Two Types of Chemical Rxns. Oxidation/Reduction. Two Types of Chemical Rxns. Two Types of Chemical Rxns. Review of Oxidation Numbers

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1 Two Types of Chemical Rxns Oxidation/Reduction Chapter Exchange of Ions no change in charge/oxidation numbers Acid/Base Rxns NaOH + HCl Two Types of Chemical Rxns Precipitation Rxns Pb(NO 3 ) 2 (aq) + KI(aq) Dissolving Rxns CaCl 2 (s) Two Types of Chemical Rxns 2. Exchange of Electrons changes in oxidation numbers/charges Fe(s) + CuSO 4 (aq) FeSO 4 (aq) + Cu(s) Remove spectator ions Fe + Cu 2+ Fe 2+ + Cu Protons Electrons Review of Oxidation Numbers Oxidation numbers the charge on an ion or an assigned charge on an atom. Al Cl 2 P 4 Mg 2+ Cl Review of Oxidation Numbers Calculate the oxidation numbers for: HClO Cr 3+ S 8 Fe 2 (SO 4 ) 3 Mn 2 O 3 SO 2 3 KMnO 4 NO 3 HSO 4 1

2 Oxidation 1. Classical Definition addition of oxygen 2. Modern Definition an increase in oxidation number Fe + O 2 Fe 2 O 3 Fe + O 2 Oxidation (limited oxygen) CO + O 2 CO 2 CH 3 CH 2 OH CH 3 CHO CH 3 COOH Fe + O 2 (excess oxygen) Oxidation Reduction C + O 2 (limited oxygen) 1. Classical addition of hydrogen 2. Modern decrease (reduction) in oxidation number N 2 + 3H 2 2NH 3 (Haber process) C + O 2 (excess oxygen) RC=CR + H 2 H H (unsaturated fat) (saturated fat) Oxidizing/Reducing Agents Oxidation and Reduction always occur together. Oxidizing Agents Get reduced Gain electrons Reducing Agents Get oxidized Lose electrons Oxidizing/Reducing Agents 0 Got oxidized, reducing agent +2 Cu + O 2 CuO 0 2 Got reduced, oxidizing agent 2

3 Identify the Oxidizing/Reducing Agents in the following (Calculate the ox. numbers also). Cu + S 8 Cu 2 S H 2 + O 2 H 2 O Cu + AgNO 3 Cu(NO 3 ) 2 + Ag H 2 O + Al + MnO 4 Al(OH) 4 + MnO 2 Balancing Redox Reactions HalfReaction Method Break eqn into oxidation half and reduction half Easy Examples: Al + Fe 2+ Fe + Al 3+ Cu + Zn 2+ Cu 2+ + Zn Mg + Na + Mg 2+ + Na Balancing Redox Reactions What s really happening: Cu 2+ + Zn Cu + Zn 2+ Balancing Redox Reactions Steps for more complicated examples 1. Balance all atoms except H and O 2. Balance charge with electrons 3. Balance O with water 4. Balance H with H + =================================== 5. (Add OH to make water in basic solutions) 3

4 Example 1: Balancing Redox Reactions MnO 4 + C 2 O 4 2 Mn 2+ + CO 2 1. Separate into half reactions MnO 4 Mn 2+ C 2 O 4 2 CO 2 Balancing Redox Reactions MnO 4 Mn e + MnO 4 Mn 2+ 5e + MnO 4 Mn H 2 O Balancing Redox Reactions C 2 O 4 2 CO 2 C 2 O 4 2 2CO (1 e per carbon) C 2 O 4 2 2CO 2 + 2e 8H + + 5e + MnO 4 Mn H 2 O Balancing Redox Reactions C 2 O 2 4 2CO 2 + 2e (X 5) 8H + + 5e + MnO 4 Mn H 2 O (X 2) Balancing Redox Reactions (Acidic Solutions) Cr 2 O Cl Cr 3+ + Cl 2 14 H + + Cr 2 O Cl 2Cr Cl 2 + 7H 2 O 5C 2 O CO e 16H e + 2MnO 4 2Mn H 2 O 16H + +5C 2 O MnO 4 2Mn CO 2 + 8H 2 O Cu + NO 3 Cu 2+ + NO 2 Cu + 4H + + 2NO 3 Cu NO 2 + 2H 2 O Mn 2+ + BiO 3 Bi 3+ + MnO 4 14H + + 2Mn BiO 3 5Bi MnO 4 + 7H 2 O 4

5 Balancing Redox Reactions (Basic Solutions) Just add enough OH to each side to make all of the H + to water. NO 2 + Al NH 3 + Al(OH) 4 5H 2 O + OH + NO 2 + 2Al NH 3 + 2Al(OH) 4 Cr(OH) 3 + ClO CrO Cl 2 2Cr(OH) 3 + 6ClO 2CrO Cl 2 + 2OH + 2H 2 O Balance in Both Acidic and Basic Solutions F + MnO 4 MnO 2 + F 2 HNO 2 + H 2 O 2 O 2 + NO H is +1 Voltaic (Galvanic) Cells F + MnO 4 MnO 2 + F 2 8H + + 6F + 2MnO 4 2MnO 2 + 3F 2 + 4H 2 O 4H 2 O + 6F + 2MnO 4 2MnO 2 + 3F 2 + 8OH HNO 2 + H 2 O 2 O 2 + NO 2HNO 2 + H 2 O 2 O 2 + 2NO + 2H 2 O Voltaic(Galvanic) Cells redox reactions that produce a voltage Spontaneous reactions ( G<0) Voltage of the cell (E o cell) is positive Batteries Electrolytic cells redox reactions that must have a current run through them. G>0 and E o cell is negative. Often used to plate metals Voltaic (Galvanic) Cells History Galvani (died 1798) uses static electricity to move the muscles of dead frogs Volta (1800) Created the first battery Voltaic (Galvanic) Cells Voltaic cell 1. Anode Oxidation site 2. Cathode Reduction site (RC cola) 3. Salt bridge completes the circuit 5

6 Voltaic (Galvanic) Cells Voltaic (Galvanic) Cells Cell Notation Zn Zn 2+ (aq) Cu 2+ (aq) Cu Anode Zn Zn e Cathode Cu e Cu Cell Cu 2+ + Zn Cu + Zn 2+ Hydrogen Electrode 1. Standard Electrode 2. Voltage(potential) = 0 Volts 2H + (aq) + 2e H 2 (g) H 2 (g) 2H + (aq) + 2e 0 volts 0 volts 3. Often used in electrodes (like ph) Standard Reduction Potentials Always written as a reduction If cell is positive, produces a voltage Rules Flipping an equation changes the sign of E Multiplying an equation does not change the magnitude of E Calculating Cell Potential A cell is composed of copper metal and Cu 2+ (aq) on one side, and zinc metal and Zn 2+ (aq) on the other. Calculate the cell potential. Zn e Zn 0.76 V Cu e Cu V flip the zinc equation Zn Zn e V Cu e Cu V Zn + Cu 2+ Zn 2+ + Cu V 6

7 Example 1 What is the cell emf of a cell made using Cu and Cu 2+ in one side and Al and Al 3+ in the other? Write the complete cell reaction. Example 2 Calculate the standard emf for the following reaction. Hint: break into halfreactions. 2Al(s) + 3I 2 (s) 2Al 3+( aq) + 6I (aq) ANS: 2.2 V Example 3 A voltaic cell is based on the following half reactions. Example 4 Calculate the standard emf for the following reaction. In + (aq) In 3+ (aq) + 2e Br 2 (l) + 2e 2Br (aq) V Cr 2 O H + + 6I 2Cr I 2 + 7H 2 O If the overall cell voltage is 1.46 V, what is the reduction potential for In 3+? Two half reactions in a voltaic cell are: Zn 2+ (aq) + 2e Zn(s) Li + (aq) + e Li(s) a) Calculate the cell emf. b) Which is the anode? Which is the cathode? c) Which electrode is consumed? d) Which electrode is positive? e) Sketch the cell, indicating electron flow. Given the following halfreactions: Pb e Pb Ni e Ni a. Calculate the cell potential (E o ). b. Label the cathode and anode. c. Identify the oxidizing and reducing agents. d. Which electrode is consumed? e. Which electrode is plated? f. Sketch the cell, indicating the direction of electron flow. 7

8 Strengths of Oxidizing and Reducing Agents larger the reduction potential, stronger the oxidizing agent Wants to be reduced, can oxidize something else. lower the reduction potential, stronger the reducing agent Would rather be oxidized F 2 + 2e 2F V Stronger Cl 2 + 2e 2Cl V Oxidizing. Agents.... Li + + e Li 3.05V Example 1 Which of the following is the strongest oxidzing agent? Which is the strongest reducing agent? NO 3 Cr 2 O 7 2 Ag + Example 2 Which of the following is the strongest reducing agent? Which is the strongest oxidizing agent? I 2 (s) Fe(s) Mn(s) Example 3 Spontaneity Can copper metal (Cu(s)) act as an oxidizing agent? Voltaic Cells Positive emf Spontaneous Can produce electric current Batteries Electrolytic Cells Negative emf Not spontaneous Must pump electricity in Electrolysis 8

9 Example 1 Are the following cells spontaneous as written? EMF and G o G = nfe a) Cu + 2H + Cu 2+ + H 2 b) Cl 2 + 2I 2Cl + I 2 c) I 2 + 5Cu H 2 O 2IO 3 + 5Cu + 12H + d) Hg I Hg + I 2 n = number of electrons transferred E = Cell emf F = 96,500 J/Vmol (Faraday s Constant) Positive Voltage gives a negative G (spont) Ex 1 Calculate the cell potential and free energy change for the following reaction: 4Ag + O 2 + 4H + 4Ag + + 2H 2 O Ex 2 Calculate G and the EMF for the following reaction. Also, calculate the K. 3Ni Cr(OH) OH 3Ni + 2CrO H 2 O ANS: V, 170 kj/mol ANS: +87 kj/mol, 0.15 V, 6 X EMF and K G o = RTlnK ( G o = nfe o ) nfe o = RTlnK lnk = nfe o (assume 298 K) RT log K = ne o Example 1 Calculate G, cell voltage and the equilibrium constant for the following cell: O 2 + 4H + + 4Fe 2+ 4Fe H 2 O ANS: 177 kj/mol, V, 1 X

10 Example 2 If the equilibrium constant for a particular reaction is 1.2 X 10 10, calculate the cell potential. Assume n = 2. Concentration Cells: Nernst Equation G = G o + RT lnq nfe = nfe o + RT lnq E = E o RT lnq (assume 298 K) nf E = E o log Q n Can adjust the voltage of any cell by changing concentrations Using the Nernst Eqn Suppose in the following cell, the concentration of Cu 2+ is 5.0 M and the concentration of Zn 2+ is M. Calculate the cell voltage. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) V E = E o V log Q n E = 1.10V V log [Cu][Zn 2+] 2 [Cu 2+ ][Zn] E = 1.10V V log [Zn 2+] 2 [Cu 2+ ] E = 1.10V V log [0.050] 2 [5.0] E = V Example 1 Calculate the emf at 298 K generated by the following cell (E o = 0.79 V) where: [Cr 2 O 7 2 ]= 2.0 M, [H + ]=1.0 M, [I ]=1.0 M and [Cr 3+ ]= 1.0 X 10 5 M. Cr 2 O H + + 6I 2Cr I 2 + 7H 2 O Example 2 Calculate the emf at 298 K generated by the following cell (E o = 2.20 V) where: [Al 3+ ]= M and [I ]=0.010 M. 2Al(s) + 3I 2 (s) 2Al 3+( aq) + 6I (aq) ANS: 0.89 V ANS: V 10

11 Example 3 If the voltage of a ZnH + cell is 0.45 V at 298 K when [Zn 2+ ]=1.0 M and P H2 =1.0 atm, what is the concentration of H +? Note that atm can be used just like molarity. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Example 4 What ph is required if we want a voltage of V and [Zn 2+ ]=0.10 M and P H2 =1.0 atm? Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) ANS: 5.2 X 10 6 M ANS: 5.84 X 10 5 M, ph = 4.22 Batteries Alkaline Batteries Lead Acid Battery 12 Volt DC Discharges when starting the car, recharges as you drive (generator). Running reaction backward. PbO 2 (s) + Pb(s) +2HSO 4 (aq) + 2H + (aq) 2PbSO 4 (s) +2H 2 O(l) Basic Zinc can acts as the anode 2MnO 2 (s)+2h 2 O(l)+2e 2MnO(OH)(s) + 2OH (aq) Zn(s) + 2OH (aq) Zn(OH) 2 (s) + 2e Rechargable uses NiCd Corrosion Iron rusts in acidic solns (not above ph=9) Water needs to be present Salts accelerate the process O 2 + 4H + + 4e 2H 2 O Fe Fe e (The Fe 2+ eventually goes to Fe 3+, Fe 2 O 3 ) 11

12 Preventing Corrosion Paint Sometimes oxide layer(al 2 O 3 ) Galvanizing (coating Fe with Zn) Fe e Fe E = 0.44 V Zn e Zn E = 0.76 V Zinc is more easily oxidized (Zn Zn e E = V) Example 1 Cathodic protection (sacrificial anode) Magnesium used in water pipes Magnesium rods used in hot water heaters An iron gutter is nailed using aluminum nails. Will the nail or the iron gutter corrode first? Fe e Fe Al e Al E = 0.44 V E = 1.66 V Al will corrode first (Al Al e,E = V) Ex 2 Which of the following metals could provide cathodic protection to iron: Electrolysis and Electroplating Plating of silver on silverware Al Cu Ni Zn 12

13 Electrolytic cells Must run electricity through them Running a voltaic cell backwards Used to produce sodium metal Na + (aq) + e Na (s) 2.71 V Cl 2 (g) + 2e 2Cl (aq) V As a voltaic cell 2Na(s) 2Na + (aq) + 2e V Cl 2 (g) + 2e 2Cl (aq) V 2Na(s) + Cl 2 (s) 2NaCl(aq) V As an electrolytic cell 2Na + (aq) + 2e 2Na (s) 2.71 V 2Cl (aq) Cl 2 (g) + 2e 1.36 V 2NaCl(aq) 2Na(s) + Cl 2 (s) 4.07 V Quantitative Electrolysis Electric current = Amperes 1 ampere = 1Coloumb I = Q 1 second t 1 F = 96,500 C/mol One mole of electrons has a charge of 96,500 C One electron has a charge of X C Example 1 What mass of aluminum can be produced in 1.00 hour by a current of 10.0 A? Al e Al Q= I t Q = (10.0 A)(3600 s) = 36,000 C Example 2 Moles of e = (36,000C)(1 mol e ) = mol e (96,500 C) Al e Al mol Al e Al mol mol Al 3.36 g Al What mass of magnesium can be produced in 4000 s by a current of 60.0 A? Mg e Mg ANS: 30.2 g Mg 13

14 Example 3 What current is required to plate 6.10 grams of gold in 30.0 min? How long would it take to plate 50.0 g of magnesium from magnesium chloride if the current is A? Au e Au (6.10 grams Au)(1mol Au) = mol Au (196.97g Au) Q = I t I = Q/t (Actually I = dq/dt) Au e Au mol Au I = 8966 C = 5.0 amps 1800 s Au e Au mol e mol Au ( mol e)(96,500 C) = 8966 C (1 mol e) Given the following: Ag + (aq) + e Ag(s) V Fe 3+ (aq) + e Fe 2+ (aq) V a. Write the reaction that occurs. b. Calculate the standard cell potential. c. Calculate G rxn for the reaction from the cell potential. d. Calculate for the reaction. e. Predict the sign of S rxn. f. Sketch the cell, labeling anode, cathode, and the direction of electron flow. Do SO 3 and SO 3 2 have the same molecular shape? How about SO 2? 14

15 16. a) Not redox b) I oxidized (1 to +5), Cl reduced (+1 to 1) c) S oxidized (+4 to +6), N reduced (+5 to +2) d) Br oxidized (1 to 0), S reduced (+6 to +4) 20 a. Mo e Mo b. H 2 O + H 2 SO 3 SO e + 4H + c. 4H + + 3e + NO 3 NO + 2H 2 O d. 4H + + 4e + O 2 2H 2 O e. 4OH + Mn 2+ MnO 2 + 2e + 2H 2 O f. 5OH + Cr(OH) 3 CrO e + 4H 2 O g. 2H 2 O + 4e + O 2 4OH 22. a. 3NO 2 + Cr 2 O H + 3NO 3 + 2Cr H 2 O b. 2HNO 3 + 2S +H 2 O 2H 2 SO 3 + N 2 O c. 2Cr 2 O CH 3 OH + 16H + 4Cr 3+ 3HCO 2 H + 11H 2 O d. 2MnO Cl + 16H + 2Mn Cl 2 + 8H 2 O e. NO 2 + 2Al + 2H 2 O NH AlO 2 f. H 2 O 2 + 2ClO 2 + 2OH O 2 + 2ClO 2 + 2H 2 O 26. a) Al oxidzed, Ni2+ reduced b) Al Al3+ + 3e Ni2+ + 2e Ni c) Al anode, Ni cathode d) Al negative, Ni positive e) Electrons flow towards the Ni electrode f) Cations migrate towards Ni electrode 34 a) Cd is anode, Pd is cathod b) E red = 0.63 V 36 a) 2.87 V b) 3.21 V c) V d) 0.636V 38 a) 1.35 V b) 0.29 V 41 a) Mg b) Ca c) H 2 d) H 2 C 2 O 4 42 a) Cl 2 b) Cd 2+ c) BrO 3 d) O a) Ce 3+ (weak reductant) b) Ca (strong reductant) c) ClO 3 (strong oxidant) d) N 2 O 5 ( oxidant) 46 a) H 2 O 2 strongest oxidizing agent b) Zn strongest reducing agent 50.a) 3.6 X 10 8 b) c) V 54 a) 4 X b) 2 X10 65 c) 7.3 X a) 2.35 V b) 2.48 V c) 2.27 V 64. a) V b) V 88. a) 173 g b) 378 min 90.E = 1.10 V W max = 212 kj/mol Cu W = 1.67 X 10 5 J 15

16 1a) 14H + + Cr 2 O Fe 2Cr Fe H 2 O b) 2Br + F 2 2F + Br 2 c) 4OH + 2Cr(OH) 3 + ClO 3 2CrO Cl + 5H 2 O 2b) V c) 89.4 kj/mol d) 4.4 X 1015 e) V 3) F 2 is str. oxidizing agent, Li, str. reducing agent 4) b) 78 minutes c) 1.19 g d) g In a measuring cup: 5 ml of oil 5 ml of ethanol 5 ml of 50% NaOH solution (approximately 30 drops). Place in beaker Heat the mixture, stirring with popsicle stick. Remove from heat. After ~5 minutes, add 10 ml of saturated salt solution. Collect some of the solid and test the ph of your soap. Compare the ph to that of commercial bar soap and liquid detergent solution. See if it lathers. 16

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