6.1 Oxidation Reduction

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1 BC Science Chemistry 12 Chapter 6 Oxidation Reduction and Its Applications Answer Key September 20, Oxidation Reduction Warm Up Name Formula Charge on Metal Ion iron(iii) chloride FeCl 3 +3 iron(ii) acetate Fe(CH 3 COO) 2 +2 manganese(ii) bromate Mn(BrO 3 ) 2 +2 manganese(iv) oxide MnO 2 +4 potassium permanganate KMnO 4 +7 Practice Problems: 1. H 2 O H = +1 O = Cs 2 O 2 Cs = +1 O = CO 3 C = +4 O = Na 2 Cr 2 O 7 Na = +1 Cr = +6 O = BaH 2 Ba = +2 H = NH 4 N = -3 H = S 8 S = 0 8. Al 2 (SO 4 ) 3 Al = +3 S = +6 O = -2 Quick Check: 1. 6 e- 2. nothing (no e- transferred) Quick Check 1. You oxidized (LEO) Partner - reduced (GER) 2. Which partner acted as an oxidizing agent? my partner (they got reduced) Edvantage Interactive

2 A reducing agent? me (I got oxidized) 3. What happened to your oxidation state? Increased (I was oxidized) What about your partner? Reduced got smaller (they were reduced) Practice Problems: Agents C(s) + 2 H 2 (g) CH 4 (g) O.A. = C(s) R.A. = H 2 (g) Sr(s) + 2 FeBr 3 (aq) 2 Fe(s) + 3 SrBr 2 (aq) O.A. = Fe 3+ R.A. = Sr CO(g) + Cl 2 O 5 (s) 5 CO 2 (g) + Cl 2 (s) O.A. = Cl 2 O 5 R.A. = CO PH 3 (g) P 4 (g) + 6 H 2 (g) O.A. and R.A. = PH Ba(s) + 2 H 2 O(l) Ba(OH) 2 (s) + H 2 (g) O.A. = H 2 O R.A. = Ba 6.1 Activity Characteristic Oxidizing Agent Reducing Agent Causes another species to be oxidized reduced Is itself during reaction. reduced oxidized Its oxidation state is during reaction. decreased increased Causes electrons to be... lost gained electrons. gains loses Belongs to family. halogen alkali METAL Has a electronegativity. high low 6.1 Review Questions Elements that get oxidized (act as reducing agents) form_(a)_+ ions when they react. This means reducing agents are generally_(b) metals. Reducing agents may also be _(c)_negatively charged ions. The most active reducing agents likely belong to the (d) alkali metal_ family on the periodic table. The most active oxidizing agents must belong to the _(e) halogen family. 2. CaI 2 b) OF 2 c) C 6 H 12 O 6 d) Rb 2 O 2 e) S 2 O 3 2- f) BeH 2 g) BrO - h) Cl a) A species that gets reduced/causes other species to be oxidized/gains e- b) A species that gets oxidized/causes other species to be reduced/loses e- Edvantage Interactive

3 c) Increased electronegativity meakes a stronger O.A. and a weaker R.A. 4. For each of the following reactions, indicate the species being oxidized and reduced and show the oxidation states above their symbols a) 2 KBrO 3 (s) 2 KBr(s) + 3 O 2 (g) Oxidized: KBrO 3 Reduced: KBrO b) Sr(s) + 2 CuNO 3 (aq) Sr(NO 3 ) 2 (aq) + 2 Cu(s) Ox d: Sr Red d: Cu c) 2 F 2 (g) + O 2 (g) 2 OF 2 (g) Oxidized: O 2 Reduced: F d) NH 4 NO 3 (s) N 2 O(g) + 2 H 2 O(l) Ox d: NH 4 NO 3 Red d: NH 4 NO 3 (solid) 5. Determine the oxidizing and reducing agent in each of the following reactions. Then indicate the number of electrons transferred by one atom of the reducing agent (a) 2 Sn(s) + O 2 (g) 2 SnO(s) OA: O 2 RA: Sn No e - : (b) 2 V(s) + 5 I 2 (g) 2 VI 5 (s) OA: I 2 RA: V No e - : (c) Sr(s) + 2 HCl(aq) SrCl 2 (aq) + H 2 (g) OA: H+ RA: Sr No e - : 2-8/ (d) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) OA: O 2 RA: C 3 H 8 No e - : 6⅔ 6. a) Write a balanced redox equation (in net ionic form) to show what has occurred in the beaker over time: Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) b) What is the oxidizing agent? Cu 2+ (aq) The reducing agent? Fe(s) c) How many electrons were transferred in the equation? 2e- 7. Give the oxidation state of the underlined element in each species: a) P b) (NH 4 ) 2 Zr(SO 4 ) 3 +4 c) Na 2 C 2 O 4 +3 d) N 2 H 5 Cl -2 e) MnO Edvantage Interactive

8. a) Write a balanced equation to represent the redox reaction occurring in the beaker. (see above) b) Oxidized? Cu metal Reduced? Ag + ion c) Reducing agent? Cu metal Oxidizing agent?

4 8. a) Write a balanced equation to represent the redox reaction occurring in the beaker. (see above) b) Oxidized? Cu metal Reduced? Ag + ion c) Reducing agent? Cu metal Oxidizing agent? Ag + ion d) 2e- a) Alkali metal family (IA) b) Halogens (VIIA) or (17) 9. a) ClO4 -, ClO3 -, ClO2 - b) Cl2, Cl Balancing Oxidation-Reduction Equations Warm Up Does this equation appear to be balanced? _yes or no_ Which species is oxidized and consequently acts as a reducing agent? Zn Which species is reduced and consequently acts as an oxidizing agent? Cu + How many electrons are lost by each atom of reducing agent being oxidized? _2_ How many electrons are gained by each ion of the oxidizing agent as it is reduced? _1_ Rewrite the equation in its balanced form: Zn(s) + 2 Cu + (aq) Zn 2+ (aq) + 2 Cu(s) Edvantage Interactive

5 Practice Problems: 1. Sm Sm e - oxidation 2. 8e H + + NO 3 - NH H 2 O reduction 3. 2e H IO 4 - IO 3 + H 2 O 2 OH - 2OH - 2e H 2 O + IO 4 - IO 3 + 2OH - reduction 4. 5 H 2 O + S 2 O SO H + + 8e - oxidation 5. 10e H BrO 3 Br H 2 O 12 OH - 12 OH - 10e H 2 O + 2 BrO 3 Br OH - reduction Practice Problems 1. 8e H + + H 3 AsO 4 AsH H 2 O (Zn Zn e - ) x 4 4 Zn + 8 H + + H 3 AsO 4 AsH H 2 O + 4 Zn 2+ Species Reactants Products Hydrogen Zinc 4 4 Oxygen 4 4 Arsenic 1 1 Charge H 2 O + C 2 H 5 OH CH 3 COOH + 4 H + + 4e - ( 2e H NO 3 N 2 O H 2 O) x 2 4 H NO 3 + C 2 H 5 OH CH 3 COOH + 2 N 2 O H 2 O Species Reactants Products Hydrogen Nitrogen 4 4 Oxygen Carbon 2 2 Charge 0 0 Edvantage Interactive

6 Practice Problems Disproportionation H 2 O + HXeO 4 4- XeO H + + 2e - 2e H HXeO 4 Xe + O H 2 O - 2 HXeO 4 4- XeO 6 + Xe + O H + 2 OH - 2 OH HXeO OH - 4- XeO 6 + Xe + O H 2 O Species Reactants Products Hydrogen 4 4 Xenon 2 2 Oxygen Charge e H BrO 3 Br H 2 O (2 Br - Br 2 + 2e - ) x 5 12 H BrO Br - 6 Br H 2 O 12 OH - 12 OH H 2 O + 2 BrO Br - 6 Br OH - Species Reactants Products Hydrogen Bromine Oxygen Charge e H + + CH 3 COO - 2 CH H 2 O 2H 2 O + CH 3 COO - 2 CO 2 + 7H + + 8e - 2 H CH 3 COO - 2 CH CO 2 2 OH - 2 OH - 2 H 2 O + 2 CH 3 COO - 2 CH CO OH - Species Reactants Products Hydrogen Carbon 4 4 Oxygen 6 6 Charge Activity Key CN ClO 3 - CNO - + Cl 2 Δ ON of C = +2, of Cl = (-5 x 2) = -10 Edvantage Interactive

7 2H CN ClO 3-5 CNO - + Cl 2 + H 2 O Species Reactants Products Hydrogen 2 2 Nitrogen 5 5 Oxygen 6 6 Carbon 5 5 Chlorine 2 2 Charge Fe + As 2 O 3 2 AsH 3 + Fe 3+ Δ ON of Fe = +3, of As = (-6 x 2) = Review Questions 12 H Fe + As 2 O 3 2 AsH Fe H 2 O 12 OH = 12 OH H 2 O + 4 Fe + As 2 O 3 2 AsH Fe OH - Species Reactants Products Hydrogen Arsenic 2 2 Oxygen Iron 4 4 Charge 0 0 1) Reactants Products Add e.g e - to products a e - to reactants b e - to reactants c e - to products d e - to products 2) a. 2 e NO H 2 O N 2 O OH - (reduction) b. 2 Cr H 2 O Cr 2 O H e - (oxidation) c. 8 e - + ClO H 2 O Cl OH - (reduction) d. S 2 O H 2 O 2 SO H e - (oxidation) 3) Edvantage Interactive

8 a. 14 e H ClO 4 - Cl H 2 O (reduction) b. H 2 O + 2 FeO Fe 2 O H e - (oxidation) c. 2 H 2 O + N 2 O 4 2 NO H e - (oxidation) 4) a. 4 e H + + CrO 4 2- Cr(OH) H 2 O (+ 6 OH - ) (+6 OH - ) 4 H 2 O + 4 e - + CrO 4 2- Cr(OH) OH - (reduction) b. 2 S 2 O 3 2- S 4 O e - (oxidation) c. 4 e H + + IO Cl - ICl H 2 O (+ 6 OH - ) (+6 OH - ) 4 e H 2 O + IO Cl - ICl OH - (reduction) 5) a. 3 x (Sn 2+ Sn e - ) 2 x (3 e H + + MnO 4 - MnO H 2 O) 8 H MnO Sn 2+ 3 Sn MnO H 2 O (+ 8 OH - ) (+8 OH - ) 4 H 2 O + 2 MnO Sn 2+ 3 Sn MnO OH - b. 4 x (V 2+ V 3+ + e - ) 4 e H H 2 SO 3 S 2 O H 2 O 2 H H 2 SO V 2+ 4 V 3+ + S 2 O H 2 O c. 10 e H IO 3 - I H 2 O 5 x (2 I - I e - ) 12 H IO I - 6 I H 2 O (+12 OH - ) (+12 OH - ) (6 H 2 O + 2 IO I - 6 I OH - ) 2 3 H 2 O + IO I - 3 I OH - Edvantage Interactive

9 d. 4 x (6 e H + + ClO 3 - Cl H 2 O) 3 x (2 H 2 O + N 2 H 4 2 NO + 8 H e - ) 3 N 2 H ClO 3-4 Cl NO + 6 H 2 O e. 2 x (3 e H + + NO 3 - NO + 2 H 2 O) 3 x (Zn Zn e - ) 8 H NO Zn 3 Zn NO + 4 H 2 O (+8 OH - ) (+8 OH - ) 4H 2 O + 2NO Zn 3Zn NO + 8OH - f. 8 x (H 2 O + ClO 3 - ClO H e - ) 16 e H ClO 3 - Cl - + Cl H 2 O 2 H ClO 3-8 ClO Cl - + Cl 2 + H 2 O g. 16 x (3 e H + + MnO 4 - MnO H 2 O) 3 x (9 H 2 O + SnS 3 O 3 3 SO Sn H e - ) 48 e H MnO 4-16 MnO H 2 O 27 H 2 O + 3 SnS 3 O 3 9 SO Sn H e - 10 H SnS 3 O MnO 4-16 MnO H 2 O + 9 SO Sn 4+ (+10 OH - ) (+10 OH - ) 5 H 2 O + 3 SnS 3 O MnO 4-16 MnO OH SO Sn 4+ h. 20 e H Mg 3 (AsO 4 ) SiO 2 As MgSiO H 2 O 10 x (H 2 O + C CO + 2 H e - ) 2 Mg 3 (AsO 4 ) SiO C 10 CO + As MgSiO 3 6) Balance a ON= (+4) ON= (-1)x2=(-2)--0 SeO F - Se + F 2 6H + + SeO F - Se + 2F 2 + 3H 2 O Edvantage Interactive

10 b ON= (-3) ON= (+2)x2=(+4) ReO Sb 2 O 3 ReO 2 + Sb 2 O 5 4 H ReO Sb 2 O 3 4 ReO Sb 2 O H 2 O (+4 OH - ) (+4 OH - ) 2 H 2 O + 4 ReO Sb 2 O 3 4 ReO Sb 2 O OH - c ON=(+4) ON=(-3) Pd + I - + NO 3 - PdI NO 16 H NO Pd + 18 I - 3 PdI NO + 8 H 2 O d ON= (+2) ON= (-2) Pb(OH) BrO - PbO 2 + Br - 2 H + + Pb(OH) BrO - PbO 2 + Br H 2 O (+2 OH - ) (+2 OH - ) Pb(OH) BrO - PbO 2 + Br - + H 2 O + 2 OH - 7) 2 K + + Cr 2 O CH 3 CH 2 OH + H + + Cl - CH 3 COOH + K + + Cl - + Cr Cl - 2 x (6 e H + + Cr 2 O Cr H 2 O) 3 x (H 2 O + CH 3 CH 2 OH CH 3 COOH + 4 H e-) 16 H Cr 2 O CH 3 CH 2 OH 3 CH 3 COOH + 4 Cr H 2 O + H 2 O 16 HCl + 2 K 2 Cr 2 O CH 3 CH 2 OH 3 CH 3 COOH + 4 CrCl H 2 O + 4 KCl 8) Na + + Cl H + + SO MnO 2 2 Na + + SO Mn Cl - + H 2 O + Cl 2 2e H + + MnO 2 Mn H 2 O 2 Cl - Cl e - 4 H + + MnO Cl - Mn H 2 O + Cl 2 4 NaCl + 2H 2 SO 4 + MnO 2 2 Na 2 SO 4 + MnCl H 2 O + Cl 2 Edvantage Interactive

11 6.3 Using the Standard Reduction Potential (SRP) Table to Predict Redox Reactions Warm Up 1. For example: both are lists of equilibria, both contain numerical values, both show one type of chemical converting to another (both show only one type of reaction), reactants get stronger as you go up on the left side of both, products get stronger as you go down on the right side of both. 2. reductions; oxidizing agents 3. alkali metals 4. fluorine; halogens Practice Problems Determining Whether a Spontaneous Redox Reaction Will Occur 1. yes; 2I + Br 2 2Br + I 2 2. no; both are oxidizing agents 3. yes; 2Ag + + Sn Sn Ag 4. no; I 2 isn t a strong enough oxidizing agent to spontaneously react with Cl Practice Problems Determining the Outcome of a Single Replacement Reaction 1. no; I 2 isn t a strong enough oxidizing agent to spontaneously react with F 2. yes; 3Cu Al 2Al Cu 3. yes; Cl 2 + 2Br Br 2 + 2Cl (* Question revised from complimentary copy) 4. no; Al 3+ isn t a strong enough oxidizing agent to spontaneously react with Sn Quick Check 1. Sr 2. Fe I 4. Sn B V V Practice Problems Determining the Predominant Redox Reaction 1. Sn 4+ + Ni Sn 2+ + Ni 2+ (* Question revised from complimentary copy) 2. 3Cu Al 2Al Cu 3. 2Cu + Cu 2+ + Cu 4. Cu + Br 2 2Br + Cu 2+ Edvantage Interactive

12 Practice Problems Determining a Chemical s Concentration via Redox Titration 1. a. Cr 2 O Fe H + 2Cr Fe 3+ +7H 2 O b L x mol = 3.00 x 10 4 mol Cr 2 O 7 2 L 3.00 x 10 4 mol Cr 2 O 2 7 x 6 mol Fe 2+ = 1.80 x 10 3 mol Fe mol Cr 2 O 7 [FeCl 2 ] = [Fe 2+ ] = 1.80 x 10 3 mol Fe 2+ = M L g Na 2 C 2 O 4 x 1 mol Na 2 C 2 O 4 = 7.46 x 10 3 mol Na 2 C 2 O g Na 2 C 2 O x mol C 2 O 4 x 2 mol MnO 4 = 2.99 x 10 3 mol MnO mol C 2 O 4 [KMnO 4 ] = [MnO 4 ] = 2.99 x 10 3 mol MnO L = M 3. a L x mol = x 10 2 mol Cr 2 O 7 2 L x 10 2 mol Cr 2 O 7 2 x 3 mol CH 3 CH 2 OH = x 10 2 mol CH 3 CH 2 OH 2 mol Cr 2 O 7 2 [CH 3 CH 2 OH] = x 10 2 mol CH 3 CH 2 OH = 1.97 M L b mol x 46.0 g x 1 ml = 115 ml CH 3 CH 2 OH = 11.5% L mol g L beverage Activity: Making an SRP Table 1. Oxidizing Reducing Agents Agents Q + e Q D 3+ + e D 2+ P + e P M e M L + + e L C e C 2. Q 3. C 2+ Edvantage Interactive

13 Review Questions 1. a. no b. yes 2. Either Fe 3+, Hg 2 2+, Ag +, or Hg a. yes; 3Mg + 2Al 3+ 3Mg Al b. no c. yes; Hg Ag Hg + 2Ag + 4. Metals bottom right of SRP Table tend to give electrons give e to chemicals above them on the left Non-metals top left of SRP Table tend to take electrons take e from chemicals below them on the right 5. a. yes; Fe + Sn 2+ Fe 2+ + Sn b. yes; F 2 + 2Br Br 2 + 2F c. no 6. a. 2I + 2Fe 3+ I 2 + 2Fe 2+ b. Br 2 + 2Fe 2+ 2Br + 2Fe Sn NO 3 + 8H + 2NO + 4H 2 O + 3Sn No. The Fe 3+ would oxidize and dissolve the Al container. 9. a. reverse b. forward 10. Mg + 2H + Mg 2+ + H The H + in acids is not a strong enough oxidizing agent to oxidize silver but the nitrate ion in nitric acid is Na + H 2 O 2Na + + H 2 + OH The OH turns the phenolphthalein pink. The H 2 is ignited by the energy released by the exothermic reaction. 13. a. Ag + Edvantage Interactive

14 b. Mg 14. Ga 3+ ; Al is being oxidized by gallium ions therefore Ga 3+ is a stronger oxidizing agent (higher on the left in the SRP table) than Al 3+ and so Ga 3+ has a greater reduction potential than Al a V b V 16. a. Fe + Co 2+ Fe 2+ + Co b. Cu + + Cr 2+ Cr 3+ + Cu (* Question revised from complimentary copy) c. Cu 2+ + Sn 2+ Sn 4+ + Cu (* Question revised from complimentary copy) 17. Zn + 2Fe 3+ 2Fe 2+ + Zn a. C 3+ b. D SRP Table C 3+ + e C 2+ A e A B e B D + + e D L x mol = 6.56 x 10 3 mol Cr 2 O 7 2 L 6.56 x mol Cr 2 O 7 x 6 mol I = x 10 2 mol I 2 1 mol Cr 2 O 7 [KI] = [I ] = x 10 2 mol I L = 2.62 M g H 2 C 2 O 4 2H 2 O x 1 mol H 2 C 2 O 4 2H 2 O = 8.33 x 10 4 mol H 2 C 2 O 4 2H 2 O g H 2 C 2 O 4 2H 2 O 8.33 x 10 4 mol H 2 C 2 O 4 x 2 mol MnO 4 = 3.33 x 10 4 mol MnO 4 5 mol H 2 C 2 O 4 [KMnO 4 ] = [MnO 4 ] = 3.33 x 10 4 mol MnO L = M 21. a. 2Cr 2 O C 2 H 5 OH + 16H + 4Cr H 2 O + 2CO 2 Edvantage Interactive

15 b L x mol = x 10 4 mol Cr 2 O 7 2 L x 10 4 mol Cr 2 O 7 2 x 1 mol C 2 H 5 OH = x 10 5 mol C 2 H 5 OH 2 mol Cr 2 O x 10 5 mol C 2 H 5 OH x 46.0 g C 2 H 5 OH = 3.50 x 10 3 g C 2 H 5 OH 1 mol C 2 H 5 OH 3.50 x 10 3 g C 2 H 5 OH = % Legally impaired now, not before Sep g C 2 H 5 OH 6.4 The Electrochemical Cell Warm Up 1. ion 2. electron x Quick Check 1. Cr 2. from right to left or from the Cr electrode to the Ag electrode 3. from right to left or from the Cr/Cr 3+ half-cell to the Ag/Ag + half-cell 4. Ag Practice Problems Determining Standard Cell Potentials (Voltages) 1. Ni Ni e E = 0.26 V Br 2 + 2e 2Br E = 1.09 V Br 2 + Ni Ni Br E = 1.35 V 2. 2Al 2Al e E = 1.66 V 3Cu e 3Cu E = 0.34 V 2Al + 3Cu 2+ 2Al Cu E = 2.00 V 3. 6I 3I 2 + 6e E = 0.54 V 2Au e 2Au E = V 6I + 2Au 3+ 3I 2 + 2Au E = V Edvantage Interactive

16 Quick Check 1. Type Alkaline Cell Lead-acid Storage Battery Anode Material Zn powder packed around a brass pin Pb alloy grids packed with spongy Pb Cathode Material MnO 2 (s) and powdered C Pb alloy grids packed with PbO 2 Fuel Cell porous carbon porous carbon Electrolyte Medium a moist paste of KOH H 2 SO 4 (aq) proton exchange membrane Use e.g. portable electronics, toys, flashlights automobiles e.g. electric vehicles, space travel, Quick Check 1. hydrated iron (III) oxide or iron(iii) oxide monohydrate 2. ½ O 2 (g) + 2H + + 2e H 2 O or ½ O 2 (g) + H 2 O + 2e 2OH 3. impurities or physical stresses or [O 2 ] or [ions] or surface area 4. coating it with paint or grease or cathodic protection or galvanizing it. Review Questions 1. Anode oxidation occurs mass decreases attracts anions electrons flow away Cathode reduction occurs mass increases attracts cations electrons flow towards 2. a. The diagram should show e flowing through the wire from the Fe/Fe 2+ halfcell to the Ni/Ni 2+ half-cell. b. Fe/Fe 2+ c. Fe Fe e ; Ni e Ni d. Fe labelled anode; Ni labelled cathode Edvantage Interactive

17 e. Fe Fe e E = V Ni e Ni E = 0.26 V Fe + Ni 2+ Fe 2+ + Ni E = V f. NH 4 + flow from the salt bridge into the Ni/Ni 2+ half-cell NO 3 flow through the salt bridge into the Fe/Fe 2+ half-cell g. The Fe electrode loses mass. The Ni electrode gains mass. 3. for example: Br2(l) NaBr(aq) Pt electrode & wire 4. a. The diagram should show e flowing through the wire from the Cr/Cr 3+ halfcell to the H 2 /H + half-cell b. H 2 /H + c. Cr Cr e ; 2H + + 2e H 2 d. Cr labelled anode; Pt labelled cathode e. 2Cr 2Cr e E = 0.74 V 6H + + 6e 3H 2 E = 0.00 V 2Cr + 6H + 2Cr H 2 E = 0.74 V f. Cr 3+ flow through the porous barrier into the H 2 /H + half-cell Cl flow through the porous barrier into the Cr/Cr 3+ half-cell g. The Cr electrode loses mass. The Pt electrode s mass is unchanged. Edvantage Interactive

18 5. F 2 (g) e flow cathode F Zn 2+ Zn anode 1 M NaF 1 M ZnCl 2 d. Zn Zn e E = 0.76 V F 2 + 2e 2F E = 2.87 V Zn + F 2 Zn F E = 3.63 V e. 6. it allows ion migration that completes the circuit 7. 2Fe 2+ 2Fe e E = 0.77 V Br 2 + 2e 2Br E = V 2Fe 2+ + Br 2 2Fe Br E = V 8. a V (the E of F 2 + 2e 2F is 2.07 V greater than the E of Ag + + e Ag) b V (the E of Mg e Mg is 3.17 V less than the E of Ag + + e Ag) 9. Cr e Cr E = 0.56 V Sr Sr e Cr e Cr Sr + Cr 2+ Sr 2+ + Cr E = V (from Table A7, p.487) E = 0.56 V E = V (provided by question) 10. a V Cu Cu e Pd e Pd E = 0.34 V (from Table A7, p.487) E = V Cu + Pd 2+ Cu 2+ + Pd E = V (provided by question) Edvantage Interactive

19 b V 2Np 2Np e 3Pd e 3Pd 2Np + 3Pd 2+ 2Np Pd E = 1.90 V E = 0.83 V (answer to 10 a.) E = 2.73 V (provided by question) 1.90 V is the Standard Oxidation Potential (SOP) of Np therefore the Standard Reduction Potential (SRP) of Np 3+ is 1.90 V. 11. Reduction Half-cell Cd e Cd Pt e Pt Oxidation Half-cell Pt Pt e 1.60 V 0 V Ni Ni e 0.17 V V Ce Ce e V V Construct an SRP Table from the data provided: Pt e Pt V (1.43 V above Ni e Ni) Ni e Ni 0.26 V (from Table A7, p.487) Cd e Cd 0.43 V (0.17 V below Ni e Ni) Ce e Ce 2.36 V (1.93 V below Cd e Cd) 12. a. increased voltage (The reactant concentrations are greater than 1 M so the cell s potential will be greater than the standard cell potential.) b. 0 V c. increased voltage (The product concentrations are less than 1 M because the S 2 ions precipitate out some Fe 2+ so the cell s potential will be greater than the standard cell potential.) d. no effect 13. For e.g. Reactants biodegradable. Products are just CO 2 and H 2 O. No metal or metal ion reactants or products. 14. Weld a zinc bar to the staircase preferably above the area exposed to ocean spray. The zinc is oxidized more readily than the steel (Fe). The zinc bar transfers electrons through the staircase to the oxidizing agents in the ocean spray thus Edvantage Interactive

20 leaving the staircase intact. 15. If the zinc is oxidized it forms zinc oxide. Zinc oxide forms a hard, impenetrable coating. Should the zinc or the zinc oxide coating be scratched off in some areas, the remaining zinc coating still provides cathodic protection. 16. The reactants in electrochemical cells react through electrical contact, via an action-at-a-distance force, without ever coming into physical contact, i.e. colliding. 6.5 The Electrolytic Cell Warm Up Device Energy Conversion From To light bulb electrical energy light energy photocell light energy electrical energy speaker electrical energy sound energy microphone sound energy electrical energy heating element electrical energy heat energy thermocouple heat energy electrical energy motor electrical energy mechanical energy generator mechanical energy electrical energy electrolytic cell electrical energy chemical energy electrochemical cell chemical energy electrical energy Practice Problems Predicting the Voltage Required to Operate an Electrolytic Cell 1. 2Ag 2Ag + + 2e E = 0.80 V Ni e Ni E = 0.26 V 2Ag + Ni 2+ 2Ag + + Ni E = 1.06 V A voltage of at least 1.06 V would be required to operate this cell. Edvantage Interactive

21 2. 2F F 2 + 2e E = 2.87 V Cu e Cu E = V 2F + Cu 2+ F 2 + Cu E = 2.53 V A voltage of at least 2.53 V would be required to operate this cell. 3. 3Sn 3Sn e E = V 2Al e 2Al E = 1.66 V Quick Check 3Sn + 2Al 3+ 3Sn Al E = 1.52 V A voltage of at least 1.52 V would be required to operate this cell. 1. electrochemical cell 2. electrolytic cell 3. both 4. electrochemical cell 5. both 6. electrolytic cell 7. electrochemical cell Practice Problems Predicting the Half-Reactions That Will Occur in an Electrolytic Cell 1. a. Type 1 2I I 2 + 2e Mg e Mg b. Type 2 H 2 O ½ O 2 + 2H + + 2e 2H 2 O + 2e H OH c. Type 3 Fe Fe e 2H 2 O + 2e H OH Edvantage Interactive

22 2. DC Source Br Quick Check Anode 2Br Br 2 + 2e E = 1.09 V Cathode Ni e Ni E = 0.26 V E = 1.35 V A voltage of at least 1.35 V would be required to operate this cell. 1. the electrolytic recovery of a metal from a solution containing its ions. 2. aluminum oxide dissolves in molten cryolite (Na 3 AlF 6 (l)) 3. sodium hydroxide and chlorine 4. impressed current cathodic protection Activity: Location, Location, Location 1. For example: proximity to source of ore 3. Brazil proximity to shipping port deep water port proximity to market western USA, Japan inexpensive electricity B.C. Hydro available work force company town inexpensive labour strong labour unions stable political climate Canada stable democracy Review Questions 1. citric acid (weak electrolyte), sodium chloride, sodium citrate, and potassium dihydrogen phosphate Edvantage Interactive

23 2. DC Source (Graphite ) Cl Na + (Graphite ) Anode 2Cl Cl 2 + 2e E = 1.36 V Cathode Na + + e Na E = 2.71 V E = 4.07 V A voltage of at least 4.07 V would be required to operate this cell. 3. because the products (Na and Cl 2 ) spontaneously react with each other 4. Cell Type (1,2,3) Electrolyte Anode/Cathode Anode Products Cathode a. 1 NaCl(l) Pt / Pt Cl 2 Na b. 2 NaCl(aq) Pt / C Cl 2 H 2 + 2OH c. 2 CuBr 2 (aq) C / C Br 2 Cu d. 2 AlF 3 (aq) C / C ½O 2 + 2H + H 2 + 2OH e. 3 CuCl 2 (aq) Cu / Cu Cu 2+ Cu 5. a. Anode H 2 O ½ O 2 + 2H + + 2e E = 1.37 V (approximate) Cathode NO 3 + 4H + + 3e NO + 2H 2 O E = V E = 0.41 V b. A voltage of at least 0.41 V would be required to operate this cell. 6. 2H 2 O + 2e H OH Edvantage Interactive

24 7. In electrorefining, the metallic ions reduced at the cathode are replaced by oxidizing an impure anode. 8. Uneconomical to electrolyze Al 2 O 3 (l) because aluminum oxide has a high melting point. Impossible to produce Al by electrolyzing Al 2 O 3 (aq) because water is a stronger oxidizing agent than Al In impressed current cathodic protection the metal being protected acts as the cathode in an electrolytic cell (non-spontaneous redox reaction) whereas in galvanic cathodic protection the metal being protected acts as the cathode in an electrochemical cell (spontaneous redox reaction). 10. DC Source Ag Ag + + e Ag + + e Ag E = 0.80 V E = V E = 0.0 In theory, it should require no voltage to operate this cell because no net reaction occurs. In practice, a small voltage is required to overcome the internal resistance of the cell. 11. a. Type 1 Cell. Only one chemical, water, can be oxidized or reduced. Water is molten H 2 O. b. Any salt containing ions which are weaker reducing agents and weaker oxidizing agents than water; for example, sodium sulphate V 2Br Br 2 + 2e E = 1.09 V Cu e Cu E = V 2Br + Cu 2+ Br 2 + Cu E = 0.75 V Edvantage Interactive

2. a) CaI 2 b) OF 2 c) C 6H 12O 6 d) Rb 2O e) S 2O. Chapter 7

2. a) CaI 2 b) OF 2 c) C 6H 12O 6 d) Rb 2O e) S 2O. Chapter 7 Chapter 7 Page 433, Practice Problems 7.1.1 1. H = +1 O = 2 2. Cs = +1 O = 1 3. C = +4 O = 2 4. Na = +1 Cr = +6 O = 2 5. Ba = +2 H = 1 6. N = 3 H = +1 7. S = 0 8. Al = +3 S = +6 O = 2 Page 434, Quick Check