125ch20a_001

125ch20a_001 - 0 .E‘Iectrochemistry Oxidation-Reduction...

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Unformatted text preview: 0 .E‘Iectrochemistry Oxidation-Reduction Reactions 20.1 (a) Oxidation is the loss of electrons. (b) The electrons appear on the products side (right side) of an oxidation half-reaction. (c) The oxidant is the reactant that is reduced; it gains the electrons that are lost by the substance being oxidized. ([1 x in/mal‘ 1'5 7‘14 at Id: EM; Zgfixl‘) 20.2 _ (3) Reduction is the gain of electrons. (b) The electrons appear on the reactants side (left side) of a reduction half-reaction. (c) The reductant is the reactant that is oxidized; it provides the electrons that are gained by the substance being reduced. (WWI/“2‘ /} 75% F ) 20.3 (a) I is reduced from +5 to 0; C is oxidized from +2 to +4. (b) Hg is reduced from +2 to 0; N is oxidized from -2 to 0. (c) N is reduced from +5 to +2; S is oxidized from -2 to 0. (d) Cl is reduced from +4 to +3; 0 is oxidized from -1 to 0. 20.4 (a) No oxidation-reduction (b) N is both oXidized and reduced; it is reduced from +4 to +2 and oxidized from +4 to +5. ' (c) No oxidation-reduction (d) S is reduced from +6 to +4; Br is oxidized from -1 to 0. 20.5 (a) TiCl4(g) + 2Mg(l) —) Ti(s) + 2MgCl2(I) (b) M90) is the reductant; TiCI4(g) is the oxidant. 20-5 (a) ' 2N2H4(9)+N204(9) 9 3N2(9)+4H20(9) (b) N204 serves as the oxidizing agent; it is itself reduced. NZH4 serves as the reducing agent; it is itself oxidized. 20.7 (a) Sn"*(aq) -) Sn‘*(aq) + 2e', oxidation (b) TiO,(s) + 4H‘(aq) + 2e‘ -) Ti2*(aq) + 2H20(I). reduction (c) 2HOCl(aq) + 2H‘(aq) + 2e‘ —) Cl2(aq) + 2H20(i). reduction 439 20 Electrochemistry Solutions to Exercises 20.8 20.9 20.10 (d) (e) (a) (b) (0) (d) (e) (a) (b) (C) (d) (8) (f) (a) (b) Net: (6) Net: (6) Net: Net: (1’) La(s) + 30H‘(aq) -) La(OH)3(s) + 3e", oxidation NO3'(aq) + H200) + 23' 9 N02‘(aq) + 20H'(aq). reduction Mo‘”(aq) + 3e' -9 Mo(s), reduction H2803(aq) + H200) -) SOf‘(aq) + 4H*(aq) + Ze‘. oxidation 5:90:21”) + 8HWait!) + 3e‘ -) Fe3*(aq) + 4H20(i), reduction Cl02‘(aq) + 214200) +4e' -> Ci'(aq) +40H'(aq), reduction 02(9) + 2H20(|) + 4e‘ -) 4OH'(aq), reduction Cr2072‘(aq) + I‘(aq) + 8H+ —) 209*(aq) _+ 103'(aq) + 4H20(I) 4Mn04‘(aq) + 50H30H(aq) + 12H‘(aq) -) 4Mn2*(aq) + 5HCOQH(aq) + 11H20(|) 4As(s) + 30l03'(aq) + 3H*(aq) + 6H20(I) -) 4H3A503(aq) + 3HCIO(aq) A5203(5) + 2N03'(aQ) + 2H20(l) + 2H‘GIQ) -) 2H3A504(aQ) + N203(aQ) 2Mn0"(aq) + Br‘(aq) + H200) -)_ 2Mn02(s) + BrO3’(aq) + 20H‘(aq) 4H202(aQ) + C|207(acl) + 20H’(BQ) 9 200218!» + 402(9) + 5H20(|) Pb(OH)42‘(aq) + CIO'(aq) -) Pb02(s) + C|‘(aq) + 20H'(aq) + H200) The half reactions are: 3H20(I) + “203(5) + 4e' -) 2TIOH(s) + 4OH‘(aq) 2[2NHZOH(aq) + 20H‘(aq) -) N2(g) + 4H20(l) + 2e‘] WWW 2[Cr2072'(aq) + 14H*(aq) + 6e‘ -) 209*(aq) + 7H20(I)] 3[CH30H(aq) + H200) -) HCOZH(aq) + 4H*(aq) + 4e'] WWW 2[Mn0.‘(aq) + 8H*(aq) + 5e“ -) Mn2‘(aq) + 4H20(i)] 5[ZCI‘(aq) -) C|2(aq) + 2e'] WWW H202(aq) + 29‘ 9 02(9) + 2H*(aq) Since the reaction is in base, the H‘ can be “neutralized” by adding ZOH' to each side of the equation to give H202(aq) + 20H'(aq) + 2e" -) 02(9) + 2H20(I). The other half reaction is 2[CI02(aq)_+ e‘ -) CI02'(aq)]. H204“) + 20'02(aQ) + 20H'(3Q) ') 02(9) + 20102‘(aQ) + 2H200) 3[No,-(aq) + H200) -> N03‘(aq) + 2H’(aq) + 2e'] Cr2072‘(aq) + 14H*(aq) + 6e' -) ZCr'”(aq) + 7H20(I) MW 440 20 Electrochemistry Solutions to Exercises Voltaic Cells; Cell Potential 20.11 20.12 20.13 20.14 20.15 (a) (b) (a) (b) (a) (b) (0) (d) (e) (a) (b) (C) (d) (e) The reaction Cu2*(aq) + Zn(s) -) Cu(s) + Zn’*(aq) is occurring in both Figures. In Figure 20.3. the reactants are in contact. and the concentrations of the ions in solution aren't specified. In Figure 20.4. the oxidation half-reaction and reduction half-reaction are occurring in separate compartments, joined by I a porous connector. The concentrations of the two solutions are initially 1.0 M. In Figure 20.4, electrical current is isolated and flows through the voltmeter. In Figure 20.3, the flow of electrons cannot be isolated or utilized. In the cathode compartment of the voltaic cell in Figure 20.5. Cu2+ cations are reduced to Cu atoms. decreasing the number of positively charged particles in the compartment. Na+ cations are drawn into the compartment to maintain charge balance as Cu” ions are removed. ' The porous glass dish in Figure 20.4 provides a mechanism by which ions not directly involved in the redox reaction can migrate into the anode and cathode compartments to maintain charge neutrality of the solutions, Ionic conduction within the cell, through the glass disk, completes the cell circuit. In the anode compartment of Figure 20.5, Zn atoms are oxidized to Zn” cations. increasing the number of positiver charged particles in the compartment. NO; anions migrate into the compartment to maintain charge balance as Zn“ ions are produced. Cd(s) a Cd2*(aq) + 2e‘; Ni’*(aq) + 2e' -) Ni(s) Cd(s) is the anode; Ni(s) is the cathode.- Cd(s) is negative (-); Ni(s) is positive (+). Electrons flow from the Cd(—) electrode toward the Ni(+) electrode. Cations migrate toward the Ni(s) cathode; anions migrate toward the Cd(s) anode. Ag*(aq) + 1e" -) Ag(s); Ni(s) -> Niz'*(aq) + 2e‘ Ni(s) is the anode. Ag(s) is the cathode. Ni(s) is negative; Ag(s) is positive. Electrons flow from the Ni(-) electrode toward the Ag(+) electrode. Cations migrate toward the Ag(s) cathode; anions migrate toward the Ni(s) anode. 20 Electrochemistry Solutions to Exercises 20.16 W porous barrier or salt bridge 20.17 (a) Eleotmmotive force. emf, is the driving force that causes electrons to flow through the external circuit of a voltaic cell. It is the potential energy difference between an electron at the anode and an electron at the cathode. (b) One volt is the potential energy difference required to impart 1 J of energy to a charge of1 coulomb. 1' V = 1 J10. (0) Call potential. E“... is the emf of an electrochemical cell. 20.18 (a) The negative and (the anode) of a D-size battery has the higher potential energy for electrons. Electrons in the battery flow from (-) anode to (+) cathode. from high potential energy to lower potential energy. (b) E," for a D-slze battery is positive. because electrons flow through the cell spontaneously to provide electrical current. If E”. were negative or zero. the battery would provide no current. (c) A standard cell potential is a cell potential (emf) measured at standard conditions, 1 M aqueous solutions and gases at 1 atm pressure. 20.19 (a) 2H*(aq)+2e' “42(9) (b) A standard hydrogen electrode is a hydrogen electrode where the components are at standard conditions. 1 M H*(aq) and H2(g) at 1 atm. (c) The platinum foil in an SHE serves as an inert electron carrier and a solid reaction surface. 442 20 Electrochemistry 20.20 (a) (b) (C) 20.21 (a) (b) 20.22 (a) '(b) 20.23 (a) Solutions to Exercises “2(9) -> 2H*(aq) + 2e- The platinum electrode serves as a reaction surface: the greater the surface area. the more H2 or H’ that can be adsorbed onto the surface to facilitate the flow of electrons. «H2(g), 1 atm A standard reduction potential is the relative potential of a reduction half-reaction measured at standard conditions. 1 M aqueous solutions and 1 atm gas pressure. E”; = 0V for a standard hydrogen electrode. The reduction of Ag‘(aq) to Ag(s) is much more energetically favorable, because it has a substantially more positive E rgd (0.799 V) than the reduction of Sn=*(aq) to Sn(s) (-0.136 V). It is not possible to measure the standard reduction potential of a single half- reaction because each voitaic cell consists of two half-reactions and only the potential of a complete cell can be measured. The standard reduction potential of a half-reaction is determined by combining it with a reference half—reaction of known potential and measuring the cell potential. Assuming the half-reaction of interest is the reduction half—reaction: 0 EM = Em(cathode) - Endanode) = End(unknown) - E W(reference); E r;d(unknown) = E 0;" + Er; (reference). Na‘(aq) + e‘ -) Na(s) ° = -2.71 V Ca”(aq) 4- 2e‘ 4 Ca(s) ° = -2.87 V The reduction of Ca"(aq) to Ca(s) is the more energetically unfavorable reduction because it has a more negative E” value. The two haif—reactions are: T1“(aq) + 2e' -) Tl*(aq) 2[Cr2‘(aq) -) Cr°‘(aq) + e‘] Ii ‘9 cathode En; anode E fed -0.41 443 20 Electrochemistry Solutions to Exercises (b) (C) 20.24 (a) (b) 20.25 (a) (b) (C) E " =2":d (cathode)-Er;d (anode); 1.19V=E';d-(-0.41V); E " =1.19v-0.41v=0.7av red if" 16' Anode (-) CathOde (+) Inert Inert Electrode (Pt) Salt Electrode (Pt) Bridge Solution Contains T|‘(aq). 113*(aq) Note that because Cr“(aq) is readily oxidized. it would be necessary to keep Oxygen out of the ieft—hand cell compartment. Solution Contains Oman). CP‘(aq) anions <— cations ————-> Pdle'(aq) + 2e‘ 9 Pd(s) + 4Cl' Cd(s) -) Cd”(aq) + 2e' cathode E r; = ? anode E red = .0403 v E a" = Ergd (cathode) - Ergd (anode); 1.03 v = Ergd _ (.0403 V); E ° =1.03v—0.403 = 0.63v e'r" je- Anode (-) Cathode (+) Cd Salt Pd Bridge Solution Contains anions Solution Contains ‘— - Cdz‘YaQ) cations PdC|42‘(aQ). Cl (80) —> 142(9) -> 2H*(aq) + 2e' E n; = 0.000 v 12(5) + Ze‘ -) 21'(aq) . E"; = 0.536 v E” = 0.536 v- 0.000 v = 0.536 v ' Ni(s) -> Ni2‘(aq) + 2e' .5“; = 028 v 2[Ce"‘(aq) + 19' -) Ce-‘flaqn em; = 1.61 v ° = 1.61 v- (-0.28 V) = 1.09 v Cr(s) -) Cr"*(aq) + 3e' E rgd = -0.74 V 3[Cr°‘(aq) + 1e" -> Cr2*(aq)] E "zd = -0.41 v E = ~0.41V-(-0,74 V) = 0.33 V 444 20 Electrochemistry Solutions to Exercises (d) 3[Cd(s) —> Cd"(aq) + 2e'] E n; = -o.4oa v 2[AI3*(aq) + 35' -> Al(s)] Er; = -1 .63 v ° = -1.66 v - (-0.403 V)] = -1.26 v 20.26 (a) 5(9) + 2e- -> 2F'(aq) Er; = 2.87 v 2CI'(aq) -) c149) + 2e‘ E fgd = 1.359 v E“ =2.87V-1.359V= 1.51 V (b) Zn(s) -> zn2*(aq) + 2e‘ E W; = -0.763 v aa2*(aq) + 2e" -> Ba(s) E éd = -2.90 v ° = -2.90 V - (-0.763 V) = -2.14 V (c) Fe2‘(aq) + 2e' -> Fe(s) E r; d = .0440 v 2[Fez*(aq)-)Fe3*(aq)+1e'] E ° = 0.771 v red E" = -0.440 V- 0.771 V = -1.21 V (d) ngz‘(aq) + 2e' -> 2Hg(t) Emf’d = 0.789 v 2[Cu*(aq) a Cu2‘(aq)+1e'] E n; = 0.153v E” =0.789V-0.153V= 0.636V 20.27 (a) 3[Ag+(aq)+1e- -> Ag(s)] Em} =o.799 Cr(s) -> Cr'"(aq) + 3e' Er; = -o.74 3Ag‘(aq) + Cr(s) -) 3Ag(s) + CP‘Yaq) E° = 0.799 - (10.74) = 1.54 V (b) Two of the combinations have essentially equal E° values. 2[Ag*(aq) + 1e‘ -) Ag(s)] Er; = 0.799 v Cu(s) -> Cuz“(aq) + 2e' E m", = 0.337 v 2Ag*(aq) + Cu(s) -> 2Ag(s) + Cu=+(aq) ° = 0.799 v - 0.337 v = 0.462 v 3[Niz*(aq) + 2e' -) Ni(_s)] E "’d = -0.28 v 2[Cr(s) -) Cr”(aq) + 3e'] E"; = -o.74 v 3Ni=+(aq) + zcr(s) -» 3Ni(s) + 2Cr5'(aq) E“ = -o.2a v - (-0.74 V) = 0.45 v 20.28 (a) 2[Au(s) +4Br'(aq) -) AUBr4'(aQ) + 315‘] E"; = -o.353v 3[2e' + IO'(aq) + H,O(l) 9 I'(aq) + 20H'(aq)] ._ E41: 0.49v _____________—.._....—___———-_———-— 2Au(s) + 83r(aq) + 310'(aq) + 3H20(l) -) 2AuBr.‘(aq) + 31‘(aq) + 60H'(aq) E = 0.49 - (-0.858) = 1.35 V 445 20 Electrochemistry Solutions to Exercises 20.29 20.30 20.31 (b) 2(Eu’*(aq) -) Eu3*(aq) + 1e‘] E"; = .043 V Sn=*(aq) + Ze" a Sn(s) Em; = -o.14 v 22mm) + Sn1‘(aq) ' -) 2Eu3”(aq) +_ Sn(s) 5° = -o.14 +0.43) = 0.29 v (a) Mn0.'(aq) + 8H’(aq) + 5e‘ -> Mn“(aq) + 4H20(I) E“; = 1.51 v (b) Because the half-reaction in part (a) is the more favorable reduction, it is the cathode reaction. (c) Snz“(aq) -) Sn‘*(aq) + 2e‘ _ Ergd = 0.154 v Balance electrons by multiplying the cathode reaction by 2 and the anode reaction by 5. 53n2*(aq) + 2Mn0“(aq) + 16H*(aq) -) 58n“(aq) + 2Mn2‘(aq) + 8H20(l) E“ = 1.51 V-0.154V= 1.356= 1.36V (d) (e). (a) The half-reactions are: 2H’l8Q) + 2e‘ 9 H2(9) Al(s) -> AP*(aq) + 3e- 57;, = -1.66 v Because it has the larger E (0.00 V vs 4.66 V). the first half-reaction is the reduction half-reaction in a voltaic cell. The standard hydrogen electrode (SHE) is the cathode and the Al strip is the anode. EmI = o.oov (b) The AI strip will lose mass as the reaction proceeds, because Al(s) is transformed to AF*(aq). - Balance electrons by multiplying the cathode reaction by 3 and the anode reaction by 2. 6H*(aq) + 2A|(s) -) 3Hz(g) + 2AI*(aq) (0) (d) The reduction half-reactions are: Cuz‘(aq) + 2e' -) Cu(s) Sn”(aq) + 2e' -) Sn(s) ° =0.00V-(-1.66V)=1.66V E“ = 0.337V E° = -O.136V (a) it is evident that Cu:+ is more readily reduced. Therefore, Cu serves as the cathode, Sn as the anode. (b) The copper electrode gains mass as Cu is plated out. the Sn electrode loses mass as Sn is oxidized. ' (c) The overall cell reaction is Cu"(aq) + Sn(s) -) Cu(s) + Sn2‘(aq) (d) E = 0.337 V - (—0.136 V) = 0.473 V 446 2 0 Eiectrochemistry Solutions to Exercises 20.32 (a) The two half-reactions are: Pb‘.‘(aq) + 2e; -> 90(5) E“ = -0.126 v 05(9) + 2e‘ -) 20l'(aq) E“ = 1.359 V _ Because E“ for the reduction 'of Cl2 is higher, the reduction of Cl2 occurs at the Pt cathode. The Pb electrode is the anode. (b) The Pb anode loses mass as Pb2'(aq) is produced. (6) 02(0) + Pbts) 9 Pb’*(aq) + 2CI'(aq) (d) ° =1.359V-(~0.126V) =1.485V Oxlclizing and Reducing Agents: Spontaneity ' 20.33 20.34 20.35 20.36 20.37 Negative. A strong reductant is likely to be oxidized. thus having a negative (a) reduction potential. (b) Right. Reducing agents are likely to be oxidized. and thus to be in a low oxidation state; the products of reduction half-reactions are in lower oxidation states than reactants. ' (a) Top. The reduction half-reactions near the top of Table 20.1 are most likely to occur. a strong oxidant is most likely to be reduced. (b) Left. An oxidant is reduced, so it is a reactant in a reduction half-reaction. in each case. choose the half—reaction with the more positive reduction potential and with the given substance on the left. (a) Cl2(g) (1.359 v vs. 1.065 V) (c) Br03'(aq) (1.52 v vs. 1.195 V) (b) N'i2*(aq) (—0.st vs. -0.403V) (d) 03 (g) (2.07 v vs. 1.776 V) In each case choose the half-reaction with the more negative reduction potential and the given substance on the right (a) Ca(s) (-2.87 V vs. .237 V) (b) A|(s) (-1.66 v Vs. -1.18 V) (c) H,(g. acidic) (0.000 v vs. 0.141 V) (d) H2803(aq) (0.17 v vs. 0559 V) If the substance is on the left of a reduction half-reaction. it will be an oxidant; if it is on the right. it wiil be a'r'eductant. The sign and magnitude of the E, determines whether it is strong or weak. (a) (b) CI‘(aq): Very weak reductant (on the right, E r; = 1.359 V) Mno.‘(aq. acidic): strong oxidant (on the ielt, E"; = 1.51 V) 447 20 Electrochemistry Solutions to Exercises 20.38 20. 39 20.40 20.41 20.42 20.43 (C) (d) Ba(s): strong reductant (on the right, Er; = -2.90 V) Zn"(aq): weak oxidant (on the left. E $2, = —0.763 V) If the substance is on the left of a reduction half-reaction. it will be an oxidant; if it is on the right, it will be a reductant. The sign and magnitude of the E rgd determines whether it is strong or weak. (a) (b) (c) (d) (a) (b) (a) (b) Na(s): strong reductant (on the right, Ergd = -2.71 V) 03(9): strong oxidant (on the left, E "fd = 2.07 V) oe°*(eq): very weak reductant (on the right, E “’6 = 1.61 V) Ca2*(aq): very weak oxidant (on the left. 'E a; = -2.87 V) Arranged in order of increasing strength as oxidizing agents (and increasing reduction potential): Cu’*(aq) < 02(9) < Cr207 '(aci) < Clzig) < Hzoztaq) Arranged in order of increasing strength as reducing agents (and decreasing reduCtion potential): H202(aq) < I'(aci) < Sn”(aq) < Zn(s) < Alts) The strongest oxidizing agent is the species most readily reduced , as evidenced by a large, positive reduction potential. That species is H202. The weakest oxidizing agent is the species that least readily accepts an electron. We expect that it will be very difficultto reduce Zn(s); indeed, Zn(s) acts as a comparatively Strong reducing agent. No potential is listed for reduction of Zn(s). but we can safely assume-that it is less readin reduced than any of the other species present. The strongest reducing agent is the species most easily oxidized (the largest negative reduction potential). Zn, E”; = «0.76 V, is the strongest reducing agent and F', E m’d = 2.87 v, is the weakest. Any of the reduced species in Table 20.1 from a half-reaction with a reduction potential more negative than 43.43 V will reduce Eu3+ to Eu". These include Zn(s). H,(g) in basic solution, etc. Fe(s) is questionable. ' Any oxidized species with a reduction potential greater than 0.59 V will oxidize RuO4 ' to Ru04'. MnO.‘ in neutral or aikaline solution is On the borderline. but any species above it on Tabie 20.1 would be strong enough. (a) (b) The more positive the emf of a reaction the more spontaneous the reaction. Reactions (a), (b) and (c) in Exercise 20.25 have positive E values and are spontaneous. ...
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125ch20a_001 - 0 .E‘Iectrochemistry Oxidation-Reduction...

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