How To Oxidize A Cell

Transcription

1 hapter a) Oxidation ½ Reaction Fe + HlHFel 4 Fe + 4HlHFel 4 Fe + 4HlHFel 4 + 3H + Homework #3 hapter 11 Electrochemistry Fe + 4HlHFel 4 + 3H + + 3e - Reduction ½ Reaction H H + H H + + e- H Balanced Reaction (Fe + 4HlHFel 4 + 3H + + 3e - ) 3(H + + e- H ) Fe(s) + 8Hl(aq) HFel 4 (aq) + 3H (g) b) Oxidization ½ Reaction I - - I 3 3I - - I 3 3I - I e - Reduction ½ Reaction IO I 3 3IO I 3 3IO - 3 I H O 3IO H + I H O 3IO H e - I H O Balanced Reaction 3IO H e - I H O 8(3I - I e - ) 3IO - 3 (aq) + 18H + (aq) + 4I - (aq) 9I - 3 (aq) + 9H O(l) Divide trough by 3 IO - 3 (aq) + 6H + (aq) + 8I - (aq) 3I - 3 (aq) + 3H O(l) c) Oxidation ½ Reaction r(ns) 4-6 r 3+ + NO O + SO 4 r(ns) 6 4- r NO O + 6SO 4 - r(ns) H O r NO O + 6SO 4 - r(ns) H O r NO O + 6SO H + r(ns) H O r NO O + 6SO H e - 1

2 Reduction ½ Reaction e 4+ e 3+ e 4+ + e - e 3+ Balanced Reaction r(ns) H O r NO O + 6SO H e - 97(e 4+ + e - e 3+ ) r(ns) 4-6 (aq)+ 54H O(l) + 97e 4+ (aq) r 3+ (aq) + 6NO - 3 (aq) + 6O (g) + 6SO 4 (aq) + 108H + (aq) + 97e 3+ (aq) d) Oxidation ½ Reaction ri 3 ro IO 4 ri 3 ro IO 4 ri H O ro IO 4 ri H O ro IO 4 + 3H + ri H O ro IO 4 + 3H + + 7e - Reduction ½ Reaction l l - l l - l + e - l - Balanced Equation (ri H O ro IO 4 + 3H + + 7e - ) 7(l + e - l - ) ri 3 (s) + 3H O(l) + 7l (g) ro - 4 (aq) + 6IO - 4 (aq) + 64H + (aq) + 54l - (aq) The solution is basic not acidic, add 64 OH - to both sides ri 3 (s) + 7l (g) + 64OH - (aq) ro - 4 (aq) + 6IO - 4 (aq) + 3H O(l) + 54l - (aq) e) Oxidation ½ Reaction Fe(N) 4-6 Fe(OH) 3 + O NO 3 Fe(N) 4-6 Fe(OH) 3 + 6O NO 3 Fe(N) H O Fe(OH) 3 + 6O NO 3 Fe(N) H O Fe(OH) 3 + 6O NO H + Fe(N) H O Fe(OH) 3 + 6O NO H e - Reduction ½ Reaction e 4+ e(oh) 3 e H O e(oh) 3 e H O e(oh) 3 + 3H + e H O + e - e(oh) 3 + 3H + Balance Reaction Fe(N) H O Fe(OH) 3 + 6O NO H e - 61(e H O + e - e(oh) 3 + 3H + ) Fe(N) 6 4- (aq) + H O(l) + 61e 4+ (aq) Fe(OH) 3 (s) + 6O 3 - (aq) + 6NO 3 - (aq)+ 58H + (aq) + 61e(OH) 3 (s)

3 The solution is basic not acidic, add 58OH - to both sides Fe(N) 6 4- (aq) + 58OH - (aq) + 61e 4+ (aq) Fe(OH) 3 (s) + 6O 3 - (aq) + 6NO 3 - (aq)+ 61e(OH) 3 (s) + 36H O(l) hapter Reactions of Interest Ni + + e - Ni u + + e - u u + + e - u E =-0.3 V E =0.5 V E =0.34 V Zn + + e - Zn E =-0.76 V In order to plate out Ni you need the Ni reaction to be the reduction ½ reaction (cathode). In addition, you also need the cell to be galvanic (E >0). The E cell of the nickel/copper cell is V or V depending on the ion of copper that is used. Therefore, neither of these cells would be a galvanic cell, resulting in copper not being an appropriate material. The E cell of the nickel/zinc cell is 0.53 V. Therefore, the nickel/zink cell would plate out Ni. 10. E is the reaction potential at the standard state. The standard state is 1 M concentration of aqueous solutions or 1 atm pressure of gasses. By convention E is set to 0 for hydrogen s reduction ½ reaction (H + + e - H ). E is the reaction potential not at the standard state. E is 0 when the cell is at equilibrium. 19. a) Reactions of interest r O H + + 6e - r H O l + e - l - 3(l + e - l - ) r H O r O H + + 6e - E = 1.33 V E = 1.36 V E = 1.36 V E = V 3l (g) + r 3+ (aq) + 7H O(l) 6l - (aq) + r O 7 - (aq) + 14H + (aq) E =1.36 V-1.33 V=0.03V 3

4 b) Reactions of Interest u + + e - u E = 0.34 V Mg + + e - Mg E =-.37 V u + + e - u E = 0.34 V Mg Mg + + e - E =.37 V Mg(s) + u + (aq) u(s) + Mg + (aq) E = 0.34 V +.37 V =.71 V c) Reactions of Interest IO H + + 5e - ½I +3H O E =1.0 V Fe 3+ + e - Fe + E = 0.77 V IO H + + 5e - ½I +3H O 5(Fe + Fe 3+ + e - ) IO - 3 (aq) + 6H + (aq) + 5Fe + (aq) ½I (s) +3H O(l) + 5Fe 3+ (aq) E = 1.0 V E =-0.77V E =1.0V-0.77V= 0.43 V Note: I does not conduct electricity; therefore, the Pt electrode is used d) Reactions of Interest Ag + + e - Ag (Ag + + e - Ag) Zn Zn + + e - Zn + + e - Zn Ag + (aq) + Zn(s) Ag(s) + Zn + (aq) E = 0.80 V E = V E = 0.80 V E = 0.76 V E = 0.80 V V = 1.56 V 4

5 1. a) l + e - l - E = 1.36 V Br - Br + e - l (g) + Br - (aq) l - (aq) + Br (aq) Note: Br is a liquid at room temperature E =-1.09 V E =1.36 V 1.09 V= 0.7 V b) 5(IO H + + e - IO 3 - +H O) E = 1.60 V (Mn + + 4H O MnO H + + 5e - ) 5IO 3 - (aq)+mn + (aq)+3h O(l)5IO 3 - (aq)+mno 4 - (aq)+6h + (aq) E = V E = 1.60V 1.51V=0.09V 5

6 c) H O +H + + e - H O E = 1.78 V H O O + H + + e - H O (aq) O (g) + H O(l) E = V E = 1.78 V V = 1.10 V d) (Fe e - Fe) E = V 3(Mn Mn + + e - ) E = 1.18 V Fe 3+ (aq) + 3Mn(s) Fe(s) + Mn + (aq) E = V V = 1.14 V. a) For a galvanic cell E must be positive therefore u + must be oxidized given: Au e - Au E = 1.50V u + +e - u + E = 0.16 V Au e - Au E = 1.50V 3(u + u + +e - ) Au 3+ (aq) + 3u + (aq) Au(s) + 3u + (aq) E = V E = 1.34 V b) For a galvanic cell E must be positive therefore d must be oxidized given: VO + + H + + e - VO + + H O E = 1.00V d + +e - d E = V 6

7 (VO + + H + + e - VO + + H O) E = 1.00V d d + +e - E = 0.40 V VO + (aq) +4H + (aq) + d(s) VO + (aq) +H O(l) + d + (aq) E = 1.40 V 4. a) u + + e - u E = 0.34 V In order for the cell to be a galvanic cell u + must be reduced, therefore, SE is oxidized and at the anode. E cell = 0.34 V V = 0.10 V b) Fe 3+ + e - Fe + E = 0.77 V In order for the cell to be a galvanic cell, Fe 3+ must be reduced, therefore, SE is oxidized and at the anode. E cell = 0.77 V V = 0.53 V c) Agl + e - Ag + l - E = 0. V In order for the cell to be a galvanic cell, Ag must be oxidized, therefore, SE is reduced and at the cathode. E cell = 0.4 V 0. V = 0.0 V d) Al e - Al E = V In order for the cell to be a galvanic cell, Al must be oxidized, therefore, SE is reduced and at the cathode. E cell = 0.4 V V = 1.90 V e) Ni + + e - Ni E = -0.3 V In order for the cell to be a galvanic, cell Ni must be oxidized, therefore, SE is reduced and at the cathode. E cell = 0.4 V V = 0.47 V 5. a) u u + +e - E = V H + + e - H E = 0.0 V No H + cannot oxidize u b) I - I +e - E =-0.54 V Fe 3+ + e - Fe + E = 0.77 V Yes Fe 3+ is capable of oxidizing I - if it is going to Fe + I - I +e - E =-0.54 V Fe e - Fe E = V No Fe 3+ cannot oxidize I - if it is going to Fe c) Ag + + e - Ag E =0.80 V H H + +e - E = 0.0 V Yes H is capable of reducing Ag + 7

8 d) r 3+ + e - r + E = V Fe + Fe 3+ + e - E =-0.77 V No Fe + is not capable of reducing r 3 + to r + 6. The oxidizing agent is the species that is reduced. Therefore, the best oxidizing reagent is the species that has the largest E value for the reduction ½ reaction. K + < H O < d + < I < Aul 4 - < IO 3-7. The reducing agent is the species that is oxidized. Therefore, the best reducing agent is the 8. hoices species that has the smallest E value for the reduction ½ reaction. Br + e - Br - H + + e - H d + + e - d La e - La F - < H O < I < u + < H - < K E =1.09 V E =0.0 V E =-0.40 V E =-.37 V a + + e - a E =-.76 V Underlined are possible answers a) The oxidizing agent is the species that is being reduced or species in the reduction reaction. Therefore Br -, H, d, and a can be eliminated because if they were on the reactants side of the reaction, therefore, they can only be in an oxidization reaction. Out of the remaining species the best oxidizing agent is the one with the largest E value in the reduction ½ reactions. Br b) The reducing agent is a species that is being oxidized or the species in the oxidizing reaction. Therefore Br, H +, La 3+ can be eliminated because they were on the reactant side of the equation, therefore, they can only be in a reduction reaction. Out of the remaining species the best oxidizing agent is the one with the smallest E value in the reduction ½ reactions. a c) MnO H + + 5e - Mn + + 4H O E = 1.51 V In order for MnO 4 - to oxidize a species the species has to be in an oxidation reaction; therefore, Br, H +, and La 3+ can be eliminated because they were on the reactant side of the equation, therefore, they can only be in a reduction reaction. Of the remaining species the E value of the reduction ½ reaction must be smaller than 1.51 V. Br -, H, d, and a d) Zn + + e - Zn E =-0.76 V Reaction of interest Zn Zn + + e - E = 0.76 V In order for Zn to reduce a species it must be in a reduction reaction. Therefore Br -, H, 8

9 d, and a can be eliminated because if they were on the reactants side of the reaction, therefore, they can only be in an oxidization reaction. Of the remaining species the E value of the reduction ½ reaction must be greater than V. Br or Ag + 9. a) Br + e - Br - E = 1.09 V l + e - l - Reactions of interest Br - Br + e - l - l + e - E =1.36 V E = V E =-1.36 V In order to oxidize Br - and not l - the E value of the reduction ½ reaction must be greater than 1.09 V but smaller than 1.36 V r O H + + 6e - r H O O + 4H + + 4e - H O MnO + 4H + + 4e - Mn + + H O E = 1.33 V E = 1.3 V E = 1.1 V IO H e - ½I + 3H O E = 1.0 V Therefore, r O - 7, O, MnO, and IO - 3 could oxidize Br - to Br but not oxidize l - to l. b) Mn + + e - Mn E = V Ni + + e - Ni Reactions of interest Mn Mn + + e - Ni Ni + + e - E = -0.3 V E = 1.18 V E = 0.3 V In order to oxidize Mn and not Ni the E value of the reduction ½ reaction must be greater than V but smaller than -0.3 V PbSO 4 + e - Pb + SO 4 - d + +e - d Fe + + e - Fe r 3+ + e - r + r 3+ + e - r Zn + + e - Zn E = V E = V E = V E = V E = V E =-0.76 V H O + e - H + OH - E = V Therefore, PbSO 4, d +, Fe +, r 3+, Zn +, and H O are capable of oxidizing Mn to Mn + but not oxidizing Ni to Ni a) u + + e - u E = 0.34 V u + + e - u + E = 0.16 V 9

10 In order to reduce u + to u but not to u + the E values of the reduction ½ reaction must be greater than 0.16 V and less than 0.34 V. The species also must be on the product side of the reduction reaction. Hgl + e - Hg + l - E = 0.7 V Agl + e - Ag + 4l - E = 0. V SO H + + e - H SO 4 + H O E = 0.0 V Therefore, Hg/l -, Ag/l -, and H SO 3 are capable of reducing u + to u but not to u +. b) Br + e - Br - E = 1.09 V I + e - I - E =0.54 V In order to reduce Br to Br - but not I to I - the E values of the reduction ½ reaction must be great than 0.54 V and less than 1.09 V. The species also must be on the product side of the reduction reaction. VO + + H + + e - VO + + H O E = 1.00V Aul e - Au + 4l - E = 0.99V NO H + + 3e - NO + H O E = 0.96V lo + e - lo - Hg + + e - Hg + Ag + + e - Ag Hg + + e - Hg Fe 3+ + e - Fe + O + H + + e - H O E =0.954 V E =0.91 V E = 0.80 V E =0.80 V E = 0.77 V E =0.68 V MnO e - - MnO 4 E =0.56 V Therefore, VO +,Au/l -, NO, lo -, Hg +, Ag, Hg, Fe +, H O, and MnO - 4 are capable of reducing Br to Br - but not I to I a) (lo - lo +e - ) E =-0.95 V l + e - l - E = 1.36 V lo - (aq) + l (g) lo (g) + l - (aq) Note: The Na + in the equation is just there as a spectator ion E =1.36 V V = 0.41 V G E e 96, V k E ln K K K 0.41V ln e 96, K b) Assume acidic conditions Reduction ½ Reaction e e K 10

11 lo l - lo l - + H O lo + 4H + l - +H O lo + 4H + + 5e - l - +H O Oxidation ½ Reaction - lo lo 3 - lo + H O lo 3 lo + H O lo H + lo + H O lo H + + e - Balanced Reaction lo + 4H + + 5e - l - +H O 5(lO + H O lo H + + e - ) 6lO (g) + 3H O(l) l - (aq) + 5lO - 3 (aq) + 6H + (aq) 41. a) Reaction 1: Unbalanced Mn(s) + NO - 3 (aq) NO(g) + Mn + (aq) Oxidation ½ reaction Mn Mn + Mn Mn + + e - Reduction ½ reaction (The problem told you that you have nitric acid as a reactant. To determine what species NO - 3 forms, use the standard reduction potentials) NO - 3 NO NO - 3 NO + H O NO H + NO + H O NO H + + 3e - NO + H O Balanced Reaction 3(Mn Mn + + e - ) (NO H + + 3e - NO + H O) 3Mn(s) + NO - 3 (aq) +8H + (aq) 3Mn + (aq) + NO(g) + 4H O(l) Reaction : Unbalanced Mn + (aq)+ IO - 4 (aq) MnO - 4 (aq) + IO - 3 (aq) Oxidation ½ reaction Mn + - MnO 4 Mn H O MnO 4 Mn + + 4H O MnO H + Mn + + 4H O MnO H + + 5e - Reduction ½ Reaction IO IO 3 IO - 4 IO H O IO H + + e - IO H O IO H + + e - IO H O Balanced Reaction 11

12 (Mn + + 4H O MnO H + + 5e - ) 5(IO H + + e - IO H O) Mn + (aq) + 5IO - 4 (aq) +3H O(l) MnO - 4 (aq) + 5IO - 3 (aq) + 6H + (aq) b) Reaction 1 3(Mn Mn + + e - ) E = 1.18 V (NO H + + 3e - NO + H O) E = 0.96 V 3Mn(s) + NO - 3 (aq) +8H + (aq) 3Mn + (aq) + NO(g) + 4H O(l) E cell = 1.18 V V =.14 V 4. a) Given 6 G E 6 e 96, k e E ln K 696, E K K 500. K e 10 e The K number is most likely too large for your calculator Reaction (Mn + + 4H O MnO H + + 5e - ) E = V 5(IO H + + e - IO H O) E = 1.60 V Mn + (aq) + 5IO - 4 (aq) +3H O(l) MnO - 4 (aq) + 5IO - 3 (aq) + 6H + (aq) E cell = 1.60 V V = 0.09 V 4 G E 10 e 96, k E ln K 1096, E K K K e H + + e- H O + 4H + + 4e - H O e (H H + + e - ) E =-0.00 V O + 4H + + 4e - H O H + O H O E = 1.3 V E =0.00 V E = 1.3 V E cell = 1.3 V V = 1.3 V 5 G E 4 e 96, k b) Two gases go to a liquid, therefore, the positional probability decreases causing ΔS to be negative. G HTS If ΔS is negative the only way for ΔG to be negative is if ΔH is negative. c) ΔG is the maximum possible work that can be done. Increasing the temperature increases the TΔS term, therefore, since this term is +, less possible work can be done. e 1

13 43. G E G H TS E H TS H S E T S H E T H If E vs. T was plotted on a graph, the intercept would equal and the slope of the line S would equal The smaller the S term is the smaller the temperature dependence of E. Therefore, cells that have small ΔS terms are relatively temperature independent. 45. a) u + u + + e - E = V u + + e - u E = 0.5 V u + u + + u E = V V = 0.36 V This reaction is spontaneous G E 1 96, V E ln E K 196, V K K 6 K e e 1.10 b) (Fe + Fe 3+ +e - ) E = V k Fe + +e - Fe E = V 3Fe + Fe 3+ + Fe E =-0.44V V=-1.1 V This reaction is not spontaneous c) HlO + H O lo H + + e - E =-1.1 V HlO + H + + e - HlO + H O E =1.65 V HlO (aq) lo - 3 (aq) + H + (aq) + HlO(aq) E = -1.1 V V =0.44 V k 96, V 85,000 G E 85 E ln E K 96, V K K 14 K e e

14 53. a) For galvanic cell E must be positive Au 3+ +3e - Au 3(Tl Tl + + e - ) E = 1.50 V E = 0.34 V Au 3+ (aq) + 3Tl(s) Au(s) + 3Tl + (aq) b) 5 E = 1.50 V V = 1.84 V G E V E ln E K 396, V K K 93 K e e.610 c) 54. (r + r 3+ + e - ) o + + e - o r + + o + r Given k K 98.14K Tl E E ln Q E ln V ln.04v Au K 98K 96, E ln K ln V K 98K 96, r.0 E E ln Q E ln 0.0V ln 0.151V r o G E 96, V k Al e - Al Pb + + e - Pb (Al Al e - ) 3(Pb + + e - Pb) E = 1.66 V E = V E = V E = V Al(s) + 3Pb + (aq) Al 3+ (aq) + Pb(s) E = 1.66V V= 1.53V alculate the final Pb + concentration: Pb + Al 3+ Initial hange -3x +x=0.60 Final x =1.60 x must equal 0.30 M, therefore, the Pb + concentration will change by 0.90 M and the final Pb + concentration equals =0.10 M. 14

15 alculate E of cell 3 Al E E ln Q E ln 3 Pb K 98K 696, E 1.53V ln 1.50V Since the same material is on both sides of the cell E = 0 In order to be a galvanic cell the concentration on the anode side of Ni + must be smaller than the concentration of Ni + on the cathode side Ni + + e - Ni Ni ( anode) E E ln K ln Ni ( cathode ) K 98K 1.0M a) E ln 1.0M 0.0V 96, 485 b) c) d) e) No electron flow K 98K 1.0M E ln.0m V 96, 485 Anode is on the left (1.0 M) the cathode is on the right (.0 M) and electrons flow from left to right on the diagram K 98K 0.1M E ln 1.0M 0.030V 96, 485 Anode is on the right (0.1 M) the cathode is on the left (1.0 M) and electrons flow from right to left on the diagram K 98K M E ln 1.0M 0.13V 96, 485 Anode is on the right ( M) the cathode is on the left (1.0 M) and electrons flow from right to left on the diagram K 98K.5M E ln.5m 0.0V 96, 485 No electron flow. 68. a) Al e - Al 1000g 1 Al 3 e 1kg 6.98 g Al 1 Al 1.0 kg Al 111 e It n F , t s 9.7h I 100.0A b) Ni + + e - Ni 15

16 69. 1 Ni e g Ni 1 Ni 1.0 g Ni e , 485 t 33s I A c) Ag + + e - Ag 5.0 Ag 5.0 e 1 e 1 Ag , 485 t 4800s 1.3h I A 60min 60s 15A 1.0h It 1h 1min n 0.56 e F 96, 485 a) o + + e - o 1 o g o 1 o 0.56 e 17 g o b) Hf e - Hf e 1 Hf g Hf 1 Hf 0.56 e 5 g Hf 4 e c) I - I +e - 1 I g I e 1 I g I e d) ro 3 + 6H + + 6e - r + 3H O 1 r 5.00 g r 1 r 0.56 e 4.9 g r 6 e 70. The question wants you to calculate the arity of the initial solution n c V We know the volume so we need to calculate the e of Ag +. The ½ reaction that we are interested in is Ag + +e - Ag If we know the es of electrons we can get the es of Ag +. alculate the es of electrons. 60s It.00A.30 min 1min n e F 96,485 alculate the es of Ag Ag e e alculate arity n c M Ag V 0.50L Ag 16

17 71. The question asked you to calculate time to plate out 10 g Bi. To calculate the time use It n F Therefore, need to determine the number of e -. To do this you must find a relationship between e - and Bi. Balance the equation. BiO + Bi BiO + Bi + H O BiO + + H + Bi + H O BIO + + H + + 3e - Bi + H O Note: If you balanced it in basic conditions you would also come out with 3 electrons Determine the number of e - 1 Bi 3 e 09.0g 1 Bi n 10.0 g Bi Determine time t 556s I 5.0A 73. For electrolysis reaction E cell is - a) athode reaction Ni + + e - Ni E = -0.3 V Anode reaction Br - Br + e - E = V b) athode reaction Al e - Al E = V Anode reaction F - F + e - E = -.87 V c) athode reaction Mn + + e - Mn E = V Anode reaction I - I + e - E = V For the aqueous solution we must also consider the reactions with H O d) athode reaction Ni + + e - Ni H O + e - H + OH - E = -0.3 V E = V Since it is easier to reduce Ni + than H O the nickel reaction will occur at the cathode. Anode reaction Br - Br + e - H O O + 4H + + 4e - E = V E = -1.3 V Since it is easier to oxidize B - than H O the bromine reaction will occur at the anode. e) athode reaction Al e - Al E =-1.66 V 17

18 H O + e - H + OH - E = V Since it is easier to reduce H O than Al 3+ the water reaction will occur at the cathode. Anode reaction F - F + e - H O O + 4H + + 4e - E = -.87 V E = -1.3 V Since it is easier to oxidize H O than F - the water reaction will occur at the anode. f) athode reaction Mn + + e - Mn H O + e - H + OH - E = V E = V Since it is easier to reduce H O than Mn + the water reaction will occur at the cathode. Anode reaction I - I + e - H O O + 4H + + 4e - E =-0.54 V E =-1.3 V Since it is easier to oxidize I - than H O the iodine reaction will occur at the anode. 76. The question wants you to determine the charge on the ruthenium. The reaction that we are interested in is: Ru n+ + ne - Ru. Therefore, we should be able to identify n by comparing the es of Ru to the es of e-. alculate the es of Ru.618 g Ru Ru alculate the es of e - 1 Ru g Ru 60s.50A 50.0 min It 1min n e F 96,485 Because the ratio between the es of e- and es of Ru is 3:1 the charge on the ruthenium must be The question wants you to calculate the volume of O and H gas produced in the electrolysis of water. The question gives you the time and the current. With this information you can calculate the e of electrons. 60s S It.50A15.0 min 1min n e F 96, 485 You need to find a relationship between the es of electrons and the es of O and H, to do this look at the equations for the electrolysis of water. (H + + e - H ) H O O + 4H + + 4e - H O H + O alculate the es of H H e e H 18

19 alculate the volume of H Latm n K 73.15K V 0.6 L H P 1.00atm alculate es of O O e e O alculate the volume of O Latm n K 73.15K V L O P 1.00atm 88. A battery is a galvanic cell. When a battery is new there are many more reactants than products in the battery and therefore, Q is small. As the battery is used it turns reactants into product and Q gets larger. Since E E ln Q as the battery is used up E gets smaller and smaller until it is 0. A battery is not a system at equilibrium otherwise E = 0. Fuel cells and batteries are both galvanic cells. The difference between them is that batteries are enclosed and therefore, eventually reach equilibrium. For fuel cells the reactants are continually supplied and therefore, never reach equilibrium. 90. The reaction that goes on in a hydrogen oxygen fuel cell is H (g) + O (g) H O(l) O + 4H + + 4e - H O E = 1.3 V (H H + + e - ) E = -0.0 V O (g) + H (g) H O(l) w max G E Because standard conditions w w max max E E = 1.3 V V = 1.3 V 5 k 4 96, V This is the w max for es of H O. Need to find the work mass for 1.00 kg of H O 1000 g 1 HO 475 k 1kg 18.0 g H O H O 1.00kg 13, 00k The work done can be no larger than -13,00 k. Usually the max work is not done and some of the energy is lost to heat. Fuel cells are more efficient in converting chemical energy into electrical energy than combustion reaction. The main disadvantage of fuel cells is their cost. 19