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Silberberg Chapter 21 Solutions _Spring 2010_

Silberberg Chapter 21 Solutions _Spring 2010_ - CHAPTER 21...

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Unformatted text preview: CHAPTER 21 ELECTROCHEMISTRY: CHEMICAL CHANGE AND ELECTRICAL WORK ____’_’___._’_——-—————-— 21.1 Oxidation is the loss of electrons (resulting in a higher oxidation number), while reduction is the gain of electrons (resulting in a lower oxidation number). In an oxidation-reduction reaction, electrons transfer from the oxidized substance to the reduced substance. The oxidation number of the reactant being oxidized increases while the oxidation number of the reactant being reduced decreases. 21.2 No. one half-reaction cannot take place independently of the other because there is always a transfer of electrons from one substance to another. If one substance loses electrons (oxidation half-reaction), another substance must gain those electrons (reduction half-reaction). 21.3 Spontaneous reactions, 19th}.s < 0, take place in voltaic cells, which are also called galvanic cells. Nonspontaneous reactions take place in electrolytic cells and result in an increase in the free energy of the cell (A653,, > 0). 21.4 a) True b) True c) True d) False, in a voltaic cell, the system does work on the surroundings. e) True f) False, the electrolyte in a cell provides a solution of mobile ions to maintain charge neutrality. increases from —1 in C1‘ to 0 in C12. b) Mn04' is reduced because the oxidation number of Mn decreases fi'om +7 in M10; to +2 in M11”. c) The oxidizing agent is the substance that causes the oxidation by accepting electrons. The oxidizing agent is the substance reduced in the reaction, so MnOf is the oxidizing agent. d) Cl" is the reducing agent because it loses the electrons that are gained in the reduction. e) From 0', which is losing electrons, to M1104”. which is gaining electrons. f) 8 HZSO4(aq) + 2 KMnOAaq) + 10 KCl(aq) —) 2 MnSOAaq) + 5 Cl;(g) + 8 H100) + 6 K2304(aq) @ a) To decide which reactant is oxidized, look at oxidation numbers. Cl" is oxidized because its oxidation number 21.6 2 CrOflaq) + 2 H206) + 6 Clo—(aq) —> 2 CrOfTaq) + 3 {311(3) + 4 OH'(aq) a) The C10; is the oxidized species because Cr increases in oxidation state from +3 to +6. b} The 00' is the reduced species because Cl decreases in oxidation state from +1 to 0. c) The oxidizing agent is (310‘; the oxidizing agent is the substance reduced. d) The reducing agent is Croz'; the reducing agent is the substance oxidized. e) Electrons transfer from CrOf to ClO‘. 1') 2 NaCrOfiaq) + 6 NaClO(aq) + 2 H300!) —>2 NaZCr04(aq) + 3 Cl;(g) + 4 NaOH(aq) 21.7 '- _ 8.) Divide into half-reactions: Clog—(cg) —> Cl'(aq) w" flag) —> 12(3) Balance elements other than 0 and H ClOflaq) —) Cl_(aq) chlorine is balanced 2 I'(aq) —) 12(5) iodine now balanced Balance 0 by adding H20 Clog-(sq) —) Cl_(aq) + 3 H200?) add 3 waters to add 3 0’5 to product 2 11an -—> 12(3) no change 21-1 Balance H by adding H+ - C103'(aq) + 6 H+(aq) —) til-(sq) + 3 H200) add 6 H+ to reactants 2 I_{aq) —) 12(3) no change Balance charge by adding e‘ Clog-(cg) + 6 HYaq) + 6 e" —> Cling) + 3 H200) add 6 e" to reactants for a —1 charge on each side 2 flag) —) 12(3) + 2 e' add 2 e‘ to products for a —2 charge on each side Multiply each half-reaction by an integer to equalize the number of electrons ClOflaq) + 6 H+(aq) + 6 e" —) Cl‘(aq) + 3 H200) multiply by 1 to give 6 e' 3 { 2 Hang) —) 12(5) + 2 e'} multiply by 3 to give 6 e- Add half-reactions to give balanced equation in acidic solution. Clog—(aq) + 6 lfiaq) + 6 flag) «-> Cl'(aq) + 3 H200) + 3 12(5) Check balancing: Reactants: 1 Cl Products: 1 Cl 3 O 3 O 6 H ’ 6 H 6 I 6 I —1 charge —1 charge Oxidizing agent is C103" and reducing agent is 1—. b) Divide into half-reactions: MnOflaq) —) Mn02(s) soaziaqi —> SUE—rag) Balance elements other than 0 and H Mn04"(aq) w—> Mn02(s) Mn is balanced SO32‘(aq) —) SO42'(aq) S is balanced Balance 0 by adding H2O Mn04_(aq) —> Mn02(r) + 2 H200) add 2 H20 to products sogz—(aq) + Hgoa) —> soiling) add 1 H20 [0 reactants Balance H by adding H+ Mn04_(aq) + 4 H+(aq) —> Mn02(s) + 2 H200) add 4 H+ to reactants soflaq) + 1120(1) —> SOB-(age) + 2 H+(aq) add 2 H* to products Balance charge by adding e‘ Mn04_(aq) + 4 H+(aq) + 3 e" —> Mn02(s) + 2 H200) add 3 e‘ to reactants for a 0 charge on each side $032‘(aq) + H2O(l) —> SOB—(cg) + 2 HTaq) + 2 e‘ add 2 e" to products for a —2 charge on each side Multiply each half-reaction by an integer to equalize the number of electrons 2{Mn04'(aq) + 4 H*(aq) + 3 e" ——> M11026) + 2 H2003” multiply by 2 to give 6 e— 3(8032_(aq) + Hzou) —> sofxaq) + 2 mag) + 2 e‘} multiply by 3 to give 6 e' Add half—reactions and cancel substances that appear as both reactants and products 2 MnOJaq) + S—lF(aq) + 3 5032'(aq) + 3—1-5943} —) 2 Mn02(s) + 4-H20(I) + 3 SUE—(am + 6-1-1163?) The balanced equation in acidic solution is: 2 Mn04_(aq) + 2 mag) + 3 5032-(aq) —> 2 Mn02(s) + H200) + 3 soiling) To change to basic solution, add OH— to both sides of equation to neutralize H+ 2 Mn04_(aq) + 2 H+(aq) + 2 OH'(aq) + 3 502,1_{aq) —) 2 Mn02(s') + H200) + 3 804211140 + 2 0H'(aq) 2 Maori—(cg) + 2 H20( 1) + 3 803215213) —> 2 Mn02(s) + H2969 + 3 SO42'(aq) + 2 OH'(aq) Balanced equation in basic solution: 2 MnO{(aq) + H200) + 3 sogz—(aq) —) 2 Mn02(s) + 3 soflaq) + 2 0H‘(aq) Check balancing: Reactants: 2 Mn Products: 2 Mn 18 0 18 O 2 H 2 H 3 S 3 S —8 charge —8 charge Oxidizing agent is Mn04_ and reducing agent is SO; '. 21-2 c) Divide into half-reactions: MnOsTaq) -—> Mn2+(aq) 11202053?) ‘-) 02(3) Balance elements other than 0 and H MnOflaq) —) Mn2+(ag) Mn is balanced H202(aq) —> 02(3) No other elements to balance Balance 0 by adding H20 - Mum-(ago —) Mn2+(aq) + 4 320(1) add 4 H20 to products H202{aq) —> 02(3) 0 is balanced Balance H by adding I-l+ Mn04'(aq) + a mag) —> Mn2+(aq) + 4 H200) add a H" to reactants H202(aq) —) 02(g) + 2 H‘(aq) add 2 H“ to products Balance charge by adding e' Mnot‘taq) + a mag) + 5 e' -> Mn2+(aq) + 4 H200?) add 5 e" to reactants for +2 on each side H202{aq) —~) 02(g) + 2 H+(aq) + 2 e' add 2 e' to products for 0 charge on each side Multiply each half-reaction by an integer to equalize the number of electrons 2{Mn04’{aq) + 3 Httag) + 5 e' —> Mn2+(aq) + 4 H20(l)} ' multiply by 2 to give 10 e- 5 { H202(aq) —-> 02(3) + 2 Whig) + 2 e'} multiply by 5 to give 10 e' Add half-reactions and cancel substances that appear as both reactants and products 2 Mn04_(aq) + -1-6efl+{aq) + 5 H202(aq) —) 2 Mn2+(aq) + 8 H2000 + 5 02(3) + 4-04-1199; The balanced equation in acidic solution 2 Mn04‘(aq) + 6 H+(aq) + 5 H202(aq) —> 2 an+(aq) + 8 H200) + 5 02(3) Check balancing: Reactants: 2 Mn Products: 2 Mn 18 O 18 0 16 H 16 H +4 charge +4 charge Oxidizing agent is Mn04‘ and reducing agent is H202. 21.8 a) 3 02(3) + 4 NO(g) + 2 H200) —) 4 N03'(aq} + 4 H+(aq) Oxidizing agent is 02 and reducing agent: N0 _ - b) 2 CrOfTaq) + 8 H200) + 3 Cu(s) —> 2 Cr(OH)3(s-) + 3 Cu(0H )2(5') + 4 0H Tag) Oxidizing agent is CrOf‘ and reducing agent: Cu c) AsOfTaq) + N02"(aq) + H 200) -—> AsOflaq) + N03'(aq) + 2 OH'(aq) Oxidizing agent is A5043— and reducing agent: N02- 21.9 a) 3 BH4‘(aq) + 4 Clog—(aq) —) 3 H2B03'(aq) + 4 CI‘(aq) + 3 H200) Oxidizing agent is C10; and reducing agent: BH4' b) 2 ctof'taq) + 3 N20(g) + 10 Httaq) —> 2 Cr3+(aq) + 6 Notg) + 5 H200) Oxidizing agent is CrOf' and reducing agent: N20 c) 3 Br2(l) + 6 OH“(aq) —> Br03‘(aq) + S Br'(aq) + 3 H200) Oxidizing agent is Br; and reducing agent is Br; “I! 3% fi) Balance the reduction half-reaction: ,1 No;(aq) —> NO(g) + 2 H200) balance 0 N03'(aq) + 4 H+(aq) —> N0(g) + 2 H20(I) balance H biog—(age) + 4 H+(aq) + 3 e‘ —> NO(g) + 2 H200) balance charge Balance oxidation half-reaction: 4 Sb(.5') -—) Sb405(3) balance Sb 4 Sb(s) + 6 H200?) -—) Sb406(s) balance 0 4 Sb(.s') + 6 H300) —> Sb405(s) + 12 mag) balance H 4 Sb(s) + 6 H200) —> Sb406(s) + 12 H+(aq) + 12 e‘ balance charge 21-3 21.11 21.12 Multiply each half-reaction by an integer to equalize the number of electrons 4{N02'(aq) + 4 H*(aq) + 3 e‘ —> NO(g) + 2 H2O(l)} Multiply by 4 to give 12 e— 1{4 Sb(s) + 6 H200) —) Sb405{s) + 12 H+(aq) + 12 e‘} Multiply by l to give 12 e‘ Add half-reactions. Cancel common reactants and products. 4 NO3_(aq) + -1-6—H+(aq) + 4 Sb(s) + 644290) —) 4 NO(g) + 8-H2OU) + Sb405(s) + Wear?) Balanced equation in acidic solution: 4 NO3'(aq) + 4 H+{aq) + 4 Sb(s) —-> 4 NO(g) + 2 H200) + Sb406(s) Oxidizing agent is NO; and reducing agent is Sb. b) Balance reduction half-reaction: Bi03'(aq) —) Bi3+(aq) + 3 H200) balance 0 Bi03'(aq) + 6 H*(aq) —) Bi3+(aq) + 3 mm!) balance H Bi03"(aq) + 6 H+(aq) + 2 e‘ —) Bi3+(aq) + 3 H200!) balance charge to give +3 on each side Balance oxidation half-reaction: . I Mnlvaq) + 4 H200) —> Mn04_(aq) balance 0 Mn2+(aq) + 4 H200) —) Mn04_(aq) + 8 H+(aq) balance H Mn2+(aq) + 4 H200) —> MnOffizq) + 8 H+(aq) + 5 e' balance charge to give +2 on each side Multiply each half-reaction by an integer to equalize the number of electrons 5{Bi03'(aq) + 6 H*(aq) + 2 e" —) Bi3*(oq) + 3 H20(!)} Multiply by 5 to give 10 e" 2{ Mn2*(aq) + 4 H200) —9 MnOJaq} + 8 H+(aq) + 5 e‘} Multiply by 2 to give 10 e' Add half-reactions. Cancel H20 and H" in reactants and products. 5 Bi03‘(aq) + 39-H'firq) + 2 Mn2+(aq) + 8-3290) —) 5 Bia"(aq) + -1—5—H20(I) + 2 MnOJaq) + -1-6—I-I+an) Balanced reaction in acidic solution: 5 Bi03‘(aq) + 14 meg) + 2 Mn2+{aq) —> 5 Bi3+(aq) + 2 H200) + 2 MnOflaq) Bi03' is the oxidizing agent and Mn2+ is the reducing agent. c) Balance the reduction half-reaction: Pb(0H)3'(aq) —> Pb(s) + 3 H200) balance 0 Pb(0H)3'(aq) + 3 H+(aq) —) Pb(s) + 3 H200) balance H Pb(0H)3_(aq) + 3 H'"(aq) + 2 e‘ —> Pb(s) + 3 H200) balance charge to give 0 on each side Balance the oxidation half-reaction Fe(OH)2(.r) + H200) ——) Fe(0H)3{s) balance 0 Fe(OH)2(s) 1- H200) —) Fe(0H)3(s) + H+(aq) balance H Fe(OH)2(s) + H20(l) —) Fe(0H)3(.s-) + H+(aq) + e" balance charge to give 0 on each side Multiply each half-reaction by an integer to equalize the number of electrons 1{Pb(OH)3'(aq) + 3 H+(aq) + 2 e' —> Pb(s) + 3 H2000} Multiply by 1 to give 2 e— 2{Fe(OH)2(s) + H200) «9 Fe(0H)3(s) + H+(aq) + e'} Multiply by 2 to give 2 e' Add the two half-reactions. Cancel H20 and H". Pb(OH)2'(aq) + 3-H+(aq) + 2 Fe(0H)2(s) + 2—1-1200) -—> Pb(s) + 3-1-1200) + 2 Fe(OH)3(s) + 2—1413?) Pb(OH)3‘(aq) + H+(aq) + 2 Fe(OH)2(s) —) Pb(s) + H200) + 2 Fe(0H)3(s) Add one OH to both sides to neutralize H'“. Pb(OH)3-(aq) + Ewmq) + 2 Fe(0H)2(s) —) Pb(s) + H200) + 2 Fe(0H)3(s) + 0H'(aq) Pb(OH)3'(aq) + 14-20(4) + 2 Fe(OH)2(3) —> Pb(.r) + H2999 + 2 Fe(OH)3(.r) + 0H'(aq) Balanced reaction in basic solution: Pb(OH)3'(aq) + 2 Fe(OH)2(s) —) Pb(s) + 2 Fe(0H)3(s) + 0H'(aq) Pb(0H)3‘ is the oxidizing agent and Fe(0H)2 is the reducing agent. 5 Fe2+(aq) + molt-(cg) + 8 H*(aq) —> Mn2+(aq) + 5 Fe3+(cq) + 4 H200) a) Balance reduction half-reaction: NOg-(aq) —> N02(g) + H200) ' balance 0 NOflaq) + 2 H+[aq) —~> N02(g) + H2000 balance H NOflaq) + 2 H“(aq) + e‘ —> NO2(g) + H200) balance charge to give 0 on each side 21-4 21.13 21.14 21.15 L/" 21.17 21.18 Balance oxidation half-reaction: - Au(.s') + 4 Cl‘(aq) —) AuCl4'(aq) balance Cl Au(s) + 4 Cl'(aq) —> AuC14'(aq) + 3 e‘ balance charge to -4 on each side Multiply each half-reaction by an integer to equaJize the number of electrons 3{N03‘(aq) + 2 H+(aq) + e‘ —-) N02{g) + H20(l)} Multiply by 3 to give 3 e' 1{Au(5) + 4 C1'(aq)- —-> AuCl4'(aq) + 3 e‘} Multiply by l to give 3 e‘ Add half-reactions. Au(s) + 3 biog—(mg) + 4 Cl‘{aq) + 6 H+(aq) —> AuClJaq) + 3 N02(g) + 3 H200?) b) Oxidising agent is NO; and reducing agent is An. c) The HCl provides chloride ions that combine with the unstable gold ion to form the stable ion, Aule. a) A is the anode because by convention the anode is shown on the left. b) E is the cathode because by convention the cathode is shown on the right. c) C is the salt bridge providing electrical connection between the two solutions. d) A is the anode. so oxidation takes place there. Oxidation is the loss of electrons, meaning that electrons are leaving the anode. e) E is assigned a positive charge because it is the cathode. f) E gains mass because the reduction of the metal ion produced the solid metal. Unless the oxidizing and reducing agents are physically separated, the redox reaction will not generate electrical energy. This electrical energy is produced by forcing the electrons to travel through an external circuit. The purpose of the salt bridge is to maintain charge neutrality by allowing anions to flow into the anode compartment and cations to flow into the cathode compartment. An active electrode is a reactant or product in the cell reaction, whereas an inactive electrode is neither a reactant nor a product. An inactive electrode is present only to conduct electricity when the half-cell reaction does not include a metal. Platinum and graphite are commonly used as inactive electrodes. a) The metal A is being oxidized to form the metal cation. To form positive ions, an atom must always lose electrons. so this half-reaction is always an oxidation. b) The metal ion B is gaining electrons to form the metal B, so it is displaced. c) The anode is the electrode at which oxidation takes place, so metal A is used as the anode. d) Acid oxidizes metal B and metal B oxidizes metal A, so acid will oxidize metal A and bubbles will form when metal A is placed in acid. The same answer results if strength of reducing agents is considered. The fact that metal A is a better reducing agent than metal B indicates that if metal B reduces acid, then metal A will also reduce acid. a) If the zinc electrode is negative, oxidation takes place at the zinc electrode: ' 211(5) —) Zn2+(aq) + 2 e' Reduction half-reaction: Sn2+(aq) + 2 e" —> Sn(s) Overall reaction: Zn(s) + Sn2*(aq) —> Zn2+(aq) + Sn(s) 13) Voltmeter 21-5 21.19 21.21 a) (red half-rxn) Ag+(aq) + 1 e' “a Ag(s) (ox half-rxn) Pb(s) —) Pb2+(aq) + 2 e' (overall rxn) 2 Ag“(aq) + Pb(s) —) 2 Ag(s) + Pb2+(aq) 13) Voltmeter a) Electrons flow from the anode to the cathode, so from the iron half-cell to the nickel half-cell, left to right in the figure. By convention, the anode appears on the left and the cathode on the right. b) Oxidation occurs at the anode, which is the electrode in the iron half-cell. c) Electrons enter the reduction half-cell, the nickel half-cell in this example. d) Electrons are consumed in the reduction half-reaction. Reduction takes place at the cathode, nickel electrode. e) The anode is assigned a negative charge, so the iron electrode is negativelyr charged. 0 Metal is oxidized in the oxidation half-cell, so the iron electrode will decrease in mass. g) The solution must contain nickel ions, so any nickel salt can be added. 1 M NiSO.1 is one choice. h) KN03 is commonly used in salt bridges, the ions being K” and _N03‘. Other salts are also acceptable answers. i) Neither, because an inactive electrode could not replace either electrode since both the oxidation and the reduction half-reactions include the metal as either a reactant or a product. j) Anions will move towards the half-cell in which positive ions are being produced. The oxidation half-cell produces Fez", so salt bridge anions move from right (nickel half-cell) to left (iron half-cell). k) Oxidation half-reaction: Fe(s) a) Fe2+(aq) + 2 e' Reduction half—reaction: Ni2+(aq) + 2 e" —) Ni(s) Overall cell reaction: Fe(s) + Niz+(aq) —> Fe2"(aq) + Ni(s) a) The electrons flow left to right. b) Reduction occurs at the electrode on the right. c) Electrons leave the cell from the left side. (1) The zinc electrode generates the electrons. e) The cobalt electron has the positive charge. 0 The cobalt electrode increases in mass. g) The anode electrolyte could be 1 M Zn(N03);. h) One possible pair would be I? and N0; 1) Neither electrode could be replaced because both electrodes are part of the cell reaction. j) The cations move from left to right to maintain charge neutrality. k) Reduction: C02+(aq) + 2 e' ——> Co(s) Oxidation: 21(5) —) 2112*(aq) + 2 e' Overall: 211(3) + Coz+(aq) —> Co(s) + Zn2+(aq) 21-6 21.23 21.24 -'-j‘."::~\1 21.25 21.26 21.27 21.28 In cell notation, the oxidation components of the anode compartment are written on the left of the salt bridge and the reduction components of the cathode compartment are written to the right of the salt bridge. A double vertical line separates the anode from the cathode and represents the salt bridge. A single vertical line separates species of different phases. Anode ll Cathode a) Al is oxidized, so it is the anode and appears first in the cell notation: A1(s)IA13+{aq)1ICr3*(aq)ICr(s) b) Cu2+ is reduced, so Cu is the cathode and appears last in the cell notation. The oxidation of SO; does not include a metal, so an inactive electrode must be present. Hydrogen ion must be included in the oxidation half- cell. Ptl802(g)lSO4 '(aq), H+(aq)llCu2*(aq)lCu(.s-) a) Mn(s) + carom) ——> Mn2+(aq) + 01(5) b) 3 Fe(s) + 2 Noaaq) + s H+(aq) —> 3 Fe2+(oq) + 2 NO(g) + 4 H200) An isolated reduction or oxidation potential cannot be directly measured. However, by assigning a standard half- cel] potential to a particular half-reaction, the standard potentials of other half-reactions can be determined relative to this reference value. The standard reference half-cell is a standard hydrogen electrode defined to have an E" value of 0.000 V. A negative Egan indicates that the cell reaction is not spontaneous, AG" > 0. The reverse reaction is spontaneous with- Egg“ 3» 0. Similar to other state functions, the sign of 5" changes when a reaction is reversed. Unlike AG", AH“ and 3°, E“ is an intensive property, the ratio of energy to charge. When the coefficients in a reaction are multiplied by a factor, the values of AG“, AH“ and 3° are multiplied by the same factor. However, E“ does not change because both the energy and charge are multiplied by the factor and their ratio remains unchanged. a) Divide the balanced equation into reduction and oxidation half—reactions and add electrons. Add water and hydroxide ion to the half-reaction that includes oxygen. - Oxidation: Sea—(cg) —) Se(s) + 2 e— Reduction: 2 SOETaq) + 3 H200) + 4 e‘ —> szofraq) + 6 OH'(aq) b) 53.». = 533mm ~E§mde 55mm = gm, 4;,“ = —0.57 V — 0.35 V = 43.92 V a) (red ha1f~n(n) 03(g) + 2 H+(aq) + 2 e" —> 03(g) + H200) (ox half—rim) Mn2+(aq) + 2 H200) —> N11110:“) + 4 H*(aq) + 2 e‘ b)E;e]1 =Eooaone_EDmnganese Futurism =E°oaone‘ E331] = 2.0? V — 0.84 V = 1.23 V The greater (more positive) the reduction potential, the greater the strength as an oxidizing agent. a) From Appendix D: F83+(aq) + e' —> Fe2+(aq) 13° = 0.77 V Br2(l) + 2 e" —) 2 Br'(aq) E,“ = 1.07 V Cu2+(aq) + e' —> 01(3) 5° = 0.34 V When placed in order of decreasing strength as oxidizing agents: Br; > Fe3+ > Cu“. 21-? b) From Appendix D: Ca2+(aq) + 2 e" —> Ca(s) 13° = —2.87 V CrZO-f—(aq) + 14 H”(aq) 6 e‘ —»> 2 Cr3"(aq) + 7 H200) E" = 1.33 V AgTaq) + e" —) Ag(s) E°=0.80V When placed in order of increasing strength as oxidizing agents: Ca1+ < Ag" at 01-10,". 21.30 a) When placed in order of decreasing strength as reducing agents: 80; > MnO; > “350.; b) When placed in order of increasing strength as reducing agents: Hg < Sn «.1 Fe "m‘ o o o . . - a 21.31% E" n = Ecamode — Ean E Values are found in Appendix D. Spontaneous reactions have Ecell > 0. k ,1; a) Oxidation: Co(s) —> C02+(aq) + 2 e" E“ = —0.23 V .1 Reduction: 2 H+(aq) + 2 e‘ —) H1(g) E" = 0.00 V Overall reaction: Co(s) + 2 H+(aq) —) C02+(aq) + 112(3) £36,] = 0.00 V — (—0.23 V) = 0.23 v Reaction is spontaneous under standard state conditions because Egan is positive. 13) Oxidation: 2{Mn2+(aq) + 4 H106) —) MnOflaq} + 8 H*(aq) + 5 e'} E" = +1.51 V Reduction: 5{Br2(l) + 2 e' —) 2 Br‘(aq)} E" = +1.07 V O...
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