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by: Mrs. Kiara Medhurst

GeneralChemistry3 CHM123

Mrs. Kiara Medhurst
GPA 3.71


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This 44 page Class Notes was uploaded by Mrs. Kiara Medhurst on Thursday October 29, 2015. The Class Notes belongs to CHM123 at Wright State University taught by DavidGrossie in Fall. Since its upload, it has received 15 views. For similar materials see /class/231125/chm123-wright-state-university in Chemistry at Wright State University.


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Date Created: 10/29/15
Thermodynamics ENTROPY FREE ENERGY AND EQUILIBRIUM Spontaneous Processes A Spontaneous process one that proceeds on its own without any external in uence hi I I Laws of Thermodynamics I I In any process spontaneous or nonspontaneous the total energy of a system and its surroundings is constant I In any spontaneous process the total entropy of a system an its surroundings always increases I The entropy of a perfectly ordered crystalline substance at 0 K is zero I Spontaneous Processes Spontaneous reaction always moves a system toward equili rlum Spontaneity or forward or reverse reaction depends on temperature pressure cernpesinen erme reac iein mixture Q lt K reaction proceeds in tne forward direction Q gt K reaction proceeds in tne reverse direction Spontaneous Processes Spontaneity ofa reaction is no indication ofthe speed ofthe reaction AI S OLUTION Botn a and o are spontaneous processes To deterrnine it c is spontaneous or HOL We rnust deterrnine Qrortne reaction it Q lt K tnen tne reaction Wiii proceed spontaneousiy in tne forward direction it Q lt K tnen tne reaction WiH proceed spontaneousiy in tne reverse direction I EXAMPLE I Determine which ofthe following processes are spontaneou a 39 are nonspontaneous a Tne odor of a dead skunkfiiis your car as you drive b o Tne nandie of a rnetai spoon in a pot of not soup is not c Tne reaction or N2 and H2 to produce NH3 at 300 ilt and partiai pressures ori 25 atrn 3 00 am and I 2 50 am respectiveiy KP 4 4 x to5 at 300 ilt S OLUTION The equii oriurn expression tortnis reaction is N2 9 3 H2 9 a 2 NW 9 P Q no 250am 06667 PHZPNZ 300am125am Q lt K Tne reaction is spontaneous in tne forward direction A Brief Review Spontaneous reactions e accompanied by conversion of potential energy to hea Some spontaneous reactions are endothermic Enthalpy alone does not account for the direction of spontaneous change A Brief Review Molecular systems tend to move spontaneously to a state of maximum randomness or disorder I im randomnm rimmed iiummewi Imam wimpy A Brief Review Spontaneous reactions Must also consider Entropy nature s tendency to move to a condition or axlmum randomness or disor er A Brief Review Entropy 8 molecular randomness or disorder State function A Al S a lrlal Slrlltlal ASlS positive increase in randomness or disorder or a 5 Stem sdlid gt liduid li uldgt gas a reacnun results in an increase in tne numoerer gaseuus mulecules dissulutlurl Elf certalrl SEIlULES ASlS negative decrease in randomness or disorder em or a syst EXAMPLE Predict the sign of AS in the system for a 392 S a 392 9 b H209 H200 c PbSs 2HN03aq gt HZSg PbN032 at h Ill Entropy and Probability Systems tend to a state of maximum mness because the random state which is more probable can be achieved in more ways Probabilities of ordered and random states are proportional to the number of ways that the st can be ac ieved hIuL l SOLUTION a AS is positive There is an increase in randomness when the number of moles of gas is increased AS is negative There is a decrease in randomness when gaseous molecules are condensed into a liquid AS is positive There is an increase in randomness when the number of moles of gas is increased 5339 o Entropy and Probability Boltzmann the entropy of a particular state is related to the number ofways that the state can be ac 39eve I S kan e number erways hattne state can be achieved K 1 sex iDz JK Entropy and ProbabilityExample Gas expands spontaneously because the state of greater volume is more probable For an rdeal gas V HRT P AS R 1n VM P V AS R mm mm Bang ne entrepy er a gas rncreases wnen rts pressure decreases at censtart temperature ne entrepy er a gas decreases wnentne pressure rncreases at censtant temperature Fm cmsmd I Entropy and Temperature Hand combinahons Probability R Ya1 ush 000154 Entropy associated with molecular motion Stxmght ush 36 0 00001385 Four ofakmd 624 000024010 Full house 3 744 0 00144058 lush 5108 0 00196540 ht 10200 0 00392465 Three of aloud 54912 0 02112845 123552 0 04753902 P w W aquot 139098240 0 42256903 lllmlt rlntihltll I M lnmlt ll Nothmg 1302540 0 501177394 h quot1 Entropy and Probability Entropy and Temperature Random molecular motion increases as the temperature ofa substanc increases increase rn tne ayerage kinetic energy ortne molecules tutal energy rs distributed arnengtne rngryrgual mulecules rn a numberufw s Boltzmannr ne rnere Ways Mtnattne energy can be distributed tne greatertne rangernness ertne state and tne nrgnertne ertrepy At absolute Zero every substance is a Solid Where partrcles are ngrdly fixed rn a crystaurne structure Entropy and Temperature Standard Molar Entropies S the entropy of1 mol ofthe pure substance at 1 atm pressure and a speci ed temperature usually 25 C Absolute entropies 7 measured with respect to an absolute rererence point Entropy and Temperature Standard Molar Entropies S Third law ofthermodynamics The entropy ofa perfectly ordered crystalline substance at 0 K is zero quotmr Wm WM wixmm r nwn Standard Entropies of Reaction i Standard entropy of reaction AS AS0 S products S reactants 1 multiplythe AS value for each substance by the balanced equation Increase in entropy whenever a molecule breaks into two or more pieces i stoichiometric coef cient of that substance in the l Calculate the AS rxn for BaCOas HZSOAU gt BaSOAS H200 0029 l l I I SOLUTION The ASDM can be calculated using information found in appendix B in your text BaCO3s stow gt BaSOys Hzow C02g AS BaSOA 132 JmolK IAS H20 599 JmolK AS co 2135 JmolK IAS BaCOa 112 JmolK AS HZSOA1569 JmolKl ASDM 132 JmolK 699 JmolK 2135 JmolK 112 JmolK 1559 JmolK 14ee JmolK l l Entropy and the Second Law of Thermodynamics Criterion for spontaneity AG 1 AGAH TAS 2 AG gt 0 nonspontaneous reaction 3 AG lt 0 spontaneous reaction 4 AG 0 reaction is at equilibrium Entropy and the Second Law of Thermodynamics First law of thermodynamics In any process spontaneous or nonspontaneous the total energy of a system and its surroundings is constant Helps keep track of energy ow between system and the surroundings Does not indicate the spontaneity of the process Entropy and the Second Law of Thermodynamics Second law of thermodynamics In any spontaneous process the total entropy of a system and its surroundings always increases Provides a clearcut criterion of spontaneity Direction of spontaneous change is alwa s Y determined by the sign of the total entropy change 3 Asmul Assyslzm t Asmmnangs AS M gt 0 spontaneous reaction AS M lt 0 nonspontaneous reaction AS M 0 reaction at equilibrium Entropy and the Second Law of Thermodynamics All reactions proceed spontaneously in the direction that increases the entropy of the system plus surroundings l l I I Entropy and the Second Law of l i Thermodynamics To determine Asma need to know Assystern and Assurroundmgs the entropy of reaction Assystem I calculate from standard molar entropies at constant pressure AHnm ASsun T AH Astotal ASign Trxn Free Energy l l I Free energy AG AH TAS TAS part of the system39s energy that is already disordered 1 AH TAS art of the system39s energy that is still ordered and therefore free to cause spontaneous change by ecoming disordere State function I Free Energy I Relationship between free energy and spontaneity NM 57 imswm AHe ms TAStotal AG and AS M have opposite signs In any spontaneous process at constant temperature and pressure the free energy of the system decreases l l I Free Energy Relationship between free energy and spontaneity In any spontaneous process at constant temperature and pressure the free energy of the system decreases Temperature acts as a weighting factor that determines the relative importance of the enthalpy and entropy changes see Table 172 page 734 in text positive to negative wt AG AHiTAS can estimate the temperature at which AG changes from h Free Energy To estimate the temperature at which AGO changes from a positive to a negative value setAGO AHO TASO 0 AH0 T EXAIVJPLE I Pure chromium is obtained by reducing CrZO3 with aluminum I Cr203s 2Als gt 2Crs AI203s Is the reaction spontaneous at 25 C Determine AH and AS for the reaction SOLUTION l l I 1 calculate AH M I I AHSm AHFA1203AHCF203 I SOLUTION 1 calculate AS rxn I A50 SFA1203 SFCT203 I SOLUTION To determine if the reaction is spontaneous we need to calculate AG for the reaction AGO AHO TASO AGO 536 kJ 298 K11089 kJK 867 kJ Since the value for AGO is negative the reaction is spontaneous Standard FreeEnergy Changes for Reactions Standard freeenergy change AGO the change in free energy that occurs when reactants in their standard states are converted to products in their standard states AG an extensive property and refers to the number of moles indicated in the chemical equation Calculate AG from the standard enthalpy change AH and the standard entropy change AS A6 AH TAS l l Problem Determine AGO for the reaction 2 NaCIs H2804 gt Na2SOAs 2 HCIg Given the following information AHO kJmol so JmolK NaCI s 4112 721 H2804 I 8140 1569 Na2804 s1387 l 1496 HCI g 923 1868 Solution Step 1 Determine AH 2 NaCIs H2solu gt Na2SOAs 2 HCIg Afon AH NaZSOAAHHClrAHf Nac1AH st001 Solution l l I Step 2 Determine AS 2 NaCIs HZSON a Na2804s 2 HCIg A5 s NaZS 04 5 HCl 7 5 NaCl s st00 Solution l Step 3 Calculate AGO AGO AHO TASO Standard Free Energies of I Formation Standard free energy of formation AGO the freeenergy change for formation of 1 mol of the substance in its standard state from the most stable form of the constituent elements in their standard states For an element in its most stable form at 25DC AGEr 0 l l I Standard Free Energies of Formation AG f measures a substance39s thermodynamic stability with respect to its constituent elements AG is negative substance is stable and does not decompose to its constituent elements AG is positive substance is thermodynamically able may not decompose to its constituent elements ifthe rate of decomposition is slow Standard Free Energies of Formation Can use to calculate standard freeenergy changes for reactions AG AG products7 AG reactants EXAIVJPLE Calculate AG rxn for the following reaction CzH4g HBrg CszBrg Given the following information CZH4 g 684 HBr g 534 CZH5Br 619 SOLUTION FreeEnergy Changes and Composition of the Reaction IVIixture To calculate the freeenergy change for a reaction when the reactants and products are present at nonstandardstate pressures and concentrations AG AG RT1nQ Q the reaction quotient an expression having the same form as he equil brium cons an expression concentrations are not necessarily at equil brium EXAIVIPLE l l l I Calculate AG for the reaction in the i above example if PCZH4 157 atm PHBr 201 atm PCZHSBr 083 I atm SOLUTION I Substituting the value for AG rxn obtained in the above example we have i l I Free Energy and Chemical 1 Equilibrium changes as the reaction progresses I The total free energy of a reaction mixture toward equilibrium Free Energy and Chemical i Equilibrium I AG AG RT111Q When the reaction mixture is mostly reactants RT In Q ltlt 0 AG lt 0 total free energy decreases as the reaction proceeds spontaneously in the forward direction Free Energy and Chemical i Equilibrium I AG AG RT1nQ When the reaction mixture is mostly products total free energy decreases as the reaction proceeds spontaneously in the reverse direc ion I Free Energy Curve Shows how the total free energy of a reaction mixture changes as the reaction progresses Goes through a minimum somewhere between pure reactants and ure products at which point the system is at equilibrium Relationship between free energy i and the equilibrium constant At equilibrium AG 0 AG RT1nK m mm pmquot AG lt 0 ln Kgt 0 equilibrium mixture is mainly products AG gt 0 ln Klt 0 equilibrium mixture is mainly reactants AG 0 In K 0 equilibrium mixture contains comparable amounts 0 reactions and products Relationship between AG and K EXANIPLE Calculate K for the reaction CzH4g HBrg CszBrg I l SOLUTION i The standard potential for the following galvanic cell is 0 ZnSIZn2aQIIEU3aQi EU2aQIPtS Calculate the standard reduction potential for the Eu 3Eu2 halfcell Tne standard thEnIlal rcirtne rcillcivnng galvanic cell is El 4n v Z SlZVl 2aqllEU 3aqi EU ZWHJlWS Calculate tne standard reducticin pcitertial rcirtne EdiEur2 halfrcell Since the Eu3lEu2 halfcell is not listed in the table of standard potentials we must nd it by difference The halfcells involved are Zns a Zn2aq 2e39 E 076 v Eu3aq 2e A Eu2aq E E52 Elquot EEu or EEu E52 E1quot 04ov 07ev 4136 v Considerthe following substances l2s Fe2aq and CrZO72aq Which is the strongest oxidizing agent Which is the weakest oxidizing agent Consider thefollowlng substances l2s Fe2 aq and Cr207239aq wnicn istne strongest oxidizing agent wnicn is theweakest oxidizing agent 2s 054 V Fe2aq 045 V Cr2072aq 133 V 39 CrZO72aq will be the strongest 39 Fe2aq will be the weakest Consider a galvanic cell that uses the reaction 39 2AQaQ SnS a 2A98 SnZaQ Calculate the potential at 25 C for a cell that has the following ion concentrations Ag 0010 M Sn2 0020 M Cunsidera galvanic celltnat usestne reacticin ZAg ad Sns s ZAgs smad Calculate tne putentlal atzs crcira cell tnat nas tne rcillcivnng icin ccincertraticins Ag El EIlEI M Snz El mm M Solution Use the Nerst equation 1 Emu 0947w10gm a 087V 0 010 Z Use standard reduction potentials to calculate the equilibrium constant at 25 C for the reaction 39 NiS 2AQaq a Ni2aq AQS Use standard reddetlen petentlalste ealedlate tne eddlllbndrn eenstart at 25 c rertne reaetlen Nls ZAg aq a M ad ZAgs 00592 E logK n logK 358 00592 K10358 647 x1035 How many grams of silver will be obtained when an aqueous silver nitrate solution is electrolyzed for 200 min with a constant current of 240 A Wmany grams err sllvervvlll be ubtalned Wnen an addedds sliver nltrate Hu Sulutlurl lS Electrulzed rer 2n n mlrl vvltn a eenstant edrrent er 2 4n A7 60 X lmole leolAgxlm mgAg3ZZgAg m lAg Z4OSgtltZOOm1ngtlt s 1mm 96500c lmole Electrolytic Cells I an electric current drives a I l I How Batteries Work lI III Redox Reaction PbN032m1 Ni S Pb S NiN032 m1 Galvanic Cells I a spontaneous chemical reaction generates an alecmc current leidation a loss of electrons an increase in oxidation number I Reduction a gain of electrons a decrease in oxidation number I Represent oxidation and reduction aspects of the reaction with halfreactions I Oxidizing agent species that causes oxidation to occur and is itself reduced I Reducing agent species that causes reduction to occur and is itself oxidized ill IllaL Combined Half Reactions If a spontaneous reaction is carried out in a eaker Oxidizing agent and reducing agent are in direct contact Electrons are directly transferred Enthalpy of reaction is lost to the surroundings for an exothermic reaction hIuL Example ofa Separated Half Reaction 32 H H Zn s Cu2aq gt Zn2aqj Cuts I For the reaction 7 consists of two halfcells e electrodes strips of zinc and copper 7 oxidation and reduction halfreactions occur at 7 electric current flows through the wire Separated Half Reactions I If spontaneous reaction is carried out in a galvanic ce chemical energy released by the reaction is converted to electrical energy I Example Describe how you would construct a galvanic ce based on the following reaction Pb2aq Zn 5 gt Pb s Zn2aq Solution Let s start by taking the overall cell reaction and breaking it into two half reactions Pb1aq 2 e39 gt Pbs Zn v gt Zn1aq 2 e39 l Solution Therefore the anode immersed in a suiutiun euntammg an39 inns such as me nitrate The cathode eumpartment Wuuid consist at a strip at head ii nitrate Solution Looking at the two half Pb1aq 2e39 gt Pbs reactions we Zn y Zn1aq 2939 oxidized Solution The two haifrceiis Wouid be Connected to each other With a Sail Solution Amth move from the cathode compartment towards the anode Whlle cattohs mlgrate from the anode Cam ode Anodlc Cell Cathodic Cell Shorthand Notation for Galvanic Cells Foanls Cuz laq an laq Culs Zn 1 w aqll leuz an I Cu ts Additional vertical line due to presence of additional phase List the gas immediately adjacent to the appropriate electrode Shorthand Notation for Galvanic Cells I Single vertical line I represents a phase boundary I Double vertical line I I represents a salt bridge Reactants in each half cell are written first followed by products ectrons move through the external circuit from left to right Example Give the shorthand notation for a galvanic cell that employs the overall reaction PbN032mI Nis rho NiN032 lI Give a brief description of the cell Solution Pb aq 2 e Pb s Ni s gt Ni2aq 2e Ni s W aq Pb2 aq Pb s This cell would consist of a strip of nickel as the anode dipping into an aqueous solution of NiN032 and a strip of Pb as the cathode dipping into an aqueous solution of PbN032 The two halfcells would be connected by a salt bridge and a wire Electromotive Force emf the driving force electrical potential that pushes the negatively charged electrons away from the anode and pulls them toward the cathode 1 Also called the cell potential E or the cell voltage 2 Potential ofa galvanic cell is a positive quantity Coulomb the amount of charge transferred when a current of 1 ampere A flows for 1 s 1J1CX1V i i Cell Potential I measured with a voltmeter Driving Forces of a Chemical Reaction cell potential E and freeenergy change AG n number of moles of electrons transferred in the reaction Ffaraday the electrical charge on 1 mol of electrons AG and E have opposite signs Spontaneous reaction has a positive cell potential but negative AG I I Standard Cell Potential E0 I The cell potential when both reactants and products are in their standard states I Solutes at 1 M concentration Gases at a partial pressure of 1 atm Solids and liquids in pure form T 25 C AGO nFE Standard Reduction Potentials Eiell Egnode Ecoathode Can t measure potential of a single electrode Measure a potential difference by placing a voltmeter between two electrodes Develop a set of standard halfcell potentials choose an arbitrary standard halfcell as a reference point and assign an arbitrary potential express the potential of all other halfcells relative to the reference halfcell Standard Hydrogen Electrode SHE Corresponding half reaction assigned an arbitrary potential ofexactly 0 V 2Haq 1M2e39 H2g1atm E 0V Shorthand notation for S H E H 1 M IH2 1 atm I Pt s Determining Standard Potentials for Halfcells I construct a galvanic cell in which the halfcell of interest is paired up with the standard hydrogen electrode Determining Standard 1 Potentials for Halfcells I l Standard oxidation potential the corresponding halfcell potential for an oxidation halfreaction l Standard reduction potential the corresponding halfcell potential for a reduction halfreaction I Whenever the direction of a halfreaction is reversed the sign of E must be reversed The standard oxidation potential and the standard reduction potential always ave e same magnitude but they have opposite signs I Construct a table of standard reduction potentials Conventions Used in Constructing a Table of Halfcell Potentials I The halfreactions are written as reductions oxidizing agents and electrons are on the reactant side reducing agents are on the product side I The halfcell potentials are standard reduction potentials also known as standard electrode potentials strongest oxidizing agents are located in the upper le of the table b strongest reducing agents are in the lower right of the table l l Conventions Used in Constructing a Table of Halfcell Potentials I The halfreactions are listed in order of decreasing standard reduction potential strongest oxidizing agents are located in the upper le o e table strongest reducing agents are in the lower right of the table I Ordering of halfreactions correspond to ordering of the oxidation reactions in the activity series the more active metals at the top ofthe activity series have the more ositive oxidation potential more negative reduction potential Using Standard Reduction Potentials I Table of standard reduction potentials summarizes an enormous amount of chemical information Can arrange any two or more oxidizing or reducing agents in order of increasing strength Predict the spontaneity or nonspontaneity of thousands of redox reac ions combine halfreac ions of interest may need to multiply halfreactions by some factor to ensure that electrons cancel 7 do not multlply values of E0 for the halfrreactlons by that factor and useEleu Excmdz 5 ume l I q Standard Reduction Potentials HalfReaction E volts Liaq e39 gt Lis Kaq e39 gt Ks 292 CaWaq 2639 gt Cas 276 Naaq e39 gt Nas 271 Zn1aq 2639 gt Zns 076 Cu aq 2639 gt Cus 034 03g 2Haq 2639 gt 02g H200 207 F2g 2639 gt 2F39aq 287 Eo Values Are Independent of l l the Amount of Reaction AG nFE AGO is an extensive property because it depends on the amount of substance change the amount of substance that reacts AGO changes by the same amount as does n the number of electrons transferred E0 AG nF remains constant Spontaneity I Spontaneity of a reaction determined by knowing the location ofthe oxidizing and reducing agent in the table An oxidizing agent can oxidize any reducing agent that lies below it in the table ED for overall reaction must be positive l l i i I I Example Write the balanced net ionic equation and calculate E for the following galvanic ce AI s iAI3 aq i Cu2 aq i Cu s Solution To write the balanced net ionic equation we need to make sure that the electrons cancel out on both sides Therefore we need to multiply the top reaction by 2 and the bottom reaction by 2As gt 2AI3aq 66 E 166V 3 Cu aq 6 e a 3 Cu 5 E0 034 V Solution AI 5 is the anode and therefore undergoes oxidation while Cu 5 is the cathode and Cu therefore undergoes reduction The half reactions and their cell potentials are AI 5 a AI aq 3 e Eo 166 V Cu aq 2 e a Cu 5 E0 034 V Notice that the sign for E0 for the AlA3 half reactions has been reversed Solution 2As gt 2 AI3aq 6e 0 166 V 3Cu2aq 66 a 3Cus E0 034V Notice that although we multiplied the coefficients in both halfreactions by the factor of 2 and 3 respectively we did not multiply the values of E0 by these factors This is because E0 values are independent of the amount of reaction The Nernst Equation and on the composition ofthe reaction mixture 1 AGAGquotRTan 39 AG inFE AGI inFE I l Cell potentials depend on temperature I 39 nFEnFE RTln Q EELM nF I I The Nernst Equation Enables us to calculate cell potentials under nonstandardstate conditions I I EERT1nQ nF EEu 10 products 7 00592 n reactantsy I Nernst equation in volts at 25 C Example reaction t t I Calculate Ecell for the following cell I 2 Crs 3 Pb2aq gt 2 Cr3aq 3 Pbs PW 015 M Cr3t 050 M Solution The halfreactions for this equation are 2Cr s a 2 Cr 6 e39 E0 074 V 2Pb2 aq 6 e a 3 Pb s E0 013 V l I The Nernst equation for this reaction is 3 Ecell Eell 00659210g EEZ t l Solution l I Substituting the information given and solving for EceH gives I 2 00592 05 Ece110617 6 logio L059 153 Electrochemical Determination l of pH I Consider a cell with a hydrogen electrode as the anode and a second reference electrode as the cathode I Pt sH2 1 atm H M Ireference I cathode Ecell 00592pH Eref Electrochemical Determination l l of pH I I pH is a linear function of the cell potential pH M 00592 can measure the pH ofa solution by measuring EEEll I Actual pH measurements use a glass electrode with a calomel electrode as the reference Example The following cell has a potential of 049 V Calculate the pH of the solution in the anode compartment l l l I Pts lH2g1 atm lHpH lC aq1MHgZC2s ngI Solution The cell reaction is H92C2S H2g 2HQU 2016quot 2WW and the standard cell potential can be calculated from the data in Appendix D The pH can be calculated using the equation pH Ecell Eref 049 028 235 00592 00592 Standard Cell Potentials and Equilibrium Constants Standard freeenergy change for a reaction is related to both the standard cell potential and the equilibrium constant AGO nFEO and AGO RT In K Can combine the two equations E0 RT1 K 2303RT n nF 1 K nF 0g 39 39 39 00592 Simplified form Eu 10 K n Standard Cell Potentials and Equilibrium Constants Positive value ofEquot corresponds to K gt 1 Negative value of E corresponds to K lt 1 D d Kfrom concentration data K IELJ L A BY K from thermochemical data 111K Kfrom electrochemical data an quot1 EXAMPLE l l the following reaction at 25 C I Calculate the equilibrium constant for 58208139aq 2s 6H20bgt1080f39aq 2O339aq 12H aq S OLUTTON The halfreactions for this reaction are SZOEZ39aq 2e gt 2304239aq E 201 v 25 BHZOU a Zlo aq 12H aq 10e39 E7 71 20V E2611 201 7120 081V bl III IILIII hIIIL I Batteries I Most important practical application of galvanic cells is their use as ba eries I Features required in abattery depend on the application I General ieatures a Compact a d lightweight a Physically rugged and inexpensi e 7 Provide a stable source of power for relatively long periods ort39rne I SOLUTION The value ofnforthis reaction is 10 We can now solve or 10081 l K 137 0g 00592 Lead Storage Battery I Used as a rellable source of power autornoblles for more tnan threerqu cerllu I l2 v baneryr Slgtlt 2 v cells connected ln senes I Anoder a senes of lead gnds packed Wltn Spongy lead for startlng arters of a I Catnode e a senes of gnds packed Wltn lead dloxlde dlpped lnto an aqueous solutlon oft 12803 38 WW Lead Storage Battery Electrode halfrreactlons and the overall CeH reactlon Anude Pbs HSOaq gt Pasqrs H aq 2 e u 296 v Ca hude Pb02s 3H aq HSOAaq 29 wagers morn E7 r 628 v Pbs Pb02s 2H aq also aq gt Foams morn E r 924v PbSOo adheres to the some or the electrodes u r 5mg an ext nar source or drrectcurrentto echarge by u er I drrvethe cerr reactrorr rrr the reverse nonspontaneous tilIILIII Alkaline Dry Cell I modified Version ofLeclanch r replace acidic NHClwith NaOH or KOH e electrm lerezctinnsr oxidation nflinc and reduction 39de ormanganese than I produces higher power and more stable current and volta e 7 more ef cient inn transport in the alkaline elect nlyte DryCell Batteries Leclanche cell I Anodee Zn metal can r I Cathode e inert graphite rod surrounded by a paste of solid Mno2 and carbon black I Electrolyte e a moist paste of NHC1 and ZnClZ e surrounds the Mod containingpaste e acidic 7 causes corrosion of the Zn anode Zn gt N 510l lllltlllltlvl Mercury Insulator Battery I used in watches heart pacemakers and other devices I small size I anode Zn same as dry cell I cathode steel in contact with HgO in an alkaline medium of KOH and ZnOH2 Nickel Cadmium Batteries used in calculators and portable power tools Rechargea e l Anode r cadmlum metal Cathode e NlOOH supported on rllckel metal I Solld products of electrode reactlorl adhere to the SH rface of the electrodes 7 alluwshatterytu herecnargeu ill quotI IL Fuel cells I a galvanic cell in Which one of the reactants is a traditional fuel Reactants are not seltcontalned Wltnln the cell Bestrkrlowrl e hydrogenroxygen fuel cel hIuL Lithium Batteries I light Weight high voltage rechargeable battery Anode e lltnlurn metal Catnod rmetal oxlde or sulnde that can lrlcorporate Ll Electrolyte e lltnlurn salt ln an orgarllc solvent Used ln cell phones laptop computers a d cam s an Win unmannlan Corrosion Corrosion the oxidative deterioration ofa metal Wellknown example of corrosion conversion of I to rust e rust is deposited at a location physically separated from e its Proposed mechanism for formation ofrust an electrochemical process in which iron is oxidized 39n one region ofthe surface and oxygen is reduced in another region 7 FE reacts Wllh Oz and l5 DdelZEd El FE Corrosion I Explains Why cars rust more rapidly When road salt is used to melt snow and ice 7 dissolved salt in water greatly increases the conductivity of the electIoly e I O2 able to oxidize all metals except a few 7 OZHZO halfreaction 1ies above the MWM halfreaction Electrolysis and Electrolytic Cells I Electrolytic cell an electric current is used drive a nonspontaneous reaction Proce ses occurring in galvanic and electrolytic cells are the reverse of each other l I Electrolysis the process of using an electric current to bring about chemical change I Prevention of corrosion shield the metal surface from oxygen and moisture 7 Durable surface coating metals such as o 39 ti c n or zin 7 Galvanizing coating by dipping into a bath of molten zinc 7 Cathodic protection protecti nosionby conne 39n itt that is more easily oxi g dized ng a metal from 0 a second metal Electrolytic Cell Two electrodes that olip into an electrolyte and are connected to a battery or some other source of direct electric current 7 battery 7 an eleetron purnp pushing eleetrons into one eleetroole andpulllng thern out ofthe other eleetroole I Anode 7 electrode Wh ere oxldatlon takes place pusmve slgh the battery pulls electmhs nut of t I Cathode 7 electrode Where reductlon takes place mega Me Sign the battery pushes Electrons lrttu t l Electrolytic Cell quotK mm am as 2 NHL y ZNn l I Electrolysis of aqueous NaCl l I Electrode reactions in an aqueous solution can differ from those for a molten salt I Cathode reaction can involve the reduction of Na or the reduction of water reduction of water preferred Equot7 less negative for H20 than Eater Na 2H20 I 26 gt H2g 20H aq i Electrolysis ofmolten NaCl I Cathode attracts Nat Na e39 gt Nal I Anode attracts CI39 2CI39 gt CI2g e39 quotV Electrolysis of aqueous NaCl I Anode reaction can involve the actual reaction the oxidation of CI39 due to overvoltage 2cr aq gt 029 26 oxidation of CI39 or the oxidation of water Electrolysis ofaqueous NaCl I Overvoltage amount of voltage needed above the calculated standard reduction or oxidation potential for electrolysis to occur ed when the Hal reacllo has a substantla oarner ror electron transfer slow rate surrnourts oarrler reacnun proceeds at satlsractory rate veNoltage needed ror Solutlorl or depOSlthrl or metals 7 large oveNoltage neededrorrorrnatlon or 02 or H2 7 can t predlct need experlmental evldence lr cell potentlals are Slrnllar Electrolysis of water Electrolysls or any aqueous Solutlon reoulrestne ce or an electrolyte to carry tne current ln Solutlon I lrtne lens or tne electrolyte are less easlly oxldlzed an Water lS tnen Water Wlll react at e5 Anode 2H20 l 029 4H aa 4639 Cathode 4H O 4e gt 2H2 40H Overall cellreactlon ZHZOU gt 029 2H29 917 Electrolysis ofaqueous NaCl l Overall cell reaction 2craq 2H20I gt Cl2g 2H2g 20H39aq Na is a spectator ion and react with the OH39 NaOH Commercial Applications of Electrolysis I Manufacture ofsodium produced commercially in a Downs cell by electrolysis a molten mixture of NaCl and CaClZ I Manufacture ofchlorine and sodium hydroxide electrolysis of aqueous NaCl I Manufacture ofaluminum HallHeroult process Downs Cell I Hall Heroult process i mm L52 uantitative As ects of ElectrolySIS ofaqueous NaCl Q p 1 i ElectrolySIS i I The amount ofsubstance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell 7 Foiiows directiyfrom the stoichiometry of the reaction andthe atomic mass of the product I i I Moles of electrons passed through a cell are determined from the electric current and the time that the current ows Moles ofe39 chargeC x m 95500 c an my zllg wqr mommy Quantitative Aspects of Electrolysis SOhlthIl i l Remember that a coulomb is an A s or that an ampere is Cs 2C139gt C12 2e39 I Sequence of conversion used to calculate the mass or volume of product produced y passing a known current for a fixed period of time current moles rams or 7 and gt charge gt moles ofer gt or gt liters or m01CS 0f CICCtrOnS 2 I time product product I Think of electrons as reactants in a balanced equation and proceed as with any other stoichiometry problem 425 gtlt350mingtlt60s 1m 1 7x 11m lcl Mg C1 328gc12 5 1mm 95500c Zmole39 lmolCl39 Example I I How many grams of CI2 would be produced in the electrolysis of molten NaCI by a current of 425 A for 350 min EXAMPLE The pH of a 0200 M solution of nicotinic acid AcidBase HC6H4NOZ is 278 Determine the value ofKa for this acid Examples SOLUTION SOLUTION To determine Ka first write the balanced To determine Ka we need to know the equation for the dissociation equilibrium concentrations of the species in the equilibrium and the equilibrium equation that defines mixture We can determine the concentration of a He O r from the pH HCSHANO2 aq H20 1 a HsOaq H301 antilog pH antilog 2780 166 C6H4NOZ39 aq X 10393 M SOLUTION SOLUTION Dissociation of one HC6H4NO2 molecule 0 The HC6H4NOZ concentration at equilibrium produces one H304r and C6H4NOZ39 ion the H304r is equal to the initial concentration minus the and C6H4NOZ39 concentrations are equal amount of HCSHANO2 that dissociates H301 C6H4N0239 166 x 103 M HC6H4NOZ 0200 166 x103 0198 M SOLUTION H30C H NO39 166x10 3p55x10 3 is K 139x 10 HC HANO 0198 EXAMPLE Calculate the concentration of all species present HsO C4H7Oz39 HC4H7OZ and OH39 in and the pH of 0250 M butyric acid HC4H7OZ SOLUTION 1 The species present initially are HC4H7O2 acid and H20 acid or base SOLUTION 2 The possible protontransfer reactions are HC4H702aq H20 1 gtH30a 7 C4H70239aq Ka 1 5 x 10395 H200 H2OD H30aq OH39aq 7 0x103914 SOLUTION 3 Since Ka gtgt Kw the principal reaction is dissociation of HC4H7O2 SOLUTION 4 Make atable Principal reaction HC H702 04gt gt H30 CoHvoz39 Initial concaatmtion o 250 o o 7x 02507 x x SOLUTION I5 Substitute the equilibrium concentrations into the equilibrian expression K 15X1075 WAXL 39 HC4H7OZ 0 2507 Assume that x is negligible compared With the initial concentration ofthe acid theref 250 x z 0250 n ore 0 Us1 g this Value in the denom1nator solVe for x x2 m 15 x 10 50250 SOLUTION 6The big equilibrium concentrations are HC4H702 0250 00019 0248 M H30 C4H7O239 19 x10393 M SOLUTION 7 The concentration of OH39 is obtained from the dissociation of Water K 10 10 14 OH39 X 4 53x 10 12 H30 19 X 10 SOLUTION 8 pH log 19 x103 272 EXAMPLE Calculate Kb for lactic acid HC3H503 given Ka 14 X 10398 Write the equilibrium reaction for Kb Solution K KWKagtltKb Kb7lt 39 1gtlt10 14 11 Kb 4714x10 14 gtlt10 C31150339aq H20 gt HCBHEOS OH39 EXAMPLE Determine whether aqueous solutions of the following salm are acidic neutral or basic Write the hydrolysis reaction for those solutions which are either acidic or basic NH4Br K3PO4 Csl CH33NHF SOLUTION To determine the acidity of an aqueous solution of a salt we must determine if the salt is derived from 1 a strong acidstrong base reaction 2 weak acidstrong base reaction 3 strong acidweak base reaction or 4 weak acidweak base reaction NH4Br Derived from the weak base NH3 aq and the strong acid HBr aq Acidic NHr is the weak conjugate acid ofNHs NHAJr aq H20 gtNH3 aq H304r aq K3P04 Derived from the strong base KOH aq and the weak acid HPO4239 aq Basic PO4339 is the weak conjugate base of HPO4239 PO4339aq H20 gt HPO4239aq OHquot Csl Derived from the strong base CsOH aq and the strong acid Hl aq Neutral CH33NHF Derived from the Weak base CH33N and the Weak acid HF aq To determine the aci 39ty of 39s solution We must calculate the Ka of the cation and the Kb ofthe anion ofthe salt and compare the two CH33NHF For 10x10 14 10x10 14 m K 15 10 CH33NH Kb 65x10 S X 10 10 10 10 M ForF39 KbX X A29x10 Kg 35x10 Ka gt Kb therefore the solution is acidic The hydrolysis reaction is CH33NHatI H20 1 CH33N 011 H30 011 Example Write the stepwise dissociation for arsenic acid HsAsO4 and determine the pH of a 2500 X 10394 M solution Ka1 562 X 10393 Kaz 170 gtlt10397Ka3 395 x10 Solution Write the stepwise dissociation for arsenic acid H3AsO4 H20 gt H3O HZASOA39 HZASOA39 H20 gt H3O HAsOAZ39 HAsOAZ39 H20 gt H3O AsOf39 Solution Determine the principle reaction H3A504 H20 H3o HZAsOA39 Solution Solution HKO HHzAsOX xxx Ka562x10 3 4 HzAsOA 25x10 7x X 12 X 103 This is large compared to Solution x2 7 000562x 000562000025 Quadratic equation gives x 24 x 10 pH36 Example What is the molar solubility of MgOH2 in a buffer solution that has a pH of 900 kSp 56 x 103912 Solution With a pH of 900 0H 100 x 105 MgOH2 lt gt Mg2 20H39 Ksp Mg2OH2 or Mg Ksp OH2 Mg 56 x1012100 x1052 0056 M


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