Chem 126 Solutions Chapter 18

Chem 126 Solutions Chapter 18 - Chem 126-104 HW 6 Answers...

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Unformatted text preview: Chem 126-104 HW 6 Answers to problems for Chapter 18: Electrochemistry [18.14 (a)‘ PP}: Applying Rule 3, F is assigned an oxidation state of -1. This gives y +5(-l) = 0 and y = +5 for the ox1dation state of the phosphonis in PPS. Hence, oxidation state of P = +5, F = -1. (b) Na2CrO4: Applying Rule 2, Na is assigned an oxidation state of +1. Applying Rule 5, each 0 is assigned an oxidation state of - 2. This gives 2(+l) + y + 4(-2) = 0 and y = +6 for the oxidation state of the chromium in N32Cl'04. Hence, oxidation state of Na = +1, Cr - +6, 0 = -2. (c) N05 : Applying Rule 5, 0 is assigned an oxidation state of -2. This giyes y + 2(-2) = -l and y = +3 for the oxidation state of the nitrogen in N0; . Hence, oxidation state of N = +3. 0 = -2. I 18.32 (a) Oxidation half-reaction: 531,020 4 (aq) —> 2002 (g) +211+ (aq) +2e‘] Reduction half~reactiom 2[Mno; (aq)+8H+(aq)+5e' —-> Mn2+(aq) +4H20(l)] . . Balanced redox equation: 2Mno; (aq)+6H+ (aq)_+5H,c,o4 (aq) —> 2Mn2+(aq)-l-8H20(£)+10C02(g)] (b) Oxidation half—reaction: 2Br‘ (aq) —> Br; (1) + 2e" Reduction half-reaction: Cl2 (g) +2e— —) 2Cl— (aq) Balanced redox equation: Cl2 (g) +ZBr‘ (aq) —> 2Cl‘ (aq) + 3:2 (I) (c) Oxidation half-reaction: 3[Cu(s) —> Cu2+ (aq)+2e‘] Reduction half-reaction: I 2[No; (aq) +411+ (aq)+3e' —> N0(g) + 2190(1)] Balanced redox equation: 3Cu(s) +2N0; (aq) +8H+ (aq) —) 3012+ (aq) + 2N0(g) + 4H20(t) limo (a) Oxidation half-teaction: 2[Cro; (aq)+2H20(l) —+ co; (aq)+4H+ (aq)+3o‘] Reduction half-reaction: 3[c10‘ (aq) +2H+ (aq) + 2o’ -+ or (aq) + 1120(1)] Balanced redox equation: 2cm; (aq) + nzoa) + 300‘ (aq) —> zctoi‘ (aq)-i-ZH“ (aq)+3cr (aq) Base conversion equation: 21-1+ (aq) +20H- (aq) » 2H20(l) Final redox equation: 2Cr0; (aq) + 3ClO" (aq) + 20H' (aq) —+ 200%. (aq) +3Cl’ (aq) +H20(l) 18.36 (cont) (b) Oxidation half-reaction: Brz (aq) + salon!) —> 23:0; (aq) + 1211* (aq) +10e‘ Reduction half-reaction: 5[Br2 (aq) +2ew —> 2Br" (3:1)} Balanced rcdox equation: 6131-2 (aq) + 6H20(l) —> 28:03“ (aq) + 12H+ (aq) +10Br' (aq) Smallest coefficients: 3an (sq) + 31120“) —) Bro; (aq) +611+ (aq) +5Br" (aq) Base conversion equation: 6H+ (aq)+ 60H“ (aq) —> 61120“) Final redox equation: 331'2 (aq) +6011~ (aq) —) Bro; (aq) + SBr’ (aq) + 3H2 0(8) (c) Oxidation half-reaction: ‘ N2}! 4 (aq). —) N2 (g) +4141+ (aq) +4e" Reductionhalf—reaction: 43202 (aq)+'2n* (aq)+2e“ » 211200)] Balanced redox equation: N.,_Hll (aq)+2H,O, (aq) a» N1(g)+4H20(t) ’ [18.54 ' ' (a) ‘ 211(5) «a Zn’Yaq) + 26 g E“ = 0.76 v Fefiaq) + 2e” —» Fee) 5° = -o.44 v Zn(s) + Fe’*(aq) » Zn**(aq) + Fats) 7 E15“ = 0.32 V V ' Since the calculated voltage is positive, the reaction is spontaneous in the direction shown under standard conditions. (b) ' . Fe2+(aq) 9» Fe”(a.q) + e" . E“ = «0.77 v AgCl(s) + e’ «a Ag(s) + Cl'(aq) E" = 0.222 V Fe2*(aq) + AgCl(s) «9 Fe”(aq) + Ag(s) + arm) E” = -o.55 v Since the calculated voltage is negative, the reaction is spontaneous in the reverse direction (nonspontaneous in the forward direction) under standard conditions. (c) 2Cl‘(a'q) a Clz(g) + 2e‘ ‘ E° = -l.36 V ' Brzu )+ Ze‘ w) ZBr'(aq) E° = 1.06 V 2Cl'(aq) + Br2(l ) —-> chm) + 2mm) ‘ 13° == —o.3o v Since the calculated voltage is negative, the reaction is spontaneous in the reverse direction (nonspontaneous in the forward direction) under standard conditions. ’ [13:55: * . . ‘ + . (a) Oxidation occurs at the negative electrode in a voltaic cell. Hence, 'I‘l(s) must be oXidized and Ag tag) must be reduced. This gives Anode: ' Tl(s) u» 11*(aq) + e" ' g g: i f) 80 V - Cathode: Ag+(aq) + c” ~—-) Ag(s) — . W Cell: Tl(s) + Ag*(aq) w) 11*(aq) + Ag(s) . E" —-_1.l36 V (b) Solving v + 0.80 V = 1.136 V, gives y = 0.34 V. 18.64 4 (a) AG°=-—nFE° AG” =—(2 mole‘)[—-———-—-—9'65x 1° C](°'l3é’)=-n.2 x 104 J 1 mol e' 0 logK = "E iogK = M = 10.83 K =10‘033 = invlog10.83 = 6.8 x 10“ °“ 0.0591 “‘ 0.0591 “1 Since the calculated voltage is positive, the reaction is spontaneous in the direction shown under standatd conditions. This is consistent with a negative value for 186° and Keq > 1. 4 . 0» AG" =-—nFE° A6' = ~(1 mole')[3;m——Q)("°‘55 3) = 5.30 x 10" J 1 mole' 1 C 115° ‘ (IX-055) ‘ —931 - .10 lo K = l K =-—-~—-—~—-—=-9.31 K =10 =invlo —-9.31=4.9110 8 °“ 0.0591 ' 03 “‘ - 0.0591 °“ . .. 3 Since the calculated voltage is negativmthe teaction is nonspontaneous in the direction shown under standard conditions. This is consistent with a positive value for AG° and Keq < l. .' 4 (6) AG” =—nFE° A6’ = -(1 mote-{MX'MO J ) = 5.8 x 10‘ J 1 mol e' 1 C nE° . - - logK = logic“, = M = 40.15 K“, =10 1°15 = smog—10.15 = 7.8 x 10 1‘ °" 0.0591 I 0.0591 Sincethe calculatedvoltage is negative, the reaction is nonspontaneous in the direction shown under standard conditions. “This is consistent with a positive value for AG° and Keq < 1. (a) For Z06) + F6390) -+ Zn2+(aq) + Fc(s). E = E, _ 0.0591 in” 0.0591v-mo1 e' 104 1.0 x 10‘3]=10‘37 V log E = 0.32 V — n [Fe2+] 2 mol e‘ l (b) For A3016) + F°2+(BQ) 9 Ag“) + Fewaq} + (3113(1), . " F 3* ‘ - “ E=E.,_0.0591m [ e “a E=_Q‘55V__0.0591V-mole log (°-°1°)(4-°x10 3)}‘033‘, n [1762+ ] 1 mol e‘ 0.20 (c) For Br2(l ) + ZCl‘(aq) -—> Clz(g) + 2Br'(aq), 0.0591log (PCI, ){Br‘]2 “ [Cr r 2 ' 0,0591 v. - (0.50-atm 3.6 x10’3 E = -0.30 v “"4254” .______)L___.___)__ = .048 v 2 mole 0,10 E=E°_ i 18.72 Pbso..(s) + 2e‘ —> Pb(s) + sci-(m1) AG; Pb(s) » Pb2*(aq) + 2c" AG; PbSOL;(s) 3: Pb2+(aq) + SOi'Yaq) ' AG; 4 AG: = - ° = (2 mot e') w (—0.561): 6.87 x 10‘ J 1 mo] 6’ C 4 _ A6; = -nFE°= -(2 mol e") M (0.1261) = -243 x 10‘ I 1 mol e' C A63 = .111me AG? + AGg = 4.44 x 10‘ 1 AG‘ Solving AG" = .1111an 1’0er givesKn, = e RT . (4.44xm‘ J) Hem e 8.314de K 298K)= 106‘ 104 [13.74 (a) ’ Cd(s) + 20mm) -) Cd(OH)2(s) + 2e’ 78° = 0.33 v 2[NiO(OH)(s) + H20(£ ) + e' —+ Ni(OH)2(s) + 0H(aq)] 15° = 0.52 v Cd(s) + 2Ni0(OH)(s) + 2H20(l ) ~> Cd(0H)2(s) + moms) E° = 1.35 V (b) Since the reactants and products are all solids and pure liquids, the value of the reaction quotient in the Nemst equation does not change during the course of the reaction. and the voltaze therefore remains the game 18.78 (a) Anode: 2Cl' «9 Clz(g) + 2e‘ Cathode: Ca2+ + 2e’ -—> Ca(£ ) Cell: 20' + Ca” —> CI,(g) + Ca“ ) (b) At the anode the possible reactions are: I 280?,‘(aq) -> 820%"(aq) + 2c“ E° = -2.05 v 21120“ ) -‘—> 02(g) + 4W(aq) + 46' E0 = -l.23 V ‘ Since H20 still has a more posititie potential for oxidation after allowing for the additional oxygen overtroltage of 0.4 V, it will be oxidized at the anode. At the cathode the possible reactions are: Mg’*(aq) + 2e' —+ Mg(s) E" = -237 v 2H20(l ) + 2e‘ —) H2(g) + 20H(aq) E“ = -0.83 V Since H20 still has a more positive potential for reduction afler allowing for the additional hydrogen overvoltage of 0.4, it will reduced at the cathode. The net reaction for the cell will be: 211,00 ) —) (Mg) + 2H,(g). (c) At the anode the possflile reactions are: Ni(s) —) Ni2+(aq) + 2e‘ E° = 0.25 v 2Br' -—) Brzu )+ 2e‘ , E° = -l.06 V- 2H20(£) —+ 02(3) + 4H’(aq) + 4e’ E° =.-1.23 v Since Ni has a much more positive oxidation potential, the nickel anode will be oxidized. At the cathode the possible reductions are: NW“) + 2c" —> Ni(s) E° = -o.25 v 21120“ )+ 2e“ -) H2(g) + 20H(aq) E" = -0.83 V Since Ni” has a much more positive reduction potential, it will be reduced at the cathode. . V Ni will be oxidized at the anode, and Ni2+ will be reduced at the cathode. Hence, no net reaction will occur. 18.90 (a) The equation for producing Al by electrolysis of molten A1203 is A131! ) + 3c“ -> A1“ .). C C mo! e" moIAl gAI kgAI Strata : —— ~+ —— —> a —-——- . gy s hr hr v hr 4 hr .9 hr ‘ Al 980C 60s 60min lmole‘ lmolAl 26.9SgAl lkgAl Al ?k ——-= x , x x——-—-————->< -——-——— ————-—-—-=0.329 —-— ghr ’8 1mm lhr 9.55x10‘lc .3molc'x lmolAl xloz’gAl, kg!" (b) The total charge required to produce 1 kg Al is: 103 gAl lmolAl 3mole' 965x104 C =1k Alx—-—-———x——————-—-— «~——-——-- -‘-——-———————-= .7 7 Q g lkgAl 26.98gAlxlmolAlx lmole‘ 10 “00 The electrical energy consumed is: E=QxVx——l—‘iw—"—‘-:—- E=l.07xlO7Cx4'85JxM—=l4.4kw-hr 3.60x10 J 1C 3.6oxlo‘J (c) Water is easier to reduce than Al”(aq). The pertinent reduction potentials are: 2H20(t )+ 2e' —+ H2(g) + 20H'(aq) 13° = -0.83 v Aflaq) + 3e" ~> Al(s) E“ = -l.66 v The overvoltage of approximately 0.4 V for producing H2 does not make H20 more difficult to reduce than Al”. 18.92 (a) Oxidation half-reaction: 2[Fe(s) —-> Fe2+(aq) + 25] 13° = +0.44 v Reduction half-reaction: 02(3) + 4H*(aq) + 4e' ~> 2H20(l ) > E" = +1.23 v Balanced redox equation: 2Fc(s) + 02(g) + 4mm) —+ 2Fe2*(aq) + 2H20(z ) 13° = 1.67 v (b) B°=o.44v+1.23v=1.67v (c) E = E" - log - P03 [11*] At pH= 5.9 [11*] = inv 10g (-5.9) = 1.3 x 10“5 M. ' 4 5x10.5 Hence,E= 1.67 v — 0'05” log ————(————-—)————4- = 1.44v (0.21 atm)(l.3 x 10") l 18.96 The number of moles of electrons for the reaction 2Fe(CN):‘ (aq) + 21'(aq) —-> 2Fe(CN)g"(aq) + 12(3) is two; the omdafion state of each Fe changes from +3 to +2 and the Oxidation state of each 1 changes from -1 to 0. ‘ . ' I- --34.3 103 - SolvnngAG°=-nFE°forE°givesE°= A69: ( x J) nF 4 (2 mo, 6. 9.65 x 10 c] > 1 mo! c" J = 0.178—-— = 0.178 . C V '18.106- (a) M0926) 22 man) + 210; (aq) AG? 1 mm”) + 25‘ —> Pb(s) AG; 116:: —RTan'p A6; = —s.314 J - marl ~K'l x 298 K x 1n(2.6 x 10'13 ) = 7.18 x103 1 4 AG; =—nFE° AG; =42 mole’ M “"126 J]=2.43 x 104 J 1 mol 6' l C AG; =AG;+AG; AG; =7.18 x 103 J + 2.43 x104 J = 9.61 x 104 J 9 -AG" Solving AG” = ——nFE forE gives E” = “F 4 Be = ._._'?;§_L"_19__L___ = .0493 LC“ = 4.498 v - 9.65 x104 0 (2 mole ) -——-—-~-—-— 1 mol 6' (b) For Pb(103)2(s) + 2e' -—> Pb(s) + 2105 (aq), 0.0591 V- mo! 13' 2 ———-—-————-———-l 10‘ n 0:1 3‘] 0.0591 V~ me! e' 2 mo! 12' E=E°~ 2 E = —o.49s v- log(3.5 x104) = 41.353 v Fini ...
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This homework help was uploaded on 04/07/2008 for the course CHEM 101 taught by Professor Buterris during the Fall '08 term at NJIT.

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