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Inflation 5762 require the following discussion. Inflation is a term used to describe

Algebra and Trigonometry | 9th Edition | ISBN: 9780321716569 | Authors: Michael Sullivan ISBN: 9780321716569 181

Solution for problem 58 Chapter 6.7

Algebra and Trigonometry | 9th Edition

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Algebra and Trigonometry | 9th Edition | ISBN: 9780321716569 | Authors: Michael Sullivan

Algebra and Trigonometry | 9th Edition

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Problem 58

Inflation 5762 require the following discussion. Inflation is a term used to describe the erosion of the purchasing power of money. For example, if the annual inflation rate is 3%, then $1000 worth of purchasing power now will have only $970 worth of purchasing power in 1 year because 3% of the original $1000 (0.03 & 1000 " 30) has been eroded due to inflation.In general, if the rate of inflation averages r per annum over n years, the amount A that $P will purchase after n years is where r is expressed as a decimal. A = P # 11 - r2n Inflation If the inflation rate averages 2%, how much will$1000 purchase in 3 years?

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Reference Electrode ­ There is no way to measure reduction potential of an isolated half reaction ­ Only the difference is potential between 2 half cells can be added ­ In order to assign E nought red to half reactions, reference electrode chosen, where the cell potential = 0.0 V Standard H Electrode 2H++2e­­>H2 EH+=0.0 V ­reversible ­not at equilibrium ­can be oxidation or reduction Example 5: Galvanic Cell Zn/Zn2+ Zn­>Zn2++2e­ anode 2H++2e­_>H2 cathoden+2H+­>Zn2++H2 Ecell= Er­Eo or Ecell= Ec­Ea 0.76=0V­EZn2+=­0.76 EZn2+ Ered ­table 20.1 in the book -oxidized+e→ reduced ­ions elements or compounds ­Top to bottom down E nought red ­low value Ered<0, easily oxidized ­Value Ered=less likely to undergo reduction ­Sign Ered attached to H+/H2 ­E red>0, easily reduced to undergo reduction ­F­+E2­>2F­ E=2.87 (most easily reduced) ­Li++e­­>Li Erednought= ­3.05, reverse reaction will occur Using E cell to Calculate E red ­galvanic cell 2Ag++Cu­>2Ag+2Cu2+ Ecell=0.46 V ­cu 0.34 V Ag+ reduction ­Ecell= EAg+­ECu 0.46 V=EAg+­(0.34) =EAg+=0.80V Fixed Spontaneity ­ 2 half reactions ­ Ecell >0 delta G<0 reaction is spontaneous ­ More + reduction occurs as written (reduction ­ Less + Enought red forced to go in reverse oxidation ­ E cell 1st ­ Half reaction more + goes up reduction, other half decreases oxidation ­ Left + half reaction, right ­half reaction Example 6: Cr+Au­>Au3++Cr3+ 1.5+0.74=2.24 V Cr3++3e→ Cr goes in the reverse direction Au3++3e­­>Au Example 7: Cd2++e2→ Cd occurs in the forward direction Mg2++2e→ Mg occurs in the reverse direction Example 8:Co2++2e­­>Co 1.50+(0.28)=1.78 V 1 Au3++3e­­>Au Spontaneity Redox ­ Ecell>0 Galvanic cell=spontaneous ­ Electrolytic cell=non spontaneous Ecell<0 Example 9: 6I­+BrO3+6h+­>3i2+bR­+3h2O 1.44­0.54=0.9 V i2+2E­­>2i­ BrO3­+gH++6e­=>Br­+3H2O Example 10: Au+Al3+­>Au3++al ­1.66­1.5 ­3.16 V nonspontaneous Ecell and delta G ­delta G max work at STP Max work=nFEcell F=96486 C Delta G= ­nFEcell Max work=mol e­(C/mol)x(J/C)=J Example 11: Au3++Al­>Au+Al3+ Ecell= 3.16 V 3(96485)3.16 =915 kJ/mol Applications ­ Equilibrium K delta G= ­RTlnK Ecell= RT/nF (LnKc) Example 12: 6I­+BrO3­+6H+­>3I2+Br­+3H2O Ecell=0.9(6)96486/8.314(298)= 210.3 k=2.1x10^90 Ecell= Ecell nought­ RT/nF (lnQ) [Nernst Equation] where [M] and P atm Example 13: Fe3++e­_>Fe2+ 0.77 Co2++2e­­>Co ­0.28 V 1.05=0.77+0.28 2Fe3++Co­>2Fe2++Co2+ net 105­(8.314)(298)/2(96485) ln[0.250)^2)(0.0050)/(0.0100)^2)] =0.98 V Example 14: Will be on exam H2+Cu2+­>2H++Cu 0.34­0=0.34 2e­ [H+] 8.44x10^­6(0.050) ­log(6.5x10^­4)=3.19 0.42=0.34­8.314(298)/2(96485) ln[H+]^2/0.050 *Clicker* [Al3+][OH­]^12/P^3O2 Q=[OH_]^12 Electrolytic Cell ­ Nonspontaneous ­ Electrical E to force a nonspontaneous reaction to occur ­ Must be molten ­ Electrodes switch ­ Anode and oxidation cathode reduction ­ Rechargeable batteries Example 15: NaCl (80 degrees C) Inert electrodes cathode Na++e→ Na anode 2Cl-->cl2+2e- 2Na++2e­­>2Na 2Na++2Cl­­>2Na+Cl2 net Electrical Conduction ­only occurs at the surface of electrodes ­More complex ­Other competing reactions expected Example 16: K2SO4 in H2O KS2O52­ H2+O2(actual) Why oxidation and reduction potentials 2 ­2.92 2K++2e­­>2k 2H2O+2e­­>H2+2OH­ ­0.83 ­0.83+2.92 cathode Anode S2O82­+2e­­>2SO42­ 2.01 V O2+4H++4e­ ­>2H2O 1.23 2.92­2.01=­4.93 V Enought cell =­2.06 V Net 2H2O+2e­­>H2+2OH0 2H2O­>O2+2H2 2H2O­>O2+4H++2e­ K2SO4 is a charged carrier also known as an electrolyte Using Reduction Potentials to predict electrolysis products ­Cathode can be Cu2++2e­­>Cu 0.34 V or2H2O+2e­­>H2+2OH­ ­Anode can be Br2+2e­­>2Br­ 1.07 O2+4H++4e­_>2H2O 1.23 ­ more + easier to reduce , more ­ easier to oxidize - net Cu2++2Br→ Cu+Br2 Example 17: 2H2O+2e­­>H2+2OH­ Anode S+2e­­>S2­ ­0.48 V 2 H2O+S2­­>H2+S+2OH­ Kinetics of electrolysis ­ 1C= 1 A(s) 1 F=96485 C/mol*e­ ­ q=lt=nF electrochemical cells (kinetic changes) Electroplating ­ Metal deposited or lost, half reaction and stoich ­ Mass of metal (mol of metal/MM)(coefficient of e­/cofficient m)=ne Calculations 2e­+Ni2+­>Ni current 0.150A(12.2 min(60)/96485=1.138x10^­3/2(158.69 g Ni)=0.033 g Ni Example 18: 0.8 A(t)/96485 Ag++e­­>Ag 2.5gx1 mol/107.9x1 mol e­/1 mol Agx2 mol/1 mol= 0.02317 mol e­(96485 C)/8x60=46.6 min *clicker* 1 gx 1 mol e­/1 mol Ag x 2 mol/ 1molx 96485/ 1moolx 1/65.3 minx1/60 =0.250 A Applications ­Batteries and electroplating Batteries ­ Galvanic cells, + charged, linked in series to get higher voltage’ ­ Two classes ­ 1. Primary cell non rechargeable Alkaline dry cell ­ 2. Secondary rechargeable Pb storage battery Alkaline Battery ­ Zn/MnO2 battery 1.5 V ­ Basic or alkaline electrolyte ­ Not rechargeable ­ Longer life, increase in current, less expensive ­ Anode Zn+2OH0 ­> ZnO+H2O+2e­ ­ Cathode 2MnO2+H2O+2e­­>Mn2O3+2OH­ Ni Cad battery ­ Ni, Cd are toxic so the disposal is a problem 3 ­ Rechargeable ­ Increase in density, releases energy quickly, can be recharged rapidly ­ Anode Cd+2OH0­>Cd(OH)+2e­ ­ Cathode NiO2+2H2O+2e­­>Ni(OH)2+2OH­ Important properties ­shelf life, rate energy output, energy vs density, specific energy Ni­ MM batteries ­ 1.35 V ­ Rechargeable ­ Laptops ­ Advantages 50% more power per vv, is useful longer ­ Anode MM+OH­­>M+h2O+e­ ­ Cathode MM+NiO(OH)­>Ni(OH)2+M Lithium ions batteries ­ Rechargeable ­ High specific energy with low mass high energy density ­ No oxidation or reduction reaction ­ Li+ ions are moved from graphite to CO2O called intercalation ­ Transport of Li+ ­ Uncharged no Li ­ Charge Li leave LiCOO2 to graphite ­ liCoO3+C6­>Li1­x+LixC6 ­ discharges=power discharge Li1­xCoO2+LixC6­>Li1­x+yCoO2 Fuel cells ­galvanic cells with reagents ­operate with reagents ­Clean burning, no electrode ions increase temperature to run H­O fuel cell ● Cathode C2+h2O+4e­­>4OH­ ● Anode H2+2OH­­>2H2O+2e­ ● Electrolyte at 200 degrees celsius, 2 porous electrodes with Pt Application ­electroplating current transfer e­ to other metals ­Run electrolytically Uses ­ Separate metals in aq solutions, separated by electrolysis ­ Decrease in voltage that gradually increases ­ 1st Ag+e→ Ag 0.80 V ­ 2nd Cu2++2e­_.Cu 0.34 V ­ 3rd Zn2++2e→ Zn -0.76 ­ Al from bauxite ore-> NaOH to get AlO2 and CO2 to Al2O3*nH2O then AlF63- that leads to AlOF62- ->Al2O62-+AlF63→ Al ­ Electroplating metals 4 ­ Cu electrolysis that precipitates other metals There will be more kinetics of this chapter not applications Corrosion Fe ­steel is iron ­Stress leaves corrosion, protective coating ­Fe­>Fe2++2e­ anode cathode CO2+2H2O+4e­­>4OH­ ­prevents coating galvanizing Zn ( a sacrifice coating), alloys Ni and Cr ­cathodic metals that sacrifice other metals ­however needs to be replace by other metals over time such as Mg 5

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Chapter 6.7, Problem 58 is Solved
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Textbook: Algebra and Trigonometry
Edition: 9
Author: Michael Sullivan
ISBN: 9780321716569

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Inflation 5762 require the following discussion. Inflation is a term used to describe