The Advanced Placement Examination in Chemistry. Part II - Free Response Questions & Answers 1970 to Electrochemistry

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1 The Advaced Placemet Examiatio i Chemistry Part II - Free Respose Questios & Aswers 970 to 005 Electrochemistry Teachers may reproduce this publicatio, i whole or i part, i limited prit quatities for o-commercial, face-to-face teachig purposes. This permissio does ot apply to ay third-party copyrights cotaied withi this publicatio. Advaced Placemet Examiatio i Chemistry. Questios copyright by the College Etrace Examiatio Board, Priceto, NJ Reprited with permissio. All rights reserved. apcetral.collegeboard.com. This material may ot be mass distributed, electroically or otherwise. This publicatio ad ay copies made from it may ot be resold. Portios copyright by Ulimited Potetial, Framigham, MA Compiled for the Macitosh ad PC by: Harvey Gedreau Framigham High School 5 A Street Framigham, MA (fax) (home office) hgedrea@framigham.k.ma.u s hgedreau@rc.com Requests for copies of these questios ad aswers as attachmets for either the Macitosh or the PC (MS-Office files) should be set to: apchemfiles@apchemistry.com. Please iclude your ame, school, school phoe, ame of pricipal/headmaster ad school website address. Do t forget to iclude the file format you wat, Mac or PC.

2 970 Accout for the observatio that silver dissolves i molar hydroiodic acid despite the fact that the stadard electrode potetial for the chage, Ag Ag + + e- is 0.80 volt. The stadard potetial is based upo [Ag + ] =.0 M. But because AgI is isoluble (very small K SP ) the cocetratio of Ag + ever reaches.0 M. Therefore, the equilibrium is shifted i favor of the productio of the io ad the potetial uder these coditios is > 0 volts. 970 Why are solutios of thiosulfate for aalysis ot acidic? Refer to the followig stadard electrode potetials ad write the balaced equatio that would accout for this fact. E _ S + 6 OH - - S O H O + 4e v - S O OH - - SO H O + 4e v - S O 3 + H O SO + H + + 4e v - S + 3 H O S O H + + 4e v - S O H + + 4e - S + 3 H O E = +0.50v - S O 3 + H O SO + H + + 4e - E = 0.40v S O H + S + H O + SO E = +0.0v The thiosulfate decomposes ito sulfur, water, ad sulfur dioxide. 97 Quatitative chemical data are ofte based o arbitrary stadards. Discuss this statemet with the followig data for fluorie (a) The atomic weight of fluorie is 9.00 (b) E, the stadard electrode potetial, is +.65 volts for the half reactio: F + e- F - (a) The atomic mass of fluorie is 9.00 times the mass of / of the mass of a carbo- atom. Carbo- has bee give the stadard atomic mass of.000 (b) the fluorie/fluoride potetial is.65 v better tha the half reactio, H + + e- H, which has bee give a value of 0.00 v. 97 Br + Fe + (aq) Br - (aq) + Fe 3+ (aq) Electrochemistry page For the reactio above, the followig data are available: Br - (aq) Br (l) + e- E = -.07 volts Fe + (aq) Fe 3+ (aq) + e- E = volts S, cal/mole. K Br (l) 58.6 Fe + (aq) -7. Br - (aq) 9.6 Fe 3+ (aq) (a) Determie S (b) Determie G (c) Determie H S _ = R S _ (a) prod. R S _ reactats = [(9.6)()+(-70.)()]-[58.6+(-7.)()] cal = cal (b) G = -IE = -()(3060cal/v)(0.30v) = cal. (c) G = H - T S ; H = G + T S = (98)(-05.4) = kcal 973 B S + Ag + S + + Ag (a) Calculate the stadard voltage of a cell ivolvig the system above. (b) What is the equilibrium costat for the system above? (c) Calculate the voltage at 5 C of a cell ivolvig the system above whe the cocetratio of Ag + is molar ad that of S + is 0.0 molar. (a) E = [0.80v - (-0.4v)] = 0.94v (b) (c) E = log K = E = E _ RT F log K OR E = FE = RT l K = 3. 8 ; K = l Q ; Q = [ S l og + ] [ Ag + ] 0. 0 ( ) = 0. 8v 974 B A steady curret of.00 ampere is passed through a electrolytic cell cotaiig a molar solutio of Ag-

3 NO 3 ad havig a silver aode ad a platium cathode util.54 grams of silver is deposited. (a) How log does the curret flow to obtai this deposit? (b) What weight of chromium would be deposited i a secod cell cotaiig molar chromium(iii) itrate ad havig a chromium aode ad a platium cathode by the same curret i the same time as was used i the silver cell? (c) If both electrodes were platium i this secod cell, what volume of O gas measured at stadard temperature ad pressure would be released at the aode while the chromium is beig deposited at the cathode? The curret ad the time are the same as i (b) (a). 54gAg mol A g. 00amp 07.9gAg (b) = 380 sec.. 54gAg mol A g 07.9gAg mol e mol A g (c) H O O + 4 H + + 4e-. 54gAg mol A g 07.9gAg mol e mol A g Electrochemistry page amp _ sec mol e = mol Cr 3+ 3 mol e 5.0 g C r 3 = g Cr + mol Cr mol e mol A g mol O 4 mol e.4 L O mol O = L O 976 B (a) Calculate the value of G for the stadard cell reactio Z + Cu + (M) Z + (M) + Cu (b) Oe half cell of a electrochemical cell is made by placig a strip of pure zic i 500 milliliters of 0.0 molar ZCl solutio. The other half cell is made by placig a strip of pure copper i 500 milliliters of 0.00 molar Cu(NO 3 ) solutio. Calculate the iitial voltage of this cell whe the two half cells are joied by a salt bridge ad the two metal strips are joied by a wire. (c) Calculate the fial cocetratio of copper io, Cu +, i the cell described i part (b) if the cell were allowed to produce a average curret of.0 ampere for 3 miutes 3 secods. (a) E = =.0 volts (b) G = -IE = - ()(3.06kcal/v)(.0v) = kcal E = E _ =.07 v log Q = log (c) (.0 amp)(93 sec.)( farad./96500 coul) = = faraday faraday/ = mol Cu + reduced ( ) mol = mol Cu + remaiig mol / L = M fial [Cu + ] 978 B (a) Whe milliliters of a solutio of 0.00 molar AgNO 3 is mixed with 00.0 milliliters of a molar CaCl solutio, what is the cocetratio of silver io after the reactio has goe to completio? (b) Write the et cell reactio for a cell formed by placig a silver electrode i the solutio remaiig from the reactio above ad coectig it to a stadard hydroge electrode. (c) Calculate the voltage of a cell of this type i which the cocetratio of silver io is M. (d) Calculate the value of the stadard free eergy chage G for the followig half reactio: Ag + (M) + e- Ag (a) Ag + + Cl - AgCl mol Ag + added = (0.300 L)(0.00 mol/l) = = mol mol CaCl added = (0.00 L)( mol/l) = = mol mol Cl - added = ( mol) = mol mol Ag + remaiig = = mol [Ag mol ] remaiig = = 0. 5 M L (b) H + Ag + H + + Ag

4 (c) E = E _ Q = [ H + ] log Q ; E_ = 0.80 volts [ Ag + ] ( P H ) = ( 4 0 ) = 65 E = ( ) v = 0.7 volt (d) G = -IE = -()(96.5kJ/mol)(0.80v) = -77 kj or -8 kcal 980 B M(s) + Cu + (aq) M + (aq) + Cu(s) For the reactio above E = volt at 5 C (a) Determie the stadard electrode potetial for the reductio half reactio: M + (aq) + e- M(s) (b) A cell is costructed i which the reactio above occurs. All substaces are iitially i their stadard states, ad equal volumes of the solutios are used. The cell is the discharged. Calculate the value of the cell potetial E, whe [Cu + ] has dropped to 0.0 molar. (c) Fid the ratio [M + ]aq/[cu + ]aq whe the cell reactio above reaches equilibrium. (a) M(s) M + (aq) + e- X v Cu + (aq) + e- Cu(s) Cu + (aq) + M(s) M + (aq) + Cu(s) Electrochemistry page v v X (oxid. potetial) = v v = v M + (aq) + e- M(s) E = v (b) M + Cu + M + + Cu iitial:.00 M.00 M chage: M M fial: 0.0 M +.80 M E cell = E _ log Q = = log [ M + ] [ Cu + ] (c) At equilibrium, E cell = 0 log = 0. 7 v E cell = 0 = E _ E _ = log [ M + ] [ Cu + ] log [ M + ] [ Cu + ] = E _ [ M + ] 5 [ Cu + = 0 ] log [ M + ] [ Cu + ] = ( )( ) = D A solutio of CuSO 4 was electrolyzed usig platium electrodes by passig a curret through the solutio. As a result, there was a decrease i both [Cu + ] ad the solutio ph; oe electrode gaied i weight a gas was evolved at the other electrode. (a) Write the cathode half reactio that is cosistet with the observatios above. (b) Write the aode half reactio that is cosistet with the observatios above. (c) Sketch a apparatus that ca be used for such a experimet ad label its ecessary compoets. (d) List the experimetal measuremets that would be eeded i order to determie from such a experimet the value of the faraday. (a) Cu + + e- Cu (b) H O O + 4 H + + 4e- (c) D.C. power supply aode cathode O deposited Cu CuSO 4 solutio (d) curret; time; mass of cathode before ad its mass after passage of curret - or - volume of O released with its temperature ad pressure. 98 B Pt

5 Whe a dilute solutio of H SO 4 is electrolyzed, O (g) is produced at the aode ad H (g) is produced at the cathode. (a) Write the balaced equatios for the aode, cathode, ad overall reactios that occur i this cell. (b) Compute the coulombs of charge passed through the cell i 00. miutes at 0.0 amperes. (c) What umber of moles of O is produced by the cell whe it is operated for 00. miutes at 0.0 amperes? (d) The stadard ethalpy of formatio of H O(g) is - 4 kilojoules per mole. How much heat is liberated by the complete combustio, at 98K ad.00 atmospheres, of the hydroge produced by the cell operated as i (c)? (a) cathode: H O + e- H + OH - or H + + e- H aode: H O O + 4 H + + 4eor O O + 4e- overall: H O H + O (b) (0.0amp)(0.0 mi)(60 mi ) = coul mol e - coul (c) coul. mol O 4 mol e - = =0.55 mol O (d) from H f for H O: H (g) + / O (g) H O(g) H = -4 kj mol O mol H O 4 kj = kj mol O mol H O whe the H is bured. 983 C Ti 3+ + HOBr TiO + + Br - (i acid solutio) (a) Write the correctly balaced half-reactios ad et ioic equatio for the skeletal equatio show above. (b) Idetify the oxidizig aget ad the reducig aget i this reactio. (c) A galvaic cell is costructed that utilizes the reactio above. The cocetratio of each species is 0.0 molar. Compare the cell voltage that will be observed with the stadard cell potetial. Explai your reasoig. sec Electrochemistry page 5 (d) Give oe example of a property of this reactio, other tha the cell voltage, that ca be calculated from the stadard cell potetial, E. State the relatioship betwee E ad the property you have specified. (a) Ti 3+ + H O TiO + + H + + e- H + + HOBr + e- Br - + H O Ti 3+ + HOBr + H O TiO H + + Br - (b) HOBr is the oxidizig aget ad Ti 3+ is the reducig aget. (c) The observed voltage will be greater tha the E value sice: E = E _ log( 0. ) 3 (d) Idetificatio of a property from the group: G, K, ph (e) G = -IE or E = log K 985 B (a) Titaium ca be reduced i a acid solutio from TiO + to Ti 3+ with zic metal. Write a balaced equatio for the reactio of TiO + with zic i acid solutio. (b) What mass of zic metal is required for the reductio of a millilitre sample of a 0.5 molar solutio of TiO +? (c) Alteratively, the reductio of TiO + to Ti 3+ ca be carried out electrochemically. What is the miimum time, i secods, required to reduce aother millilitre sample of the 0.5 molar TiO + solutio with a direct curret of.06 amperes? (d) The stadard reductio potetial, E, for TiO + to Ti 3+ is volt. The stadard reductio potetial, E, for Z + to Z (s) is volt. Calculate the stadard cell potetial, E, ad the stadard free eergy chage, G, for the reactio described i part (a). (a) Z Z + + e- TiO H + + e- Ti 3+ + H O TIO H + + Z Z + + Ti 3+ + H O (b) L 0.5 mol TiO + mol Z L mol TiO +

6 (c) 65.4 g Z Electrochemistry page 6 mol Z = 0.88 g Z + 0.5mol TiO mol e 0.500L L mol TiO coul mol e (d) Z Z + + e- amp _ sec coul..06 amp = 53 sec. E = v TiO + + 4H + + e- Ti 3+ + H O E = v G = -IE = mo l e = J coul v mo l e E = 0.83 v J v _ coul = 986 B A direct curret of 0.5 ampere was passed through 00 millilitres of a 0.5 molar solutio of Fe (SO 4 ) 3 betwee platium electrodes for a period of.00 hours. Oxyge gas was produced at the aode. The oly chage at the cathode was a slight chage i the color of the solutio. At the ed of the electrolysis, the electrolyte was acidified with sulfuric acid ad was titrated with a aqueous solutio of potassium permagaate. The volume of the KMO 4 solutio required to reach the ed poit was 4.65 millilitres. (a) How may faradays were passed through the solutio? (b) Write a balaced half-reactio for the process that occurred at the cathode durig the electrolysis. (c) Write a balaced et ioic equatio for the reactio that occurred durig the titratio with potassium permagaate. (d) Calculate the molarity of the KMO 4 solutio. (a) (.00 hr)(3600 sec/hr) = 3960 sec. (3960 sec)(0.5 amp) = 495 coul farad 495 coul (b) Fe 3+ + e- Fe coul = faraday (c) MO H + + 5e- M H O 5 Fe + 5 Fe e (d) MO H Fe + 5 Fe 3+ + M H O mol Fe farad f arad mol MO mol Fe L = M MO D A dilute solutio of sodium sulfate, Na SO 4, was electrolyzed usig iert platium electrodes. I a separate experimet, a cocetrated solutio of sodium chloride, NaCl, was electrolyzed also usig iert platium electrodes. I each experimet, gas formatio was observed at both electrodes. (a) Explai why metallic sodium is ot formed i either experimet. (b) Write balaced equatios for the half reactios that occur at the electrodes durig electrolysis of the dilute sodium sulfate solutio. Clearly idicate which half reactio occurs at each electrode. (c) Write balaced equatios for the half reactios that occur at the electrodes durig electrolysis of the cocetrated sodium chloride solutio. Clearly idicate which half reactio occurs at each electrode. (d) Select two of the gases obtaied i these experimets, ad for each gas, idicate oe experimetal procedure that ca be used to idetify it. (a) Na + is ot reduced as easily as H O (or H + or OH ). OR If Na(s) were formed, it would rapidly react with water to reform Na +. (b) Aode: H O O + 4 H + + 4e- cathode: H O + e- H + OH - or: H + + e- H (c) aode: Cl - Cl + e- cathode: H O + e- H + OH - or: H + + e- H (d) H - pop with a lit split; O - igites a glowig split; Cl - yellowish-gree color (other suitable tests accepted) 988 B A electrochemical cell cosists of a ti electrode i a acidic solutio of.00 molar S + coected by a salt

7 bridge to a secod compartmet with a silver electrode i a acidic solutio of.00 molar Ag +. (a) Write the equatio for the half cell reactio occurrig at each electrode. Idicate which half reactio occurs at the aode. (b) Write the balaced chemical equatio for the overall spotaeous cell reactio that occurs whe the circuit is complete. Calculate the stadard voltage, E, for this cell reactio. (c) Calculate the equilibrium costat for this cell reactio at 98K. (d) A cell similar to the oe described above is costructed with solutios that have iitial cocetratios of.00 molar S + ad molar Ag +. Calculate the iitial voltage, E, of this cell. (a) S S + + e- aode reactio Ag + + e- Ag (b) Ag + + S Ag + S + (c) (d) E = [0.80v - (-0.4v)] = 0.94v E = log K OR FE = RT l K log K = 3 = 3. 8 ; K = E = E _ RT F E = l Q ; Q = [ S + ] [ Ag + ] l og ( 0. 0) = v 989 B The electrolysis of a aqueous solutio of potassium iodide, KI, results i the formatio of hydroge gas at the cathode ad iodie at the aode. A sample of 80.0 millilitres of a 0.50 molar solutio of KI was electrolyzed for 3.00 miutes, usig a costat curret. At the ed of this time, the I produced was titrated agaist a 0.5 molar solutio of sodium thiosulfate, which reacts with iodie accordig to the equatio below. The ed poit of the titratio was reached whe 37.3 millilitres of the Na S O 3 solutio had bee added. I + S O 3 - I - + S 4 O 6 - (a) How may moles of I was produced durig the electrolysis? (b) The hydroge gas produced at the cathode durig the electrolysis was collected over water at 5 C Electrochemistry page 7 at a total pressure of 75 millimetres of mercury. Determie the volume of hydroge collected. (The vapor pressure of water at 5 C is 4 millimetres of mercury.) (c) Write the equatio for the half reactio that occurs at the aode durig the electrolysis. (d) Calculate the curret used durig the electrolysis L 0. 5 mols O 3 mol I L = mol S (a) O 3 = mol I (b) P H = P total - P HO = (75-4) mm Hg = 78 mm Hg mol H = mol I PV = RT ; V = RT/P ( mol )( L _ atm mol _ K )( 98 K ) V = ( ) atm = = 0.07 L (c) At aode: I I + e- (d) mo l I mol e mol I amp _ sec mol e mi 60 sec. 3 mi. = amp 99 D Explai each of the followig. (a) Whe a aqueous solutio of NaCl is electrolyzed, Cl (g) is produced at the aode, but o Na(s) is produced at the cathode. (b) The mass of Fe(s) produced whe faraday is used to reduce a solutio of FeSO 4 is.5 times the mass of Fe(s) produced whe faraday is used to reduce a solutio of FeCl 3. (c) Z + Pb + ( molar) Z + ( molar) + Pb The cell that utilizes the reactio above has a higher potetial whe [Z + ] is decreased ad [Pb + ] is held costat, but a lower potetial whe [Pb + ] is decreased ad [Z + ] is held costat. (d) The cell that utilizes the reactio give i (c) has the same cell potetial as aother cell i which [Z + ] ad [Pb + ] are each 0. molar. (a) Cl is more easily oxidized tha water

8 water is more easily reduced tha Na + (b) Fe + req. farad / mol Fe(s) or farad / mol Fe(s) Fe 3+ req. 3 farad / mol Fe(s) or farad / 3 mol Fe(s) for equal umbers of farad /:/3 ::.5: (c) usig LeChatelier s priciple if [Z + ], reactio shifts, cell potetial if [Pb + ], reactio shifts, cell potetial or usig the Nerst Equatio E cell = E - RT lq, where Q = [Z + ]/[Pb + ], if [Z + ], Q<, E cell > E if [Pb + ], Q>, E cell < E (d) [Z + ]/[Pb + ] does ot chage regardless of the values, E cell = E or [Z + ]/[Pb + ] = ; l Q = 0; E cell = E 99 B A ukow metal M forms a soluble compoud, M(NO 3 ). (a) A solutio of M(NO 3 ) is electrolyzed. Whe a costat curret of.50 amperes is applied for 35.0 miutes, 3.06 grams of the metal M is deposited. Calculate the molar mass of M ad idetify the metal. (b) The metal idetified i (a) is used with zic to costruct a galvaic cell, as show below. Write the et ioic equatio for the cell reactio ad calculate the cell potetial, E. Electrochemistry page 8 (a) mol e. 50amp amp _ sec 60 sec. mi mi mol M mol e =. 7 0 mol 3.06 g/ mol = g/mol; metal is Cd (b) Z Z + + e- Cd + + e- Cd Cd + + Z Cd + Z + E = v E = 0.40 v E = v (c) G = -IE = -()(96.5 kj/v)(0.36v) = -69 kj (d) E cell = E _ E cell = 0. 36v log [ Z + ] [ Cd + ] log 0. 0 = 0. 39v 993 D A galvaic cell is costructed usig a chromium electrode i a.00-molar solutio of Cr(NO 3 ) 3 ad a copper electrode i a.00-molar solutio of Cu(NO 3 ). Both solutios are at 5 C. (a) Write a balaced et ioic equatio for the spotaeous reactio that occurs as the cell operates. Idetify the oxidizig aget ad the reducig aget. (b) A partial diagram of the cell is show below. K + C l - M e t a l M Z i c C r C u. 0 M M ( N O ) 3. 0 M Z S O 4 (c) Calculate the value of the stadard free eergy chage, G, at 5 C for the reactio i (b). (d) Calculate the potetial, E, for the cell show i (b) if the iitial cocetratio of ZSO 4 is 0.0-molar, but the cocetratio of the M(NO 3 ) solutio remais uchaged.. 0 M C r ( N O ) 3 3 (i) Which metal is the cathode?. 0 M C u ( N O ) 3 3 (ii) What additioal compoet is ecessary to make the cell operate? (iii) What fuctio does the compoet i (ii) serve?

9 (c) How does the potetial of this cell chage if the cocetratio of Cr(NO 3 ) 3 is chaged to molar at 5 C? Explai. (a) Cr + 3 Cu + Cr Cu Cr = reducig aget; Cu + = oxidizig aget (b) (i) Cu is cathode (ii) salt bridge (iii) trasfer of ios or charge but ot electros (c) E cell decreases. 996 D use the Nerst equatio to explai aswer Sr(s) + Mg + Sr + + Mg(s) Cosider the reactio represeted above that occurs at 5 C. All reactats ad products are i their stadard states. The value of the equilibrium costat, K eq, for the reactio is at 5 C. (a) Predict the sig of the stadard cell potetial, E, for a cell based o the reactio. Explai your predictio. (b) Idetify the oxidizig aget for the spotaeous reactio. (c) If the reactio were carried out at 60 C istead of 5 C, how would the cell potetial chage? Justify your aswer. (d) How would the cell potetial chage if the reactio were carried out at 5 C with a.0-molar solutio of Mg(NO 3 ) ad a 0.0-molar solutio of Sr(NO 3 )? Explai. (e) Whe the cell reactio i (d) reaches equilibrium, what is the cell potetial? (a) (+); K > OR reactio is spotaeous OR E for Sr + is more positive OR E for Sr is more egative OR E = +0.5 v (b) Mg + RT ϒ (c) icrease; E = l Q, whe Q = the as T icreases so does E (d) icrease; E cell = E l Q, whe Q = the E cell becomes larger. Electrochemistry page 9 OR a explaatio usig LeChâtelier s Priciple. (e) E cell = B I a electrolytic cell, a curret of 0.50 ampere is passed through a solutio of a chloride of iro, producig Fe(s) ad Cl (g). (a) Write the equatio for the half-reactio that occurs at the aode. (b) Whe the cell operates for.00 hours, 0.5 gram of iro is deposited at oe electrode. Determie the formula of the chloride of iro i the origial solutio. (c) Write the balaced equatio for the overall reactio that occurs i the cell. (d) How may liters of Cl (g), measured at 5 C ad 750 mm Hg, are produced whe the cell operates as described i part (b)? (e) Calculate the curret that would produce chlorie gas from the solutio at a rate of 3.00 grams per hour. (a) Cl e - Cl (b) 0.50 amp 700 sec = 800 coulomb mol e coul coul = mol e - mol Fe g Fe 0.5 g Fe = mol Fe mol e - = mol Fe Fe + + e - Fe ; FeCl (c) Fe + + Cl Fe + Cl mol Cl (d) mol e - mol e - = mol Cl ( mol ) L atm _ ( 98 K ) mol K R T ( ) atm (e) V = P = = 0.3 L 3.00 g Cl hr OR hr 3600 sec mol Cl g Cl mol e amp _ sec mol Cl mol e - =.7 amp

10 998 D mol Cl = 0.66 g Cl 0.33 g Cl hrs. hrs. = hr amp X = 0.33 g Cl 3.00 g Cl ; X =.7 amp Aswer the followig questios regardig the electrochemical cell show. (a) Write the balaced et-ioic equatio for the spotaeous reactio that occurs as the cell operates, ad determie the cell voltage. (b) I which directio do aios flow i the salt bridge as the cell operates? Justify your aswer. (c) If 0.0 ml of 3.0-molar AgNO 3 solutio is added to the half-cell o the right, what will happe to the cell voltage? Explai. (d) If.0 gram of solid NaCl is added to each halfcell, what will happe to the cell voltage? Explai. (e) If 0.0 ml of distilled water is added to both halfcells, the cell voltage decreases. Explai. Aswer (a) Ag + + e - Ag E = v Cd e- - Cd + E = v Ag + + Cd Ag + Cd + E = +.0v (b) Aios flow ito the cadmium half-cell. As the cell operates, Cd + catios icrease i umber ad eed to be balaced by a equal umber of aio charges from the salt bridge. (c) Cell voltage will icrease. A icrease i silver io cocetratio will result i faster forward reactio ad a higher cell potetial. (d) Cell voltage will decrease. As the salt dissolves, the Cl io will cause the Ag + io to precipitate as Electrochemistry page 0 AgCl ad decrease the [Ag + ]. This will result i a slower forward reactio ad a decrease i cell potetial. Sice cadmium chloride is a soluble salt, it will ot affect the cadmium half-cell log [Cd + ] [Ag (e) E cell =.0v + ] ; while both cocetratios are.0m, the cell potetial is.0v. But if each solutio s cocetratio is cut i half, log [.5] [.5] the, E cell =.0v =.9v 000 B. Aswer the followig questios that relate to electrochemical reactios. (a) Uder stadard coditios at 5 C, Z(s) reacts with Co + (aq) to produce Co(s). (i) Write the balaced equatio for the oxidatio half reactio. (ii) Write the balaced et-ioic equatio for the overall reactio. (iii) Calculate the stadard potetial, E, for the overall reactio at 5 C. (b) At 5 C, H O decomposes accordig to the followig equatio. H O (aq) H O(l) + O (g) E = 0.55 V (i) Determie the value of the stadard free eergy chage, G, for the reactio at 5 C. (ii) Determie the value of the equilibrium costat, K eq, for the reactio at 5 C. (iii) The stadard reductio potetial, E, for the half reactio O (g) + 4 H + (aq) + 4 e - H O(l) has a value of.3 V. Usig this iformatio i additio to the iformatio give above, determie the value of the stadard reductio potetial, E for the half reactio below. O (g) + H + (aq) + e - H O (aq) (c) I a electrolytic cell, Cu(s) is produced by the electrolysis of CuSO 4 (aq). Calculate the maximum mass of Cu(s) that ca be deposited by a direct curret of 00. amperes passed through 5.00 L of.00 M CuSO 4 (aq) for a period of.00 hour.

11 (a) (i) Z(s) e - Z + (aq) (ii) Z(s) + Co + (aq) Z + (aq) + Co(s) (iii) oxid: Z(s) e - Z + (aq) Electrochemistry page E = +0.76V red: Co + (aq)+ e- Co(s) E = 0.8V +0.48V (b) (i) G = IE = ()(96500)(0.55) = 06 kj (ii) K eq = e G/RT = e ( 0650/(8.3)(98)) = (iii) H O(l) + O (g) H O (aq) E = 0.55 V O (g) + 4 H + (aq) + 4 e - H O(l) E =.3 V O (g) + 4 H + (aq) + 4 e - H O (aq) E = 0.68 V divide the equatio by but keep the E the same sec (d).00 hr hr mol e coul. 00 D 00. amp coul. amp sec mol Cu g Cu mol e- = 9 g Cu mol Cu Aswer the followig questios that refer to the galvaic cell show i the diagram above. (A table of stadard reductio potetials is prited o the gree isert ad o page 4 of the booklet with the pik cover.) (a) Idetify the aode of the cell ad write the half reactio that occurs there. (b) Write the et ioic equatio for the overall reactio that occurs as the cell operates ad calculate the value of the stadard cell potetial, E cell. (c) Idicate how the value of E cell would be affected if the cocetratio of Ni(NO 3 ) (aq) was chaged from.0 M to 0.0 M ad the cocetratio of Z(NO 3 ) (aq) remaied at.0 M. Justify your aswer. (d) Specify whether the value of K eq for the cell reactio is less tha, greater tha, or equal to. Justify your aswer. (a) zic; Z(s) Z + (aq) + e (b) Z(s) + Ni + (aq) Z + (aq) + Ni(s) E cell = (-0.5) V = +0.5 V (c) decrease E cell ; E cell = E cell log Q, Q = [Z + ] [Ni + ], whe the value of Q becomes larger tha the the log Q > ad is subtracted from the stadard potetial of the cell. (d) greater tha. All spotaeous reactios (this reactio is spotaeous because the cell potetial is larger tha 0) have a K eq that are larger tha, which favors the formatio of products. 00 B Aswer parts (a) through (e) below, which relate to reactios ivolvig silver io, Ag +. The reactio betwee silver io ad solid zic is represeted by the followig equatio. Ag + (aq) + Z(s) Z + (aq) + Ag(s) (a) A.50 g sample of Z is combied with 50. ml of 0.0 M AgNO 3 at 5 C. (i) Idetify the limitig reactat. Show calculatios to support your aswer. (ii) O the basis of the limitig reactat that you idetified i part (i), determie the value of [Z + ] after the reactio is complete. Assume that volume chage is egligible. (b) Determie the value of the stadard potetial, E, for a galvaic cell based o the reactio betwee AgNO 3 (aq) ad solid Z at 5 C. Aother galvaic cell is based o the reactio betwee Ag + (aq) ad Cu(s), represeted by the equatio below. At 5 C, the stadard potetial, E, for the cell is 0.46 V. Ag + (aq) + Cu(s) Cu + (aq) + Ag(s) (c) Determie the value of the stadard free-eergy chage, G, for the reactio betwee Ag + (aq) ad Cu(s) at 5 C.

12 (d) The cell is costructed so that [Cu + ] is M ad [Ag + ] is 0.00 M. Calculate the value of the potetial, E, for the cell. (e) Uder the coditios specified i part (d), is the reactio i the cell spotaeous? Justify your aswer. (a) (i) AgNO 3 solutio.50 g Z mol Z g Z mol Ag+ mol Z 000 ml 0.0 mol Ag + = 47 ml of silver itrate solutio required to completely react the zic, therefore, AgNO 3 is the limitig reaget. (ii) 50 ml AgNO mol AgNO ml mol Ag + mol AgNO mol Z 0.50 L (b) Ag + + e - Ag Z e - Z + mol Z mol Ag + = mol Z = M [Z + ] E = v E = v +.56 v (c) G = -FE = -()(96500)(0.46 v) = J (d) E cell = E = log [Cu+ ] [Ag + ] log = 0.38 v (e) yes; ay reactio is spotaeous with a positive voltage 004 D Required Electrochemistry page A electrochemical cell is costructed with a ope switch, as show i the diagram above. A strip of S ad a strip of ukow metal, X are used as electrodes. Whe the switch is closed, the mass of the S electrode icreases. The half-reactios are show below. S + (aq) + e S(s) X 3+ (aq) + 3 e X(s) E =? E = 0.4 V (a) I the diagram above, label the electrode that is the cathode. Justify your aswer. (b) I the diagram above, draw a arrow idicatig the directio of electro flow i the exteral circuit whe the switch is closed. (c) If the stadard cell potetial E cell is V, what is the stadard potetial, i volts for the X 3+ /X electrode? (d) Idetify metal X. (e) Write balaced et-ioic equatio for the overall chemical reactio occurrig i the cell. (f) I the cell, the cocetratio of S + is chaged from.0 M to 0.50 M, ad the cocetratio of X 3+ is chaged from.0 M to 0.0 M. (i) Substitute all appropriate values for determiig the cell potetial, E cell, ito the Nerst equatio. (Do ot do ay calculatios.) (ii) O the basis of your respose i (f) (i), will the cell potetial be greater tha, less tha, or equal to E cell? Justify your aswer. (a) ti electrode is the cathode; cathode is the site of reductio (gai i electros) ad will covert metal ios ito a metal. electro flow cathode (b) (see diagram)

13 (c) red: S + (aq) + e S(s) E = 0.4 V (d) Cr oxid: X(s) 3 e X 3+ (aq) Electrochemistry page 3 E = V E cell = V red: X 3+ (aq) + 3 e X(s) E = 0.74 V (e) 3 S + + Cr 3 S + Cr 3+ (f) (i) E cell = D (ii) greater; log ( (0.0) (0.50) 3 ) AgNO 3 (s) Ag + (aq) + NO 3 (aq) The dissolvig of AgNO 3 (s) i pure water is represeted by the equatio above.. (a) Is G for the dissolvig of AgNO 3 (s) positive, egative, or zero? Justify your aswer. (b) Is S for the dissolvig of AgNO 3 (s) positive, egative, or zero? Justify your aswer. (c) The solubility of AgNO 3 (s) icreases with icreasig temperature. (i) What is the sig of H for the dissolvig process? Justify your aswer. (ii) Is the aswer you gave i part (a) cosistet with your aswers to parts (b) ad (c) (i)? Explai. The compoud NaI dissolves i pure water accordig to the equatio NaI(s) Na + (aq) + I (aq). Some of the iformatio i the table of stadard reductio potetials give below may be useful i aswerig the questios that follow. Half-reactio E (V) O (g) + 4 H e- H O(l).3 I (s) + e- I 0.53 H O(l) + e- H (g) + OH Na + + e- Na(s) -.7 (d) A electric curret is applied to a.0 M NaI solutio. (i) Write the balaced oxidatio half reactio for the reactio that takes place. (ii) Write the balaced reductio half-reactio for the reactio that takes place. (iii) Which reactio takes place at the aode, the oxidatio reactio or the reductio reactio? (iv) All electrolysis reactios have the same sig for G. Is the sig positive or egative? Justify your aswer. (a) sig of G = ( ); sice the dissolvig of silver itrate is spotaeous, the G < 0 (b) sig of S = (+); a icrease i etropy occurs whe a solid becomes aqueous ad the products cotai more particles tha the reactats. (c) (i) sig of H = (+); a edothermic process will be favored whe the temperature is icreased. (ii) yes; G = H T S, as the temperature icreases the T S term will icrease, keepig G egative. (d) (i) I I (s) + e- (ii) H O(l) + e- H (g) + OH (iii) aode = oxidatio (iv) sig G = (+); by defiitio, a electrolysis is a o-spotaeous process ad requires the iput of eergy to get it to proceed.