{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ChE102-A10-Solutions

ChE102-A10-Solutions - Chapter 20 Eloctrochcmistry(b...

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 20'. Eloctrochcmistry (b) Oxidation: {21‘ (aq)—-} 12 (s)+20'} x2 — E°=—0.535 V Reduction: 02 (g)+4H" (aq)+4e_ —) 2H20(l) E“ = +1229 V filer; 4147(aa)+02(g)+4H*(aq)—>212(s)+2H20(l) E“ =+D.694V cell AG” = —nFE;u = -(4 mol e')(96,485 Cfmol c')(0.694 V): —2.63x1051 = —268 kJ Oxidation: {Ag (s)—> Ag+ (aq)+e—} x6 — E“ = -0.800 V Reduction: Cr1012'(aq)+14 H+ (aq)+6 e' a 2 Cr” (aq)+7 H200) E°= +1.33 V Net: 6 Ag (s)+Cr2011_ («0+ 14 H’ (aq)-—> 6Ag" (aq)+ 2Cr3+ (aq)+7H20(1) 15;“ = —0.800V+1.33V = +0.53V AG“ = —nFEfd,=#(6mol e“)(96,485CImol e' )(0.53V)= —3.lx10’J = —3.1x102 k} AG°=~nFE;fl=—RflnK; 1nK=1‘—F~§°fl;Thisbecomes an= " E“ (a) RT 0.0257 °‘“ Oxidation: [Ag(s) —) Ag+ (aq) +e'} X2 —E° = 41800 V Reduction: Sn“ (aq) + 26‘ ——> Sn2+ (aq) E" = +0.154 V Net: 2Ag(s) + Sn“(aq) --> 2Ag+ (aq ) + Sn1‘(aq) Egg” 2 ~ 0.646 V 2 mo] 6’ x(—0.646 V) q — a“ = ————-~——--— = —50.3 0.0257 0.0257 S 2+ A + 2 K... =24” =1><10‘n =m[ "[S]n[..]g ] (b) Oxidation: 2Cl'(aq)—> C12(g)+2 e" —E° =—1.358 V (e) Reduction: Mao, (s )+ 4H+ (aq ) + 2c“ —> Mn2+ (aq)+ 211200) E" = +1.23 V Net: 2 Cl'(aq)+Mn01(s)+4 H*(aq)—>Mn1+ (aq)+C12 (g)+2 1-1200) 5;“ =—0.13V 2 1 ‘ —0.13V Mn2+ PC] 1,. K... =£jMHofi K... = e—m=4x10-s= [_1{_:£f_)_} 0.0257 [Cl-:l [1.1+] Oxidation: 4 OH' (:a.q)—)»O2 (g)+ ZHZOU) +4 6‘ — E“ 5 41401 V Reduction: {OCI’(aq)+H20(l)+Ze' —> Cl'(aq)+20H‘] x 2 E“ = +0890 V Net: 20CI‘(aq)——-—-+2CI‘(aq)+Oz(g) 5;" : +0409 ”V '1 - or P 0 angq=4mole (0.489 V):,m_1 K =e"-‘=1x10”=[ ] { 2(3)} 0.0257 6“ [oer T Chapter 20: Electrochemistry (a) First we must calculate the value of the equilibrium constant from the standard cell 30. potential. Esau = 0.02571r1 K ; ln Koq = MEDeli = 2 mole X0.0020V n “1 0.0257 0.0257 To determine if the described solution is possible, we compare = 0.16; Kai = a” = 1.2 2+ 2+ 2 Km] with Q. Now 11;,I =M1i.Thus, when [vb] = 0.600 M and iv“ 1 1 [VB] = [NW] = 0.675M , the ion product, Q = mmeoo) 0'675 (0675)2 Therefore, the described situation cannot occur (Le. such a mixture could not be prepared). =0.533<1.2= Ketl . (b) In order to establish equilibrium, that is, to have the ion product (0.533) become equal to 1.2, the equilibrium constant, the concentrations of the products must increase and those of the reactants must decrease. Thus. a slight net reaction to the right (formation of products), will occur. Cell reaction: Zn(s) + Ag20(s) —> Zn0(s) +2Ag(s) . We assume that the cell operates at 298 K. AG” = AG: [ZnO (5)] + 2A6: [Ag(s)]- AG: [2h (s)]—AG: [Ag10(s)] = —318.3 kJ/mol+ 2(0.00 kJa'mol)—0.00 kJ/rnoi —(—1 1.20 kJ/mol) : -307.l kJ/mol = —nFE;u AG” _ —307.1x103erol Elli =_ — nF _. 2 mol e'fmol rxn x96,485C/mol e" =1.S91V From equation (20.28) we know it = 12 and the oveall cell reaction. First we must compute value of AG” . AG" 2 Ln FE; = —12 mol e' X9648S C 1 me] e' x2.71 V = —3.14><106 J = —:-1.14><103 kJ Then we will use this value, the balanced equation and values of 1st0 to calculate AG,“ [Moral]. 4 A1(s)+302 (g)+6H20(l)7+40H' (aq)-) 4 [A1 (0H)‘] (aq) AG" = 4:30; [Al(OI-I)‘]" — 4AG:[A1(5)] —3AG;'[oz (g )1 fleacfmzo(1)] —4AG,° [OH‘ (aq )] —3.14><103 k1 = 4AG:[A1(0H)J —4x0.00 kJ-sxooo kJ —6x-(—237.1 kJ)—4x(—157.2) = 4:30: [Al(01-I)‘ ]’ + 2051.4 kl ac: [Ai(0H)J = (—3.14x103 1o — 2051.4 kJ)+4 = -—l.30x103 kJ/mol Chapter 20: Eioctrochemistry Concentration Dependence of En" —the Nernst Equation (Q Oxidation: 1211 (s)—) Zn2+ (aq)+2e' — E” = +0.763V Reduction: {Ag+ (aq)+e' -—> Ag (3)} x2 E“ = +0.800V Net: 2n(s) + 231g+ (aq)-—) Zn“ (aq)+ 2133(5) 15;”: +1.563V 2112+ 3:13;" 510592 log [ 1 = +1.563V— 00592 10g 1'0? = +1.250V 1: [Ag’] 2 x logl'oon =fl%{1;21—'5§3= 10.6; x =,)2.§x10‘” = 5x10“5 M I . Therefore, [Ag*] = 5 x 10‘6 M 34. In each case, we employ the equation Emil = 0.0592 pH. (3) Eu" = 0.0592 pH = 0.0592X 5.25 = 0.311 V (b) pH = —1og(0.0103) = 1.937 13“" = 0.0592 pH = 0.0592 x 1.937 = 0.113 v (c) K=W=LS><104=L= x2 ' [110211302] 0.158—x 0.153 x: 0.158><1.3><10-5 —“—l.'1’><10'3 M pH = —log(1.7 x 10") : 2.77 Emu = 0.0592 pH = 0.0592x2.77 = 0.164V 3—5. We first calculate E fen for each reaction and then use the Nemst equation to calculate Em]. (a) Oxidation: [Al(s)—> 131310.13 M)+3 13‘} x2 —E" = +1.676V Reduction: {Fe2+ (0.35 M)+ 23' .., Fe (3)} x 3 13° = —0.440V Net: 2311(s)+31=e1+ (0.85 M)» 2A1“ (0.13 M)+3Fe(s) 5;“ = +1.236V A1» 2 0.13 2 0.059210; ] =1_23Wfl0.059210g( ) 31249“; ’1 [Feh T 6 (0.35)3 Emu: cell (b) Oxidation: {Ag(s‘)—> Ag“ (0.34M)+e'} x2 — E" = —0.800V Reduction: Cl2 (0.55 atm)+2 e' —) 2 CI' (0.098 M) E“ = +1.358V Net: (:12 (0.55 atm)+2 Ag(s)——> 2 Cl“(0.093191)+21~.g+ (0.34M); ng = +0.553v Cl‘ 2 A , '.l . 2 . 1 0.059210g[ ] [ 8 :l : +0_553_0-°59210gM=+0_633v n PIClz(g)} 2 0.55 Eu]: : Ergo“ = Chapter 20: Electrochemistry (a) Oxidation: Mn(s)—-)Mn2*(0.40M)+2e’ —E“ =+1.1sv ('3) Reduction: {Cry (0.35M)+1e' —> Ct'1+ (0.25 M)} X2 E“ : —0.424 V Net: 2c? (0.35M)+ Mn (s)—> zor2+ (0.25 M)+ Mn“ (0.40M) 13;" = +0.76V C2» 1 M 2+ .25 I 0.40 0.0592 I: r ][ n ]—+0.76V~9fl10g'(0—')(—2=w-78V E =E“ ——lo —— cell cell n g [Cry :lz 2 (035 )2 Oxidation: {Mg (s)—> Mg“ (0.016Mj+2ei} x3 — 5° = +2.356V Reduction:l [Al(0H)‘] (0.25M)+3 e‘ —? 40H‘ (0.042 M)+Al(s)}x2 ;E° = —2.310v Ficus, Mg(s)+2[Al(0H)‘]' (0.25M)—>3Mg2+ (0.016 M)+80H' (BEzM)+2A1(s); E“; = +0.04sv M 2+ 3 OH‘ 3 3 3 0.0592103[ g ] [ _ ]2 = +0.04 6_0.0592 log (0.016) (0:342) [[AI(OH)‘] ] 6 (0-7—5) =0.046 V+0.150V=0.l96 V Eccl] = E0 " cell fl All these observations can be understood in terms of the procedure we use to balance half- equations: the ion—electron method. (a) (b) (e) The reactions for which E depends on pH are those that contain either H+(aq) or OH'(aq) in the balanced half equation. These reactions are those that involve oxoacids and oxoanions whose central atom changes oxidation state. H’(aq) will inevitably be on the left side of the reduction of an oxoanion because reduction is accompanied by not only a decrease in oxidation state, but also by the loss of oxygen atoms, as in C10; —> C102", 30f" —> SO2 , and N03" —:- N0 . These oxygen atoms appear on the right-hand side as 1120 molecules. The hydrogens that are added to the right—hand side with the water molecules are then balanced with H*(aq) on the left-hand side. If a half-reaction with H*(aq) ions present is transferred to basic solution, it may be re-balaneed by adding to each side OH’(aq) ions equal in' number to the H‘(aq) originally present. This results in H200) on the side that had H*(aq) ions (the left side in this case) and OH‘ (aq) ions on the other side (the right side.) Chapter 20: Electrochemistry (b) Reduction should occur at the cathode. The possible species that can be reduced are H20 to H2(g), K+(aq) to K(s), and SO‘ZTaq) to perhaps 802(g). Because potassium is a highly active metal, it will not be produced in aqueous solution. In order for 803‘ (aq) to be reduced. it would have to migrate to the negatively charged cathode, which is not very likely probable since like charges repel each other. Thus, H2 (g) is produced at the cathode. (c) At the anode: ZHZOU) —) 4H" (aq)+02 (g)+4c' -E° = —l.229V At the cathode: {2H” (aq)+ 2e' —> H2 (g )} x2 E“ : 0.000V Net cell reaction: 2H20(l) —> 2H2 (g)+02 (g) E35" = —l.229V A voltage greater than 1.229 V is required. Because of the high overpotential required for the formation of gases, we expect that a higher voltage will be necessary. £3; (a) The two gases that are producedare H1(g) and 01(g). (b) At the anode: 2 H200) —>4 H” (aq)+02 (g)+4e’ -E° = —1.229V At the cathode: {2}? (aq)+2e' —) H2 (g)] X2 E” = 0.000V Net cell reaction: 2H10(1)—> 2H2 (g)+ 02 (g) E;,, = —1.229v @ The electrolysis of NaZSOAaq) produces molecular oxygen at the anode. —1‘5"{O2 (ail-120}: —l.229 V . The other possible product is 32032'(aq) . It is however, unlikely to form because it has a considerably less favorable half-cell potential. —E° {520: (aqysof‘ (aq)}= —2.01 v . H2(g) is formed at the cathode. 36005 x 2.83C x 1 mole x lmol OE =0.0990 “10102 1h ls 96485C 4mole The vapor pressure of water at 25"C, from Table 12 2, is 23 8 mmHg. V— nRT _ 0 0990 1110le. 08206 Latm mol l K4 X298 K P (742—233) mmng-er— 760 mmHg mo] 01 = 3.75h x = 2.56 L 02(g) fi: (3) Zn2+(aq)+2e' —> 211(5) massonn=42.5minX 60sH187C lmole xlmolan65.39an:162an (b) 21—(aq)—-> 12(s)+2e' lmoll2 X2mole'x96,435CH1_Slfl=202min t'm ded=2.79 I x—— ——"— —- i once g 2 253.8g12 Imollz 111-10] e‘ 1.75C 60$ Chapter 20: Elocuochenfistry 66. (a) (b) (b) (C) Cu“(aq) +2e' —> Cu(s) 2.68 Cx lmol e~ xl mo] Cu“ >(1000mmol l 3 96,485 C 2 mol 0' lmol = 3.92 mmol Cu2+ 2+ decrease in [Cub] = W = 0.00922 M 425 mL final [01“] = 0.366 M —0.00922 M = 0.357 M mmo] Cu2+consumed = 282 sx mmol Ag’consurned = 255 mL(0.196 M —0.175 M) = 5.30 mmol Ag“ lmolAg+ lmole“ X96485C)< ls MK 1000 mmol Ag”r 1 mol Ag+ 1 mol 0‘ 1.84 C = 2815 = 2.8x102 5 time needed = 5.36 mmol Ag+ x 1 mol Ag 1 mo] 0' X 96,485 C char 6:1.206 A X——-f~——mx =1079 C g g g 107.87 gAg lmol Ag lmol 0’ current: 1079 C = 0.7642 A 1412 s Anode, Oxidation: 21-1200) —> 4H" (aq)+ 48" +02 (g) —E“ = —1.229V Cathode, Reduction: {Ag+ (aq)+e' —> Ag (3)] x4 E” = +0.800V Net: 2H20(l) +42%+ (aq)—) 4H+ (aq)+o2 (g)+4Ag (s) 15;;ll = —0.429 V charge = (25.8639 — 25.0732) g Agxflxflxw 2 702.8(3 107.87g Ag 1 m0] Ag 1 mo} 0‘ 702.8C 1!] x 2.00 h 36005 current = : 0.0976 A The gas is molecular oxygen. 702.8 Cxfl M 0.08206L mm x _ nRT _ 96485 C 4mole' molK P 755 mmngla—[m 760 011an (234273) K V 1 =0.0445 L Ozx%—I§L—= 44.5 mLofOz Chapter 20: Electrochemj. slry Integrative and Advanced Exercises 69. Oxidation: V3+ + H20—> V02+ + 2 H+ + e' w E°fl Reduction: Ag+ + c' —>Ag(s) E" = +0800 V Net: v” + H20+ Ag’ “memo“ + 2 H‘ + Ag(s) 5;], = 0.439 v 0.439 V = —E°a +0800 V E“; = 0.800 V —0.439 V = 0.361 V Oxidation: V2+ —>V3* +e' ' —E",, Reduction: V02+ + 2 H+ +c' -—>V3+ +H20 £3" = +0.36] V Net: V2+ + VO2+ + 2 H+ —)2 V3+ + H20 E25" = +0616 V 0.616 V = — E0» +0.36] V E°b = 0.361 V —-0.616 V = —0.255 V Thus, for the cited reaction: V3+ + e‘ -—-—)V1’ E" = ~—O.255 V a The cell reaction for discharging a lead storage battery is equation (20.24). Pb(s) + P1302 (s) + 2 H‘“ (aq) + 2HSO"(aq)——-—)PbSO4(s) + 2 H200) The half—reactions with which this equation was derived indicates that two moles of electrons are transferred for every two moles of sulfate lion consumed. We first compute the amount of H280: initially present and then the amount of H2504 consumed. 5.00 1 initial amount H230. = 1.50 MM = 7.50 mol sto‘ 1L soln 1- . ‘ 1 1H SO mount H2304c0n5umed260 h xflxflxflxflxfléfi 111 Is 96485C 2mole‘ 1molSO4 _ = 0.56 mol H 280‘ finalIHZSOJ= 7.50mol—056 mol 24.63M 1.50 L 11. The cell reaction is 2 Cl"(aq) + 2 H20(l)———)2 0H" (aq) + H2 (g) + Cl 2(g) We first determine the charge transferred per 1000 kg C12. charge = 1000 kg C12 x1000 g x flxflxw = 2.72x109 C 1kg 70.90 g Cl2 lmol C12 1 mol e‘ 9 1] 11d 5 (a) encrgy=3.4SVx2.72x10 Cx x 29.38X10 kJ IV -C 1000 J w. (b) energy=9.38x109 Jxl 3x “1 Xfl-=2.61X103 kWh 1 J 3600 s 1000 W - h - - 72. We determine the equilibrium constant for the reaction. Oxidation : Fe(s) ——> Fe“ (aq) + Ze- —E° = +0440 V Reduction: {Cr3+ (aq) + e— —‘>Cr2+ (aq)} x 2 E” = 41424 V Net: Fe(s) + 2 c:3+ (aq) —-—-)Fc2‘(aq) + 2 Crz" (aq) E26" = +0.016 V Chapter 20: Electrochemistry 82. (a) Anode: 211(5) ———) Zn“(aq) + 2e“ —E°= +0763 V Cathode: {AgCl(s)+e'(aq) —9 Ag(s)+Cl'(lM)} x 2 E°=+0.2223 V Net: Zn(s)+2AgCl(s) —a-> Zn:+(aq)+2Ag(s)+2Cl‘(lM) E3,“ = +0985 v (b) The major reason why this electrode is easier to use than the standard hydrogen electrode is that it does not involve a gas. Thus there are not the practical difficulties 0 involved in handling gases. Another reason is that it yields -a higher value of Ed" , thus, this is a more spontaneous system. (c) Oxidation: Ag(s)———~> Ag+ (aq) + e" ' v-E" = —0.800 V Reduction: AgCl(s) + e“ -—) Ag(s) + Cl" (aq) E = 40.2223 V _____________,__#_______————- Net: AgCl(s)———-—> Ag+ (aq) + Cl" (aq) E22“ = —0.5'?8 V The net reaction is the solubility reaction, for which the equilibrium constant is Kip. AG” = —nFE° = —RT ln Kw 648 C lrnole‘x9 5_><(—0.578 V)x H nFE° 1 mol e l V.C in KS = = 1 = —- 22.5 P RT 8.3145 I mol‘l K' X29815 K K = em :13 x 10‘” 5P This value is in good agreement with the value of 1.8 x 1'0:10 given in Table 18-1. (a) Ag(s) —) Ag+(aq)+e' as“: —0.300v A Is +e'——> A s +cr E" = 0.2223v AgCl(s) —> Agttaqncnaq) E°m..=-0.s771v + , —3 E :50 -0-0592 loglA—g-Jfl =05??? v-0-059210g W = .0400v cell cell 1 1 _ l 1 (b) 1000 mL of 0.0100 M Crofi‘ + 1000 mL of 1.00 x 10'3 M Ag+ 01.0.11: 110.0 mL) Concentration of C1042'after dilution: 0.0100 Mx10.00 mL/l 10.00 mL = 0.000909 M Concentration of Ag” after dilution: 0.00100 Mx100.0 mL1'110.00 roL : 0.000909 M Agzcroas) _..»<__. 2Ag*(aq) + mos-(3.1) Initial 0.000909 M 0.000909 M Change(100% rxn) 41000909 M -0.000455 M New initial 0 M 0.000455 M Change I +2: +1: Equilibrium 21: 0.000455 M+x z0.000455 M 1.1 x 10'” = (2x)2(0.000454) x = 0.0000246 M Note: 5.4% of 0.000455 M (assumption may be considered valid) (Answer would be x = 0.0000253 using method of successive approx.) [Ag+] = 2x = 0.0000492 M (0.0000506 M using method of successive approx.) 0.0592 log t A31“ Cf 1: g 0. 5771 _ 0.0592 Em" = 5;" — log [1.00M]{4.92x10“‘M] = —0.323 V (0.306 V for method of successive approximations) Chapter 20: Electrochemistry (c) 10.00 ml. 0.0100 M Nag +1000 1111.. 0.0100 M (3103‘ + 100.0 mL 100x103 M Ag+ (th = 120.0 mi.) Concentration of NH3 after dilution: 10.0 Mx10.00 ELL/120.00 mL = 0.833 M Concentration of Crof‘aftet dilution: 0.0100 Mx10.00 mL 1110.00 in]. = 0.000333 M Concentration of Ag+ after dilution: 0.00100 Mx100.0 le1 10.00 mL = 0.000833 M In order to determine the equilibrium concentration of free Ag+(aq), we first consider complexation of Ag+(aq) by NH3(aq) and then check to see if precipitation occurs. Aaron) + mono—1% AgmHz)2+(aCI) Initial 0.000833 M 0.833 M 0 M Change(100% rxn) 0000833 M 000167 M +0.000833 M New initial 0 M 0.831 M 0.000833 M Change +x +2.1: —x Equilibrium x (0.831+2x) M (0.000833 —x) M Equilibrium (x =0) x 0.831 M 0.000833 M 1.6 x 101 = 0.000333 /x(0.831)2 x = 7.55010” M = [Ag+] Note: The assumption is valid Now we look to see if a precipitate forms: Qle = (754310" l)2(0.000833) : 4.7 x10“24 Since Q51, < K5,, (1.1 x 10‘”), no precipitate forms and [Ag+] = 7.54_>< 10'11 M 00592 log {signer}: 415771 V _ 0.0592 Em, = 0.02131 Euall = 13;“ - log {LOOMHISj x 10'”M] 34. We assume that the sz+(aq) is “ignored" by the silver electrode. that is, the silver electrode detects only silver ion in solution. Oxidation : Hz(g) —> 2 H*(aq) + 2 e‘ ~E°=0.000 V Reduction: [Ag‘ (aq) + e" —) Ag(s)} x 2 15°: 0.800 V Net: H2(g) + 2 Ag+ —-——> 2 H*(aq) + 2 Ag(s) Ema." = 0.800 V 0.0592 * 2 2 End, = E3“ — log [H +12 0503 V = 0.800 v ~ “0592 log 1'00 2 [Ag ] 2 [Ag"]: .002 2 0.300 — 0. 3 . 2 log 1 + 2 = (———fl = 10.0 100 =10+1°° — 1.0 ><10m {Ag ] 0.0592 {Agt}2 [Ag*]2 = 1.0 x10"° M2 [Ag’] = 1.0 x 105 M —5 «- mass Ag :05“) Lx LOX 10 mol Ag x lrnol Ag x 107.87 g Ag: 5.4x10“ g Ag 1 L soln 1 mol Ag" 1 mol Ag 5.4 x 10* g Ag a A = a g 1.050gsample x 100 % = 0.051% Ag (by mass) Chapter 20: Electrochemistry 87. (a) Anode: 21-1200) m) 4 e’ + 4 H+(aq) + 02(g) Cathode: 2 H20111+ 2 e' —> 2 Dilly) + H3! g1 Overall: 2 H200) ++LH26fU '9 m + 4%) + 2 H2(g) + 02(g) 2 H200) —> 2 H2(g) + 020:.) (b) 21.5 111A = 0.0215 A 01' 0.0215 C s'] for 683 5 0.0215 C 1 mol e’ 1 mol H‘ x] mol H280 x6335x x " =7.61x10'5 molI-IISO‘ S 96485C 111101 c' 2molH+ 7.61x10" mol HZSO‘ in 10.00 rnL. Hence [HISO‘] = 7.61x10“ mol1'0.01000 L 2 7.61x10"M First we need to find the total surface area 4"“ Outer circumference = 2m = 21t(2.50 cm) = 15.7 cm Surface area : circumference x thickness : 15.7 cm x 0.50 em = 7.8; cm2 mol H230. = Inner circumference = 2m = 21t(1.00 em) = 6.28 cm Surface area = circumference x thickness = 6.28 cm x 0.50 em = 3.1g cm2 Area of small circle = rrr2 = 110.00 em)2 = 3.14 errr2 Area of large circle = rrr2 = n(2.50 em)2 = 19.6 cm2 Total area = 7.35 cm2 + 3.14 cm2 + 2x(19.6 cmz) _ 2x015 em!) = 43.90 cm2 Volume of metal needed: surface area x thickness of plating = 43.9§ cm2 0.0050 em = 0.22 cm3 3x 89ng lmolNi x2mole' 96485C x cm3 >(58.1593 gNi lino] Ni 1 mole' Time = chargeftimc = 6437.5 CI 1.50 C/s : 4291.7 5 or 71.: min Charge required: 0.220111 =64flc 89. With reference to the aluminum-air battery in Figure 20-18 and equation (20.28), (3) Refer to reaction (20.28) and Appendix D. Emu" = E" (cathode) — E° (anode) = E° (reduction half—cell) — E“ (oxidation half-cell) Em? = 0.401 V — (—2.310) V = 2.711 V (b) For reaction (20.3), not" = —nFE...1° = —12 mol e’ x 96,485 Cfrnol e“ x 2.711 V = —3.14 x10‘5 v-c = —3.14 x106 J = —3.14 ><103 10 AG° = 4AGr°[A1(0H).]’(aq) - 6AGf°[H20(l)] — 4A0f°rorr1 = —3.14 x 103 kJ —3.14><10’k_1+6x(- 237.1)k1+4x( —157.2)kJ AG?[A1(OH)4]' (aq) = 4 AG,°[Al(OH)4]'(aq) =—1.30x10‘k,0mol (c) mass of Al- 4.00hx36005x 10. 0Cx lmole' leolAl x26.98gAI =13.4 Al 111 Is U96485C 3mole lmolAl g ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern