO 2(g) + H 2 O (l) Step 3: Ensure number e - lost = number e - gained Sn 0 Sn e - 4N e - 4N 4+ (oxidation) (reduction) Step 4: Transfer coefficients back into equation Sn (s) + 4HNO 3(aq) SnO 2(s) + 4NO 2(g) + H 2 O (l) Step 5: Finish balancing the equation Sn (s) + 4HNO 3(aq) SnO 2(s) + 4NO 2(g) + 2H 2 O (l)
22 Example 2 Pb (s) + PbO 2(aq) + H 2 SO 4(aq) PbSO 4(s) + H 2 O (l) Step 1: Assign oxidation numbers PbO 2(aq) + H 2 SO 4(aq) PbSO 4(s) + H 2 O (l) Step 2: Write the two half reactions Pb e - Pb 2+ Pb Pb e - (reduction) (oxidation)
23 Example 2 (continued) Pb (s) + PbO 2(aq) + H 2 SO 4(aq) PbSO 4(s) + H 2 O (l) Step 3: Ensure number e - lost = number e - gained Pb e - Pb 2+ Pb Pb e - (reduction) (oxidation) Step 4: Transfer coefficients back into equation There is a Pb 2+ in both half-reactions, so we need a 2. Pb (s) + PbO 2(aq) + H 2 SO 4(aq) 2PbSO 4(s) + H 2 O (l) Step 5: Finish balancing the equation Pb (s) + PbO 2(aq) + 2H 2 SO 4(aq) 2PbSO 4(s) + 2H 2 O (l)
24 Your Turn! What is the oxidation half reaction for the unbalanced reaction Ca (s) + O 2(g) CaO (s) a. Ca e - Ca b. Ca 2+ Ca + 2e - c. Ca + 2e - Ca 2+ d. Ca Ca e -
25 Your Turn! What is the reduction half reaction for the unbalanced reaction a. 2O e - O 2 b. 2O 2- O 2 + 4e - c. O 2 + 4e - 2O 2- d. O 2 2O e - Ca (s) + O 2(g) CaO (s)
26 Your Turn! How many electrons are transferred in the reaction a. 1 b. 2 c. 3 d. 4 e. 5 2 Ca (s) + O 2(g) 2 CaO (s)
27 Balancing Ionic Redox Equations Ion-Electron Strategy for Balancing Redox Equations 1. Write the two half-reactions that contain the elements being oxidized and reduced. Use entire molecule or ion. 2. Balance elements other than oxygen and hydrogen. 3. Balance hydrogen and oxygen. Acidic Solutions a. Add H 2 O to balance oxygen. b. Add H + to balance hydrogen.
28 Ionic Redox Equations (continued) 3. Balance hydrogen and oxygen (continued). Basic Solutions a. Balance as if in acid.then add as many OH - ions to each side of the equation as there are H + ions in the equation. b. Combine OH - with H + to form H 2 O. c. Rewrite the equation, canceling equal numbers of water molecules that appear on opposite side of the equation.
29 Ionic Redox Equations (continued) 4. Add electrons (e - ) to each half-reaction to bring them into electrical balance. 5. Since the loss and gain of electrons must be equal, multiply each half-reaction by the appropriate number to make the number of electrons the same in each half-reaction. 6. Add the two half-reactions together, canceling electrons and any other identical substances that appear on opposite sides of the equation.
30 Example 1 (in acid) Sn 2+ (aq) + IO 4 - (aq) Sn 4+ (aq) + I - (aq) 1. Write the two half-reactions 2. Balance elements other than oxygen and hydrogen. oxidation: Sn 2+ Sn 4+ reduction: IO 4 - I -
31 Example 1 (in acid) Sn 2+ (aq) + IO 4 - (aq) Sn 4+ (aq) + I - (aq) 3. Balance hydrogen and oxygen. a. Add H 2 O to balance oxygen. b. Add H + to balance hydrogen. oxidation: Sn 2+ Sn 4+ reduction: IO H + I - + 4H 2 O
32 Example 1 (in acid) Sn 2+ (aq) + IO 4 - (aq) Sn 4+ (aq) + I - (aq) 4. Add electrons (e - ) to each half-reaction to bring them into electrical balance. 5. Ensure number e - lost = number e - gained oxidation: 4Sn 2+ 4Sn e - reduction: 8 e - + IO H + I - + 4H 2 O
33 Example 1 (in acid) Sn 2+ (aq) + IO 4 - (aq) Sn 4+ (aq) + I - (aq) 6. Add the two equations together, combining like terms. oxidation: 4Sn Sn e - reduction: 8 e - + IO - - 4H 4 + 8H + I - + 4H 2 O 4Sn 2+ (aq) + IO 4 - (aq) + 8H + (aq) 4Sn 4+ (aq) + I - (aq) + 4H 2 O (l)
34 Your Turn! Which of these is a correctly balanced reduction halfreaction in acid for the following reaction? H 2 O 2(aq) + Cr 2 O 7 2- (aq) Cr 3+ (aq) + O 2(g) a. Cr 2 O 7 2- (aq) + 14H + (aq) Cr 3+ (aq) + 7H 2 O (l) + 9e - b. Cr 2 O 7 2- (aq) + 14H + (aq) 2Cr 3+ (aq) + 7H 2 O (l) + 6e - c. Cr 2 O 7 2- (aq) + 14H + (aq) + 9e - Cr 3+ (aq) + 7H 2 O (l) d. Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - 2Cr 3+ (aq) + 7H 2 O (l)
35 Your Turn! Which of these is a correctly balanced oxidation halfreaction in acid for the following reaction? H 2 O 2(aq) + Cr 2 O 7 2- (aq) Cr 3+ (aq) + O 2(g) a. H 2 O 2(aq) O 2(g) + 2H + (aq) + 1e - b. H 2 O 2(aq) O 2(g) + 2H + (aq) + 2e - c. H 2 O 2(aq) + 1e - O 2(g) + 2H + (aq) d. H 2 O 2(aq) + 2e - O 2(g) + 2H + (aq)
36 Example 2 (in base) Zn (s) + NO 3 - (aq) NH 3(aq) + Zn(OH) 4 2- (aq) 1. Write the two half-reactions 2. Balance elements other than oxygen and hydrogen. oxidation: reduction: Zn Zn(OH) 2-4 NO - 3 NH 3
37 Example 2 (in base) Zn (s) + NO 3 - (aq) NH 3(aq) + Zn(OH) 4 2- (aq) 3. Balance hydrogen and oxygen. a. Add H 2 O to balance oxygen. b. Add H + to balance hydrogen. c. Add OH - to neutralize H +. d. Combine OH - with H + to form H 2 O and simplify. oxidation: Zn + 4H 4OH H + 2 O - + 4OH Zn(OH) - Zn(OH) H 2 O reduction: NO H 2 O NH NH H 2 H 2 O 2 O + 9 OH -
38 Example 2 (in base) Zn (s) + NO 3 - (aq) NH 3(aq) + Zn(OH) 4 2- (aq) 4. Add electrons (e - ) to each half-reaction to bring them into electrical balance. 5. Ensure number e - lost = number e - gained oxidation: reduction: 4Zn + + 4OH 16OH - - 4Zn(OH) e e - NO H 2 O + 8 e - NH H 2 O + 9 OH -
39 Example 2 (in base) Zn (s) + NO 3 - (aq) NH 3(aq) + Zn(OH) 4 2- (aq) 6. Add the two equations together, combining like terms. oxidation: reduction: 16-9=7 OH - 4Zn + 16OH - 4Zn(OH) e - NO H 2 O + 8 e - NH OH - 4Zn (s) + NO 3 - (aq) +7OH - (aq)+6h 2 O (l) NH 3(aq) + 4Zn(OH) 4 2- (aq)
40 Relative Reactivity of Metals If you put a piece of copper wire in 1M AgNO 3 a reaction takes place. Cu (s) + 2AgNO 3(aq) 2Ag (s) + Cu(NO 3 ) 2(aq) If you put a piece of silver wire in 1M Cu(NO 3 ) 2 no reaction occurs. 2Ag (s) + Cu(NO 3 ) 2(aq) no reaction Therefore, copper is a more active metal than silver.
41 Activity Series of Metals Activity series: A listing of metallic elements in descending order of reactivity. Cu is above Ag, which means that Cu can replace Ag in a compound.
42 Using the Activity Series 1. The reactivity of the metals listed decreases from top to bottom. 2. A free metal can displace the ion of any metal below it in the activity series. 3. Free metals above H react with acids to liberate H Free metals below H don t react with acids. 5. Reaction conditions like temperature and pressure may affect the relative position of some of the metals.
43 Your Turn! Rank these metals from least reactive to most reactive using the data below: Cu (s) + HCl (aq) no reaction Zn (s) + 2HCl (aq) ZnCl 2(aq) + H 2(g) Mg (s) + ZnCl 2(aq) MgCl 2(aq) + Zn (s) a. Cu < Zn < Mg b. Cu < Mg < Zn c. Mg < Zn < Cu
44 Your Turn! What are the likely products of a reaction of chromium with concentrated hydrochloric acid? a. no reaction b. CrCl and H c. CrCl 3 and H d. CrCl and H 2 e. CrCl 3 and H 2
45 Your Turn! What are the likely products of a reaction of aluminum with 1M NiCl 2? a. no reaction b. AlCl 2 and Ni c. AlCl 3 and Ni d. AlCl and Ni e. AlNi and Cl 2
46 Electrolytic Cells Electrolysis is the process in which electrical energy is used to bring about chemical change. An electrolytic cell uses electricity to produce a chemical change for nonspontaneous redox reaction. Electrolysis is used to manufacture Na and NaOH, Cl 2 and H 2, as well as to purify and electroplate metals.
47 Electrolytic Cells - Cathode Cathode negative electrode Hydronium ions migrate to the cathode and are reduced. Reaction at the cathode: H 3 O + + 1e - H 0 + H 2 O H 0 + H 0 H 2 2HCl (aq) H 2(g) + Cl 2(g)
48 Electrolytic Cells - Anode Anode positive electrode Chloride ions migrate to the anode and are oxidized. Reaction at the anode: Figure 17.4 Place Holder Electrolysis of HCl Cl - Cl 0 + e - Cl 0 + Cl 0 Cl 2 Net Reaction 2HCl (aq) H 2(g) + Cl 2(g)
49 Your Turn! In the electrolysis of fused (molten) calcium chloride, the product at the cathode is a. Ca 2+ b. Cl - c. Cl 2 d. Ca
50 Voltaic Cells Voltaic cell produces electrical energy from a spontaneous chemical reaction. (Also known as a galvanic cell). When a piece of zinc is put in a copper(ii) sulfate solution, the zinc quickly becomes coated with metallic copper. This occurs because zinc is above copper in the activity series. If this reaction is carried out in a voltaic cell, an electric current is produced.
51 Voltaic Cells anode oxidation Zn 0 (s) Zn 2+ (aq) + 2e - cathode reduction Cu 2+ (aq) + 2e - Cu 0 (s) Net Ionic Equation: Zn 0 (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu 0 (s)
52 Your Turn! Towards which compartment will electrons flow in a voltaic cell? a. Toward the cathode b. Toward the anode c. It depends on the reaction
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