Redox Equilibria ( , , ) 1. Defining standard electrode poten;al and simple galvanic cells 2. Difference between

Transcription

1 Redox Equilibria ( , , ) 1. Defining standard electrode poten;al and simple galvanic cells 2. Difference between galvanic cell and electroly;c cell and predic;ng feasibility of redox reac;ons 3. Advanced redox cells, rechargeable cells and fuel cells 4. Prac;cal: Redox circus, Galvanic cells, Redox ;tra;on

2 17 Nov 2014

3 The more nega;ve the E value, the stronger reducing agent (Li) the more the equilibrium lies to the leq. Li + + e Li E = 3.03V The more posi;ve the E value, the stronger oxidising agent (Au 3+ ) the more the equilibrium lies to the right Au e Au E = +3.03V

4 Standard Hydrogen Electrode (SHE) Pt (s) H 2(g) (1 atm) H + (aq)(1 M) 2H + (aq) + 2e H 2 (g) The SHE is the electrode which all other half cells are compared against and is therefore arbitrarily assigned 0 V. In the SHE, a pla;num wire acts as an electrode in an acidic electrolyte, where [H + ] = 1 mol dm 3. Hydrogen gas at the pressure of 1 atm (10 5 Pa) is bubbled through the solu;on at a temperature of 25 C (298K). ph of SHE solu;on is 0 (ph=- log[1]=0)

5 Standard Hydrogen Electrode (SHE) Pt (s) H 2(g) (1 atm) H + (aq) (1M) or, Pt H 2 H + 2H + (aq) + 2e H 2 (g)

6 The standard electrode poten;al of a metal / metal ion combina;on is the emf measured when that metal / metal ion electrode is coupled to a hydrogen electrode under standard condi;ons. Pt H 2 H + Mg 2+ Mg

7 H 2 has a greater tendency to form H + than Cu to form Cu 2+. Copper Pt H 2 H + Cu 2+ Cu Copper Sulfate CuSO 4 (aq) 1 mol dm - 3 Salt bridge (filer paper contains KNO 3 ) has mobile ions allowing transfer of charge between each half- cell

8 Iden;fy the: strongest oxidising agent, strongest reducing agent Iden;fy the: weakest oxidising agent, weakest reducing agent

9 Oxidising agents (accept electrons) Reducing agents (donate electrons) strongest oxidising agent is O 2, strongest reducing agent is Cd weakest oxidising agent is Cd(OH) 2, weakest reducing agent H 2 O

10 A series of reduc;on half- equa;ons and their standard electrode poten;als wrifen with respect to the standard hydrogen electrode (SHE). Sn 2+ (aq) + 2e - Sn(s) E o = V O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) E=+0.40V

11 Redox cell (template) Zn 2+ /Zn 2H + /H 2 Cu 2+ /Cu 0.76 V 0 V V

12 Oxida;on Anode Redox cell (template) direc;on of electron flow Reduc;on Cathode Zn 2+ /Zn 2H + /H 2 Cu 2+ /Cu 0.76 V 0 V V

13 Oxida;on Anode Redox cell (template) direc;on of electron flow Reduc;on Cathode Zn 2+ /Zn 2H + /H 2 Cu 2+ /Cu 0.76 V 0 V V Znà Zn e Cu e à Cu

14 Oxida;on Anode Redox cell (template) direc;on of electron flow Reduc;on Cathode Zn 2+ /Zn 2H + /H 2 Cu 2+ /Cu 0.76 V 0 V V Znà Zn e Cu e à Cu Zn + Cu 2+ à Zn 2+ + Cu E =+0.34( 0.76)=+1.10V

15 E cell = E (cathode) E (anode) E cell = ( 0.76) = V

16 Oxida;on Anode Reduc;on Cathode Left half-cell (Oxidation, Anode) Right half-cell (Reduction, Cathode) Conventional representation of the cell EMF / V E right - E left Mg 2+ Mg (-2.37) Zn 2+ Zn (-0.76) Mg 2+ Mg (-2.37) Cu 2+ Cu (+0.34) Zn 2+ Zn (-0.76) Cu 2+ Cu (+0.34)

17 Oxida;on Anode Reduc;on Cathode Left half-cell (Oxidation, Anode) Right half-cell (Reduction, Cathode) Conventional representation of the cell EMF / V Mg 2+ Mg (-2.37) Zn 2+ Zn (-0.76) Mg Mg 2+ Zn 2+ Zn Mg 2+ Mg (-2.37) Cu 2+ Cu (+0.34) Mg Mg 2+ Cu 2+ Cu Zn 2+ Zn (-0.76) Cu 2+ Cu (+0.34) Zn Zn 2+ Cu 2+ Cu E right - E left

18 18 Nov 2014

19 Draw the diagram to show how you would measure E for Zn And write down the overall equaron for this cell (8 marks)

20 Draw the diagram to show how you would measure E for Zn And write down the overall equaron for this cell (8 marks) Pt H 2 H + Zn 2+ Zn

21 Draw a labelled diagram of the apparatus that could be connected to a standard hydrogen electrode in order to measure the standard electrode poten;al of the Fe 3+ / Fe 2+ electrode. State the condi;ons under which this cell should be operated. Given the Fe 3+ /Fe 2+ E value is +0.77V, determine the overall cell equa;on

22 Draw a labelled diagram of the apparatus that could be connected to a standard hydrogen electrode in order to measure the standard electrode poten;al of the Fe 3+ / Fe 2+ electrode. State the condi;ons under which this cell should be operated. Given the Fe 3+ /Fe 2+ E value is 0.77V, determine the overall cell equa;on 298K, 100kPa (1 bar) Reference electrode is anode(+) Pla;num electrode Pt H 2 H + Fe 3+, Fe 2+ Pt E =+0.77V KNO 3 salt bridge 2Fe 3+ + H 2 à 2Fe 2+ +2H mol dm - 3 H 2 SO 4 [H + ]=1 mol dm mol dm - 3 Fe 2 (SO 4 ) 3 (aq), [Fe 3+ (aq)]=1 mol dm mol dm - 3 FeSO 4 (aq), [Fe 2+ (aq)]=1 mol dm - 3

23 Oxida;on Anode Reduc;on Cathode Left half-cell (Oxidation, Anode) Right half-cell (Reduction, Cathode) Conventional representation of the cell EMF / V Mg 2+ Mg (-2.37) Zn 2+ Zn (-0.76) Mg Mg 2+ Zn 2+ Zn Mg 2+ Mg (-2.37) Cu 2+ Cu (+0.34) Mg Mg 2+ Cu 2+ Cu Zn 2+ Zn (-0.76) Cu 2+ Cu (+0.34) Zn Zn 2+ Cu 2+ Cu E right - E left

24 Oxida;on Anode Reduc;on Cathode Left half-cell (Oxidation, Anode) Right half-cell (Reduction, Cathode) Conventional representation of the cell EMF / V E right - E left Mg 2+ Mg (-2.37) Zn 2+ Zn (-0.76) Mg 2+ Mg (-2.37) Cu 2+ Cu (+0.34) Zn 2+ Zn (-0.76) Cu 2+ Cu (+0.34)

25 The half- equa;ons for two electrodes used to make an electrochemical cell are shown below. ClO 3 (aq) + 6H + (aq) + 6e à Cl (aq) + 3H 2 O(I) E = V SO 2 4 (aq) + 2H + (aq) + 2e à SO 2 3 (aq) + H 2 O(I) E = V Write the conven;onal representa;on for the cell using pla;num contacts. Write an overall equa;on for the cell reac;on and iden;fy the oxidising and reducing agents

26 The half- equa;ons for two electrodes used to make an electrochemical cell are shown below. ClO 3 (aq) + 6H + (aq) + 6e à Cl (aq) + 3H 2 O(I) E = V SO 2 4 (aq) + 2H + (aq) + 2e à SO 2 3 (aq) + H 2 O(I) E = V Write the conven;onal representa;on for the cell using pla;num contacts. Pt SO 3 2 (aq), SO 4 2 (aq) ClO 3 (aq), Cl (aq) Pt Write an overall equa;on for the cell reac;on and iden;fy the oxidising and reducing agents ClO 3 (aq) + 3SO 3 2 (aq) + à Cl (aq) + 3SO 4 2 (aq) oxidising agent ClO 3 reducing agent SO 3 2

27 Deduce the equa;on for the overall cell reac;on of a cell that has an e.m.f. of 0.78 V. Give the conven;onal cell representa;on for this cell. Iden;fy the posi;ve electrode

28 Cu 2+ + Fe à Cu + Fe 2+ Anode Cathode Fe Fe 2+ Cu 2+ Cu E cell E cell = E (cathode) E (anode) = ( 0.44) = V Reduc;on occurs at Cathode (+) (copper electrode)

29 Use data from the table to explain why Au + ions are not normally found in aqueous solu;on. Write an equa;on to show how Au + ions would react with water.

30 Spontaneous reac;on occurs where Au + is reduced by H 2 O to Au: Anode Cathode H 2 O ½ O 2 Au + Au with E cell= (+1.23)= V 2Au + + H 2 O 2Au + ½ O 2 + 2H +

31 Select the species that is the most powerful reducing agent State the colours of Cr 2 O 7 2- Cr 3+ MnO 4 - [Fe 2+ (aq)] can be determined by ;tra;ng the solu;on, with acidified KMNO 4 (aq). (i) Explain, why HCl should not be used to acidify KMNO 4 (aq). (ii) Explain why HNO 3 should not be used to acidify KMNO 4 (aq). Calculate the e.m.f. of the cell: Pt Mn 2+, MnO 4- S 2 O 8 2-, SO 4 2- Pt Deduce the equa;on when excess S 2 O 8 2- is added to Mn 2+ (aq) ions.

32 Select the species that is the most powerful reducing agent Fe 2+ State the colours of Cr 2 O 2-7 orange Cr 3+ green MnO - 4 purple [Fe 2+ (aq)] can be determined by ;tra;ng the solu;on, with acidified KMNO 4 (aq). (i) Explain, why HCl should not be used to acidify KMNO 4 (aq). (ii) Explain why HNO 3 should not be used to acidify KMNO 4 (aq). Calculate the e.m.f. of the cell: Pt Mn 2+, MnO 4- S 2 O 2-8, SO 2-4 Pt MnO 4 - will oxidise Cl - Fe 2+ will reduce NO 3-0.5V Deduce the equa;on when excess S 2 O 8 2- is added to Mn 2+ (aq) ions. 2Mn H 2 O + 5S 2 O SO MnO H +

33 In the external circuit of this cell, the electrons flow through the ammeter from right to leq. Suggest why the electrons move in this direc;on. Explain why the current in the external circuit of this cell falls to zero aqer the cell has operated for some ;me.

34 In the external circuit of this cell, the electrons flow through the ammeter from right to leq. Suggest why the electrons move in this direc;on. Cu e - Cu Decreased Cu 2+, shiqs equilibrium to the leq, making this electrode more electronega;ve and anode rela;ve to opposite electrode. Electrons flow from anode to cathode. [Nernst equaron] Explain why the current in the external circuit of this cell falls to zero aqer the cell has operated for some ;me. Eventually, Cu 2+ (aq) ions in each electrode will equalise, so no net difference in half- cell electrode poten;als

35 Discharging vs. Re- charging redox cells

36 ELECTROLYTIC CELL The power source drives the electrons from anode (+) to cathode (- ) Anode (oxidaron) Cl 2 + 2e - à 2Cl - E = Cl - à Cl 2 + 2e - E = 1.36 Cathode (reducron) Na + + e - à Na E = Na + + 2e - à 2Na E = 2.71 Overall: 2Na + + 2Cl - à 2Na + Cl 2 E = E cathode E anode = 2.71 (+1.36)= 4.07V OR, simply directly add half- cell E as shown in overall equa;on: E = ( 1.36)= 4.07V Nega;ve E for cell indicates voltage needs to be applied to induce the overall cell reac;on

37 Galvanic cell (produces EMF) Spontaneous chemical reac;on, produces electrical energy LeQ electrode is anode (oxida;on occurs), Anode is ( ) Right electrode is cathode (reduc;on occurs), Cathode is (+) Electrolysis/Recharging cell (apply EMF) Uses electrical energy and converts this to chemical energy in a non- spontaneous reac;on LeQ electrode becomes cathode (reduc;on occurs), Cathode is ( ) Right electrode becomes anode (oxida;on occurs), Anode is (+) Electrons always flow from Anode to Cathode, regardless of discharging or re- charging cell status

38 Spontaneous discharging cell Recharging cell H H Cu 2+ /Cu Ag + /Ag +0.79

39 +0.74 V 0.74 V Galvanic cell ElectrolyRc/Recharging cell

40 A rechargeable nickel cadmium cell can also be constructed. Write an equa;on for the overall reac;on that occurs when the cell is: discharging: and being recharged: An ethanol oxygen fuel cell may be also be constructed. Deduce a half- equa;on for the reac;on at the ethanol electrode. In this half- equa;on, ethanol reacts with water to form carbon dioxide and hydrogen ions. The e.m.f. of an ethanol oxygen fuel cell is 1.00 V. Calculate a value for the electrode poten;al of the ethanol electrode.

41 A rechargeable nickel cadmium cell can also be constructed. Write an equa;on for the overall reac;on that occurs when the cell is: discharging: 2NiO(OH) + 2H 2 O + Cd à 2Ni(OH) 2 + Cd(OH) 2 E cell= +1.4 V and being recharged: 2Ni(OH) 2 + Cd(OH) 2 à 2NiO(OH) + 2H 2 O + Cd E cell= V An ethanol oxygen fuel cell may be also be constructed. Deduce a half- equa;on for the oxida;on reac;on occurring at the ethanol electrode. In this half- equa;on, ethanol reacts with water to form carbon dioxide and hydrogen ions. The EMF of an ethanol oxygen fuel cell is 1.00 V. Calculate a value for the electrode poten;al of the CO 2 /ethanol electrode. C 2 H 5 OH + 3H 2 O à 2CO H e - 1.0= E CO 2 /ethanol E CO 2 /ethanol =+0.23V

42 Calculate the e.m.f. of this cell. Write an equa;on for the overall reac;on that occurs when the cell discharges Deduce one essen;al property of the non- reac;ve porous separator Suggest the func;on of the carbon rod in the cell. The zinc electrode acts as a container for the cell and is protected from external damage. Suggest why a cell oqen leaks aqer being used for a long ;me.

43 Calculate the e.m.f. of this cell (- 0.76)=+1.5V Write an equa;on for the overall reac;on that occurs when the cell discharges 2MnO 2 + 2H 2 O + Zn 2MnO(OH) + 2OH + Zn 2+ Deduce one essen;al property of the non- reac;ve porous separator Allows ions to pass Suggest the func;on of the carbon rod in the cell. electrode conductor The zinc electrode acts as a container for the cell and is protected from external damage. Suggest why a cell oqen leaks aqer being used for a long ;me. Zn coa;ng is used up by redox reac;on

44 Rechargeable cells (lead- acid accumulator) Eye protec;on (chemical splash- proof) One 150 cm3 beaker 2 4 V low voltage DC supply 1.25 V lamp in holder Two crocodile clips Two connec;ng leads Two 8 cm lead foil electrodes Stopclock White ;le 100 cm 3 of 0.5 mol dm 3 sulfuric acid: irritant and corrosive

45 The PbO 2 /PbSO 4 electrode is the posi;ve terminal of the cell and the e.m.f. of the cell is 2.15 V. Calculate the missing electrode poten;al. Write an equa;on for the overall reac;on that occurs when the cell is being discharged and recharged. Give one reason why the e.m.f. of the lead acid cell changes aqer several hours. Explain why the voltage remains constant in cell X.

46 The PbO 2 /PbSO 4 electrode is the posi;ve terminal of the cell and the e.m.f. of the cell is 2.15 V. Calculate the missing electrode poten;al. E cell= E cathode- E anode 2.15= E anode E anode =- 0.46V Discharged: PbO 2 + 2H + + 2HSO Pb 2PbSO 4 + 2H 2 O Recharged:2PbSO 4 + 2H 2 O PbO 2 + 2H + + 2HSO Pb Write an equa;on for the overall reac;on that occurs when the cell is being discharged and recharged. Give one reason why the e.m.f. of the lead acid cell changes aqer several hours. reagents used up. Explain why the voltage remains constant in cell X. concentra;on of reagents remain constant (fuel cell)

47 Hydrogen oxygen fuel cells can operate in acidic or in alkaline condi;ons. Commercial cells use porous pla;num electrodes in contact with concentrated KOH(aq). Write the conven;onal representa;on for an alkaline hydrogen oxygen fuel cell, calculate its EMF and final equa;on. Suggest why the EMF of H 2 - O 2 fuel cell, opera;ng in acidic condi;ons, is iden;cal to an H 2 - O 2 alkaline fuel cell. Overall reac;on is the same, so same EMF i.e. 2H 2 +O 2 à 2H 2 O +1.23V What is the main advantage of a fuel cell over a rechargeable cell for propelling a vehicle Hydrogen and oxygen supplied con;nuously, can be operated without stopping to recharge

48 Hydrogen oxygen fuel cells can operate in acidic or in alkaline condi;ons. Commercial cells use porous pla;num electrodes in contact with concentrated KOH(aq). Write the conven;onal representa;on for an alkaline hydrogen oxygen fuel cell, calculate its EMF and final equa;on. Pt(s) H 2 (g) OH - (aq), H 2 O(l) O 2 (g) H 2 O(l), OH - (aq) Pt(s) Anode: H 2 + 2OH - à 2e - + 2H 2 O Cathode: O 2 + 2H 2 O + 4e - à 4OH Overall: 2H 2 +O 2 à 2H 2 O +1.23V Suggest why the EMF of H 2 - O 2 fuel cell, opera;ng in acidic condi;ons, is iden;cal to an H 2 - O 2 alkaline fuel cell. Overall reac;on is the same, so same EMF i.e. 2H 2 +O 2 à 2H 2 O +1.23V What is the main advantage of a fuel cell over a rechargeable cell for propelling a vehicle Hydrogen and oxygen supplied con;nuously, can be operated without stopping to recharge

49 Hydrogen fuel cells Hydrogen fuel cells are the best alternative to petrol engine cars No pollution from the engine, the only product is water. High efficiency. Hydrogen can be made from renewable resources such as ethanol. Lightweight. Lower tax / no tax. Hydrogen fuel cells are not the best alternative to petrol engine cars Current vehicles would need a new engine. Hydrogen fuel pumps would need to be rolled out across the country. Hydrogen is explosive and this has handling and storage safety issues. It is still being researched and more issues may arise. Drivers are sceptical. As it is an emergent technology, it is unproven and the vehicles may be difficult / expensive to fix.

50 The e.m.f. of this cell is 2.90 V. Use this informa;on to calculate a value for the electrode poten;al of electrode B. Write an equa;on for the overall reac;on that occurs when this lithium cell is being i) discharged ii) re- charged.

51 E cell= E cathode E anode 2.90= 0.15 E anode E anode = 3.05V The e.m.f. of this cell is 2.90 V. Use this informa;on to calculate a value for the electrode poten;al of electrode B. Write an equa;on for the overall reac;on that occurs when this lithium cell is being i) discharged ii) re- charged. discharged: Li + MnO 2 à LiMnO 2 re- charged: LiMnO 2 à Li + MnO 2 E +2.90V E 2.90V

52

53 E cell= E cathode E anode = 0.44 ( 2.37)=+1.93V Mgà Mg 2+ + e Equilibrium shifs to lef so Mg 2+ /Mg is more electronegarve, so EMF increases E cell= E cathode E anode = 0.44 (+0.4)= 0.84V ; Hence, reverse reacron is spontaneous, so Fe is oxidised by H 2 O to Fe 2+

54 Excess KMNO 4 was added to V 2+ (aq). Determine the vanadium species present in the solu;on at the end of this reac;on, its oxida;on number, and write a half- equa;on for its forma;on from V 2+ (aq). In cell: Pt H 2 SO 3,SO 4 2,H + Fe 3+,Fe 2+ Pt Calculate the EMF. Explain how EMF will change if the concentra;on of Fe 3+ (aq) is increased. Deduce the overall equa;on for this cell reac;on. State how the EMF of this cell will change if the surface area of each pla;num electrode is doubled

55 VO 2 + ; oxidaron number of V is +5 V H 2 Oà VO H + + 3e - Excess KMNO 4 was added to V 2+ (aq). Determine the vanadium species present in the solu;on at the end of this reac;on, its oxida;on number, and write a half- equa;on for its forma;on from V 2+ (aq). In cell: Pt H 2 SO 3,SO 4 2,H + Fe 3+,Fe 2+ Pt Calculate the EMF. Explain how EMF will change if the concentra;on of Fe 3+ (aq) is increased. 2H 2 à 4H + +4e - O 2 +2H 2 O+4e - à 4OH - overall 2H 2 +O 2 à 2H 2 O ; Unchanged by equivalent surface area E cell= E cathode E anode = (0.17)= V Fe 3+ + e - Fe 2+ equilibrium shifs to right, more electroposirve Fe 3+ / Fe 2+, thus EMF increases Deduce the overall equa;on for this cell reac;on. State how the EMF of this cell will change if the surface area of each pla;num electrode is doubled