Chapter 12 Evens Answers

Chapter 12 Evens Answers - 12.8 We first determine the two...

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Unformatted text preview: 12.8 We first determine the two half—reactions: Unbalanced: FeHP03(aq) —) Fe(OH)3 (s) + PO43’ (aq) OC1‘(aq) —> C1” (aq) These are then balanced as follows: The ferrous hydrogen phosphite reaction is complicated because both iron and phosphorus are oxidized. The Fe atom begins with an oxidation number of +2 and ends with an oxidation number of +3, while phosphorus starts as +3 and ends up as +5. There is a total change of 3 electrons. FeHP03(aq)—> Fe(OH)3(s) + P043“(aq) + 3 e— Balancing the charge by addition of hydroxide for basic solution gives FeHP03(aq) + 6 OH‘ (aq) —> Fe(OH)3 (s) + PO43‘ (aq) + 3 e’ Balancing the hydrogen and oxygen atoms by addition of water gives FeHP03(aq)+ 6 OH'(aq) —> Fe(OH)3(s) + PO43’(aq) + 2 H200) + 3 e- The reduction of hypochlorite is simpler to balance. The oxidation number of chlorine changes from +1 in OCI' to —l in CI' , giving a change of 2 e . 0C1“ (aq) + 2 e‘ —> C1“ (aq) Balancing the Charge by addition of OH“ : OCl‘ (aq) + 2 e‘ —> Cl”(aq) + 2 OH— (aq) Balancing oxygen and hydrogen atoms by addition of H20: OCl‘ (aq) + H20(1)+ 2 e“ —> Cli(aq) + 2 OH‘ (aq) Combining the two equations: 3[OC1‘ (aq) + HZO(1)+ 2 e‘ ——> C1” (aq) + 2 OH‘ (aq)] 2[FeHPO3(aq) + 6 on (aq) —> Fe(OH)3 (s) + PO43" (aq) + 2 H200) + 3 e’] 2 FeHPO3 (aq) + 3 0C1" (aq) + 6 OH'(aq) —> 2 Fe(OH)3 (s) + 2 PO43'(aq) + 3 C1" (aq) + Hgoa) 12.16 (a) Cr2072'(aq) +14 n+(aq) + 6 e’ v» 2 Cr3*(aq) + 7 H200) (cathode half-reaction) 3[I1g22+(aq) —> 2 Hg2+ + 2 e'] (anode half—reaction x 3) (b) Adding half—reactions gives C50,} (aq) + 3 Hg?2+ (aq) + 14 FF (aq) —> 2 Cr3+ (aq) + 6 Hg2+ (aq) + 71-1200) The cell diagram is Pt(s)ng22+ (aq), HgWaq) H H*(aq).- 012072“ (aq), Cr3+(aq)tPt(s) 12.20 (a) Bi3+ (aq) + 3 e” ——) Bi(s) E°(cathode) = +0.20 V Zn 2* (aq) + 2 e‘ a Zn(s) E°(anode) = —0.76 V For the equation, 3 Zn(s) + 2 Bi3+ (aq) —> 3 Zn2+ (aq) + 2 Bi(s), n = 6 E°ceu = +0.20 V — (— 0.76 V) = +0.96 V AG"r = -nFE° : —6(9.6485 X 104] - V‘l ~ moi” + 0.96 V) = —5.6 ><102kJ-mol‘1 (b) 02 (g) + 4 H+(aq) + 4 6— —> 2 HZOU) E°(cathode) = +1.23 V 2 H+ (aq) + 2 e‘ —> H2(g) E°(anode) = 0.00 V Taking 2[2 H+ (aq) + 2 e” —9 H2 (g)], reversing it and adding it to the cathode: half-reaction gives 2 H2 (g) + 02 (g) —~) 2 H200) n = 4, E° = +1.23 V cell AG"r = —nFE° : —4 x 9.6485 x104 C ‘ mol‘I x 1.23 V : e475 kJ ~ mol”1 (0) 02 (g) + 2 H200) + 4 e— ——> 4 01—1-(aq) E°(cathode) = +0.40 v 2 H200) +2 e~ ~—> H2 (g) + 2 OH“ (aq) E°(anode) = —0.83 v Taking 2[HZO(I) + 2 e‘ —-> H2 (g) + 2 OH‘ (aq)], reversing it and adding it to the cathode half—reaction gives 2 H2 (g) + 02 (g) —> 2 H2 0(1) n = 4, E3“ = +1.23 V AG°r = nFE° : —475 kJ - moi“ (same as part b) (d) 3[Au+ (aq) + e" ——) Au(s)] E °(cathode) = +1.69 V Auh (aq) + 3 e' —-> Au(s) E°(anode) = +1.40 V Taking 3[Au+ + e— —> Au(s)] and adding it to the reverse of the reduction half—reaction gives 3 Au+ —> Au3+ + 2 Au(s) n = 3 E° =+1.69V—1.40V=0.29V cell AG°r = —nFE° = —3 x 9.6485 x 104 C - me?1 x 0.29 J -C‘1 = —84 kJ - mol“ 12.26 (a) Pt2+ /Pt E° = +1.20 V, Pt2+ is oxidizing agent (cathode) AgF/ Ag, F" E“ = +0.78 V, Ag is reducing agent (anode) Ag(S)lAgF(S)IF" (at!) H Pt2+(aQ)lPt(S) E19" = E°(cathode) - E°(anode) = +1.20 V + 0.78 V = +0.42 V (b) 1371' E° = +0.53 V, 13' is oxidizing agent (cathode) Cr3+ /Cr2+ E° = —0.41V, Cr2+ is reducing agent (anode) Pt(s)|Cr“ (aq), Cr3+ (aq)”? (aq)a 1; (aq) I Pt(s) E13“ = +0.53 V — (—0.41 V) = +0.94 V (c) H‘r/l-I2 E° = 0.00 V, H+ is oxidizing agent (cathode) Ni2+ / Ni E° = —0.23 V, Ni is reducing agent (anode) Ni(3) l NH (1161) l l W (MDI H2 0;)! P1(5) E3“ = 0.00 V — (—0.23 V): +0.23 V (d) 03, H" /02 E° = +2.07 V, 03 is oxidizing agent (cathode) O3 /O2 , OH‘ E° = +1.24 V, 02 is reducing agent (anode) "Pt(S)|03 (g), 02 (gNOH' (aQ)| |03 (g), 02 (g) | IT“ (MI) | Pt(S) 15°C." = +2.07 V — 1.24 V = +0.83 V 12.40 (a) anode: Cr(s) —-—> C1‘3+(aq) + 3 e“ E°(anodc) = —O.74 V cathode: Pb2+ (aq) + 2 e” —> Pb(s) E°(cathode) : -0.13 V Multiplying half—reactions gives 3(le2+ (aq) + 2 e' —> Pb(s)] 2[Cr(s) ——> Cr3+(aq) + 3 6'] overall reaction: 3 Pb2+ (aq) + 2 Cr(s) —> 3 Pb(s) + 2 Cir3+ (aq) E° =+O.61V,n=6 cell _ 0.025 693 V In [CW]2 n [Pb2+ ]3 E = +0.61 V _ EQ'EQEX In £377 6 (9.5 x10“ )~ E : +0.61 V — 0.0043 V in 1.6 ><105 = +0.61 V — 0.051 V = +0.56 V E=E° (b) HgZCll(s) + 2 e" —-> 2 Hg(l) + 2 Cl‘ (aq) E°(cathode) = +0.27 V 2 H+ (aq) + 2 e‘ ——> H2 (g) E°(anode) = 0.00 V Reversing the latter half—reaction and adding, we have Hg2C12(s) + II2 (g) a 2 Hg(l) + 2 l-l+ (aq) + 2 Cl’ (aq) Em" = +0.27 V _ 0.025 693 V 1m[1:~1+]2[Cr]2 E = E° n PH2 0.025 693 V In (10-3-5)2(0.75)2 2 2.0 E = +0.27 V — = +0.27 V + 0.22 V = +0.5 V (c) anode: Sn2+ (aq) —9 Sn4+ (aq) + 2 e“ E°(anode) = +0.15 V cathode: Fe:3+ (aq) + e“ —> Fe2+ (aq) E°(cathode) = +0.77 V Multiplying half-reactions, 2[Fe3+(aq) + e“ —+ Fe2+(aq)] Sn2+ (aq) —-> Sn4+ (aq) + 2 e" Adding, 2 Fe“ (aq) + Sn“ (aq) a 2 Fe2+ (aq) + Sn4+ (aq) 13°05" = +0.62 V _ 0.025 693 V In [Fe2+]2[Sn4+] n [1363+ ]2 [Sn2+] 0.025 693 V In (015)2(0059) 2 (015)2(0059) E = +0.62 V — 0.0129 V In 1 = +0.62 V E=E° E=+0.62 V -— (d) anode: Ag(s) + I‘ (aq) —> AgI(s) + e" E°(anode) = —0.15 V cathode: AgCl(s) + e‘ ——> Ag(s) + Cl“ (aq) E°(cathode) = +0.22 V Ag(s) + I“ (aq) —> AgI(s) + e‘ Adding, AgCl(s) + I‘ (aq) ——) AgI(s) + Cl' (aq) E16" = +0.37 V E 2 E0 _ 0.025 693 V In {or} n [1‘] 0.67 0.025 E = +0.37 V — 0.084 V = +0.29 V E = +0.37 V — (0.025 693 V) In 12.46 (a) AngrO4 (s) + 2 e‘ —> 2 Ag(s) + Cr042" (aq) (b) To calculate this value we need to determine the E ° value for the solubility reaction: AgZCrO4 (s) —> 2 Ag+ (aq) + CrOf‘ (aq) E° : ? The relationship AG° = ~RT In K = ~nFE° can be used to calculate the value of Ksp. The equations that will add to give the net equation we want, are: Ag2CrC4(s) + 2 e" —> 2 Ag(s) + Cr042‘(aq) E° = +0.446 v Ag(s) —> Ag’“ (aq) + e‘ E° = —0.80 V Notice that the second equation is reversed from the reduction reaction given in the Appendix, and consequently the E ° value is changed in sign. Adding the first equation to twice the second gives the desired net reaction, and summing the E ° values will give the E ° value for that process (note that we do not multiply the second equation’s E ° value by 2). E° = (+0.446 V) + (~— 0.80 V) = — 0.35 V nFE° _ (2)(9.65 ><104 c - mol“)(—o.35 V) _ —27 RT (8.314 J-K:I -mor1 )(298.2 K) ln Ksp = _ «12 Km -10 12.56 96.5 kC = 9.65 ><104 C e 1.00 F =1.00 mol e— (a) 1.00 mol Ag+(aq) +1.00mol e” —>1.00 mol Ag, or 108 g (b) 1.001nol C1‘ -—> 0.500 mol C12 (g) + 1.00 mol e" 0.500 mol Cl2 at 298 K = 0.500 mol X 24.45 L ~ mol‘l : 12.2 L (c) 0.500 molCu“(aq)+1.00mol e‘ —> 0.500 mol Cu(s); mass of copper = 0.500 mol Cu x 63.54 g -mol“’ = 31.8 g . on51erA a + e‘—9 s =—. . 1t tis a -reaction 1296 C 'd l3+ q 3 Al() (E° 166) W'h h' hlf ‘ as the anode reaction and one or both of the given reduction reactions, a cell with a positive potential can be constructed. Two adjacent filled teeth, simultaneously in contact with the aluminum, could behave as two independent cells at different potentials, corresponding to the two possible reduction half-reactions. Current will then flow between them, stimulating the pain sensors. The two possible cell reactions are 3 Hg,2+ (aq) + 4 Ag(s) + 2 Al(s) E° = +2.51 V ——> 2 Angg3 (s) + 2 A13+ (aq) cell 3 Sn“ (aq) + 9 Ag(s) + 2 Al(s) E°cen = +1.61 v —> 3 Ag3Sn(s) + 2 A13+ (aq) ...
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This note was uploaded on 05/04/2009 for the course CHEM 653196 taught by Professor Johnson during the Spring '09 term at UCSD.

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Chapter 12 Evens Answers - 12.8 We first determine the two...

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