Topic 4 Electrochem. The study of interchange energy chemical electrical

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1 Topic 4 lectrochem The study of interchange energy chemical electrical

2 Review of Terms Oxidation reduction (redox) reaction involves a transfer of electrons from the reducing agent to the oxidizing agent Oxidation loss of electrons Reduction gain of electrons Reducing agent electron donor Oxidizing agent electron acceptor

3 Half Reactions The overall reaction is split into two half reactions, one involving oxidation and one reduction. 8H + + MnO 4 + 5Fe 2+ Mn Fe + + 4H 2 O Reduction: 8H + + MnO 4 + 5e Mn H 2 O Oxidation: 5Fe 2+ 5Fe + + 5e

4 lectrochemical (Galvanic) Cell Device in which chemical energy is changed to electrical energy. Uses a spontaneous redox reaction to produce a current that can be used to do work.

5 1) oxidation-reduction (redox) reaction rxn: 8H (aq) Mn half-reactions: re : ox : 8H MnO 5 4 (aq) 2 2 (aq) MnO Fe - 4 5Fe 5Fe 5e Fe 2 (aq (aq ) ) 4H Mn e 2 2 O ( ) 4H 2 O

6 2) When the oxidizing agent & reducing agent are physically separated, e transfer through an external wire. generates electricity. ) salt bridge: connect two solns 4) electrodes; where the redox rxn occur anode: oxidation occur cathode: reduction occur anode e cathode

7 cell 5 ) Cell potential ( ) a) The voltage difference between the electrodes. electromotive force (emf) b) can be measured by voltmeter. c) emf of a cell depends on The nature of the electrodes [ions] Temp.

8 6) It is impossible to measure cell of a half-rxn directly, need a reference rxn SH standard hydrogen electrode: 2H 2e H H 1M, P 1atm H 2 2 ε cell red

9 7) Standard reduction potential red 2H Zn 2 2e 2e H 2 Zn Table 9.

11 Zn(s) Zn 2+ (aq) Pt(s) H + (aq) H 2 (g) rxn : rxn : cell 2H cell.76v Zn (s) cell 1.1V red (aq) - ox H H2 Cu Zn 2 (aq) 2 Cu Cu.4V (s) R - Zn L -.76V Zn 2 (aq) - 2 Zn Zn -.76V 2 (aq - 2 Zn Zn ) H 2(g) Cu (s)

12 x: p222 Pt(s) NADH(aq),NAD + (aq),h + (aq) H 2 O 2 (aq),h + (aq) O 2 (g) Pt(s)

13 Concept Check Sketch a cell using the following solutions and electrodes. Include: The potential of the cell The direction of electron flow Labels on the anode and the cathode a) Ag electrode in 1. M Ag + (aq) and Cu electrode in 1. M Cu 2+ (aq)

14 Work Work is never the maximum possible if any current is flowing. In any real, spontaneous process some energy is always wasted the actual work realized is always less than the calculated maximum.

15 emf (V) work (J) charge (C) - w q w q ΔG & q nf nf c mol e & w max ΔG standard condition : ΔG -nf

16 examples (1) Calculate ΔG Cu 2 (aq) Fe (s) Cu (s) Fe 2 (aq) rxn ΔG re - nf ox

17 (2) Predict whether 1M HNO will dissolve gold metal to form 1M Au +? NO Au cell 4H e re e - Au ox NO 2H.96V 1.5V.54V 2 O

18 () Concentration Cells Determine a) e flow direction? b) anode? cathode? c) =? at 25

19 The Nernst eqn ΔG nf ΔG RT lnq nf RT lnq RT lnq nf.592 logq n

20 Ion-Selective lectrodes: ph meter

21 at equilibrium: ΔG=, cell =, Q=K ΔG RT lnk RT nf lnk.592 n logk

22 (ex) Calculate K sp for AgCl at 25 soln: =.58V Ag Ag?.58V log Ag 1. 1M AgNO(aq) 1M NaCl(s) & AgCl(s) K sp Ag Cl 1.M

23 1) galvanic cell electrolytic cell chemical electrical Fig 18.19(a) electrical chemical Fig 18.19(b) Zn Cu 2 Zn 2 Cu Zn 2 Cu Zn Cu 2 1.1V external power 1.1V redox rxn: spontaneous nonspontaneous

24 Homework (4/7/214) 9.14, 9.19, 9.2, 9.29, 9.5,9.7

25 Formal potential AgCl (s) + e - Ag (s) + Cl V.197 V in saturated KCl (formal potentional) =.222V: S.H.. Cl- (aq, 1M) AgCl (s) Ag(s) (formal potential) =.197 V (in saturated KCl) S.H.. KCl (aq, saturated) AgCl (s) Ag(s)

26 Dependence of potential on ph Many redox reactions involved protons, and their potentials are influenced greatly by ph. x : H AsO 4 2H 2e - H AsO H 2 O ' log[.592pH.59pH [H AsO ] log [H AsO ][H - H.592 ] [H log [H ] [H log [H 2 AsO AsO AsO AsO 4 ] ] 4 ] ]

27 and the quilibrium Constant

28 log Q n.59 ] ][Cl [Fe ] [Fe log n.59 ) ( ] log[cl n.59 ] [Fe ] [Fe log n.59 ) ( ) (.549V ) AgCl( Fe ) Ag( Fe.222V Cl ) Ag( e ) AgCl(.771V Fe e Fe 2 2 AgCl,Ag,Fe Fe cell s s s s and the quilibrium Constant

29 Reference lectrodes Indicator electrode: responds to analyte concentration Reference electrode: maintains a fixed potential 2 [Fe cell.592log [Fe ] ].592log[Cl ]

30 Reference lectrodes Silver-Sliver Chloride AgCl + e - Ag(s) +Cl - =.222 V (saturated KCl) =.197 V Calomel Hg 2 Cl 2 + 2e - 2Hg(l) +2Cl - =.268 V (saturated KCl) =.241 V saturated calomel electrode (S.C..)

31 x: One beaker contains a solution of.2 M KMnO 4,.5 M MnSO 4, and.5 M H 2 SO 4 ; and a second beaker contains.15 M FeSO 4 and.15 M Fe 2 (SO 4 ). The 2 beakers are connected by a salt bridge and Pt electrodes are placed one in each. The electrodes are connected via a wire with a voltmeter in between. What would be the potential of each half-cell (a) before reaction and (b) after reaction? What would be the measured cell voltage (c) at the start of the reaction and (d) after the reaction reaches eq.? Assume H 2 SO 4 to be completely ionized and equal volumes in each beaker.

32 Ans: 5Fe +2 + MnO H + 5Fe + + Mn H 2 O Pt Fe +2 (.15 M), Fe + (. M) MnO 4- (.2 M), Mn +2 (.5 M), H + (1. M) Pt (a) Fe = o Fe(III)/Fe(II) (.59/1) log [Fe+2 ]/[Fe + ] = log (.15)/(.15 2) =.671 V Mn = o MnO4 - /Mn+2 (.59/5)log [Mn+2 ]/[MnO 4- ][H + ] 8 = /5 log (.5)/(.2)(1.) 8 = 1.52 V (b) At eq., Fe = Mn, 可以含鐵之半反應來看, 先找出平衡時兩個鐵離子的濃度, 得 Fe = log (.5)/(.1) =.79 V (c) cell = Mn - Fe = =.849 V (d) At eq., Fe = Mn, 所以 cell = V

Review: Balancing Redox Reactions. Review: Balancing Redox Reactions

Review: Balancing Redox Reactions Determine which species is oxidized and which species is reduced Oxidation corresponds to an increase in the oxidation number of an element Reduction corresponds to a