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A. Redox Reactions – Crash course videoCrash course video  electrochemistry is the branch of chemistry that studies Electrochemistry electron transfer.

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Presentation on theme: "A. Redox Reactions – Crash course videoCrash course video  electrochemistry is the branch of chemistry that studies Electrochemistry electron transfer."— Presentation transcript:

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2 A. Redox Reactions – Crash course videoCrash course video  electrochemistry is the branch of chemistry that studies Electrochemistry electron transfer in chemical reactions  is aOXIDATIONloss of electrons “OIL” eg) Mg (s)  Mg 2+ (aq) + 2e  2Cl  (aq)  Cl 2(g) + 2e  0 2+ Electrons are in the product = losing electrons Charge of ion or element (if charge increase, from its reactant, it’s a oxidation reaction 0 Charge of ion or element (if charge increase, from its reactant, it’s a oxidation reaction Electrons are in the product = losing electrons

3  is areductiongain of electrons “RIG” eg) Fe 3+ (aq) + 3e   Fe (s) Br 2(l) + 2e   2Br  (aq)  oxidation and reduction reactions occur together, hence the term redox  the reduction and oxidation reactions are called the  “adding” the half reactions together will give you the that takes place during the redox reaction half reactions net ionic equation Electrons are in the reactant = gaining electrons 3+ 0 Charge of ion or element (if charge decreases, from its reactant, it’s a reduction reaction

4  the e  lost in the oxidation half reaction the e  gained in the reduction half reaction must equal  you may have to of the half reactions to balance the e   (ions not changing) are included! multiply one or both spectator ionsNOT  the substance that is is called the ( ) (it causes the oxidation by taking e - )  the substance that is is called the ( ) (it causes the reduction by giving up e - ) reduced oxidizing agent OA oxidized reducing agentRA Video – REDOX reaction

5 Example 1 Given the following reaction, write the half reactions and the net ionic equation. Na(s) + LiCl(aq)  Li(s) + NaCl(aq) 01+1–1–0 1–1– oxredCl - is spectator Ox: Red: Net: Li + (aq) + 1e -  Li (s) Na (s)  Na + (aq) + 1e - Li + (aq) + Na (s)  Li (s) + Na + (aq) Step 1: Step 1: identify the charges for each element or ion. Step 2: Step 2: identify your oxidation (increase in number) and reduction (decrease in number) half reactions. Also identify your spectator ion. Step 3 Step 3: write your half reactions (charges must be balanced with same number of electrons. Step 4: Step 4: Ensure you have the same number of electrons on both sides (reactants and products) Step 5: Step 5: Write the net ionic equation.

6 Example 2 Given the following reaction, write the half reactions and the net ionic equation. 3 Zn(s) + 2 Au(NO 3 ) 3 (aq)  2 Au(s) + 3 Zn(NO 3 ) 2 (aq) 03+1–1–0 2+ 1–1– oxredNO 3 - is spectator Ox: Red: Net: Au 3+ (aq) + 3e -  Au (s) Zn (s)  Zn 2+ (aq) + 2e - 2 Au 3+ (aq) + 3 Zn (s)  2 Au (s) + 3 Zn 2+ (aq) [ ] 3 2 Step 1: Step 1: identify the charges for each element or ion. Step 2: Step 2: identify your oxidation (increase in number) and reduction (decrease in number) half reactions. Also identify your spectator ion. Step 3 Step 3: write your half reactions (charges must be balanced with same number of electrons. Step 4: Step 4: Ensure you have the same number of electrons on both sides (reactants and products) Step 5: Step 5: Write the net ionic equation.

7 Practice Questions

8 Read pages 558 – 564 Hand in introduction to redox lab Step 1: Step 1: identify the charges for each element or ion. Step 2: Step 2: identify your oxidation (increase in number) and reduction (decrease in number) half reactions. Also identify your spectator ion. Step 3 Step 3: write your half reactions (charges must be balanced with same number of electrons. Step 4: Step 4: Ensure you have the same number of electrons on both sides (reactants and products) Step 5: Step 5: Write the net ionic equation. Answer: C

9 B. Spontaneous Redox Reactions  chemical reactions which occur on their own, without the input of, are called  not all reactions are spontaneous additional energy spontaneous  in the table of redox half reactions (see pg 7 in Data Booklet), the is at the top left and the is at the bottom right strongest oxidizing agent (SOA) strongest reducing agent (SRA)

10  the rule states that a spontaneous reaction occurs if the agent is above the agent in the table of redox half reactions redox spontaneity oxidizing reducing

11 Try These: For each of the following combinations of substances, state whether the reaction would be spontaneous or non-spontaneous: Cr 3+ (aq) with Ag (s) I 2(s) with K (s) H 2 O 2(l) with Au 3+ (aq) Sn 2+ (aq) with Cu (s) Fe 2+ (aq) with H 2 O (l) non-spontaneous (both ways) spontaneous

12 Practice Question

13 C. Predicting Redox Reactions  we will be predicting the strongest or most dominating reaction that occurs when substances are mixed (other reactions do take place because of atomic collisions!) Steps are found at the bottom of examples. Answer: A

14 Example 1 Predict the most likely redox reaction when chromium is placed into aqueous zinc sulphate. SOA (Red): SRA (Ox): Net: Zn 2+ (aq) + 2e -  Zn(s) Cr(s)  Cr 2+ (aq) + 2e - Zn 2+ (aq) + Cr(s)  Zn(s) + Cr 2+ (aq) Cr(s) Zn 2+ (aq)SO 4 2- (aq) H 2 O(l) RAOAOA with H 2 O (l)OA/RASS spont

15 Example 2 Predict the most likely redox reaction when silver is placed into aqueous cadmium nitrate. SOA (Red): SRA (Ox): Net: Cd 2+ (aq) + 2e -  Cd(s) Ag(s)  Ag + (aq) + e - Cd 2+ (aq) + 2 Ag(s)  Cd(s) + 2 Ag + (aq) Ag(s) Cd 2+ (aq)NO 3 - (aq) H 2 O(l) RAOAOA with H + (aq) OA/RASS [ ]2 nonspont

16 Example 3 Predict the most likely redox reaction when potassium permanganate is slowly poured into an acidic iron (II) sulphate solution. SOA (Red): SRA (Ox): Net: MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O(l) Fe 2+ (aq)  Fe 3+ (aq) + e - MnO 4 - (aq) +8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 4H 2 O(l) + 5 Fe 3+ (aq) K + (aq) H + (aq)Fe 2+ (aq) H 2 O(l) OAOA with H + (aq) OA with H + (aq), H 2 O (l) OA/RA S S [ ]5 MnO 4 - (aq)SO 4 2- (aq) OAOA/ RA spont

17 D. Generating Redox Tables  you can be given data for certain ions and elements then be asked to generate a redox table like the one on pg 7 of you Data Booklet (a smaller version!)  you may have to generate a table from real or fictional elements and ions  the tables that we use are all written as half reactions reduction

18 Example 1 Generate a redox table given the following data (useful when all reactions are given: Cu 2+ (aq) Zn 2+ (aq) Pb 2+ (aq) Ag + (aq) Cu(s)   Zn(s)    Pb(s)    Ag(s)   indicates no reaction  indicates a reaction Video: REDOX tables

19 Redox Table Ag + (aq) + e -  Ag(s) Cu 2+ (aq) + 2e -  Cu(s) Pb 2+ (aq) + 2e -  Pb(s) Zn 2+ (aq) + 2e -  Zn(s) SOA SRA Put the oxidizing agents in order from strongest to weakest. Put the reducing agents in order from strongest to weakest. Ag + (aq), Cu 2+ (aq),Pb 2+ (aq),Zn 2+ (aq) Zn(s),Pb(s),Cu(s),Ag(s)

20 Redox Table + e -  Ag(s) Cu 2+ (aq) + 2e -  Hg 2+ (aq) + 2e -  Zn 2+ (aq) + 2e -  SOA SRA Example 2: Generate a redox table given the following data: Cu(s) + Ag + (aq)  Cu 2+ (aq) + Ag(s) Zn 2+ (aq) + Ag(s)  no reaction Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Hg(l) + Ag + (aq)  no reaction Ag + (aq) Cu(s) Zn(s) Hg(l) Label each as OA or Ra

21 Example 2 (continued): Put the oxidizing agents in order from weakest to strongest. Put the reducing agents in order from weakest to strongest. Zn 2+ (aq), Cu 2+ (aq),Ag + (aq),Hg 2+ (aq) Hg(l),Ag(s),Cu(s),Zn(s)

22 Redox Table Z 2 (g) + 2e -  + 2e -  2Y - (aq) + 2e -  2W - (aq) X 2 (g) + 2e -  SOA SRA Example 3: Generate a redox table given the following data: 2X - (aq) + Y 2 (g)  spontaneous reaction 2Z-(aq) + Y 2 (g)  no reaction 2Z - (aq) + W 2 (g)  spontaneous reaction Label each as OA or RA Y 2 (g) 2X - (aq) 2Z - (aq) W 2 (g)

23 Example 3 (continued): Put the oxidizing agents in order from strongest to weakest. Put the reducing agents in order from strongest to weakest. W 2 (g), X - (aq), Z 2 (g),Y 2 (g),X 2 (g) Y - (aq),Z - (aq),W - (aq)

24 Practice Question

25 E. Oxidation Numbers (States)  an is the charge an atom to have when found in a or charged  can be used when you have a where there are no to determine if oxidation or reduction is occurring  how do you use a change in the number? oxidation number appears molecular compound ion charges 1. if the number then has occurred decreasesreduction 2. if the number then has occurred increasesoxidation neutral molecule polyatomic ion Answer: A

26 Rules for Assigning Oxidation Numbers: 1. In a pure element, the oxidation number is. 2. In simple ions, the oxidation number is equal to the. zero ion charge 3. In most compounds containing hydrogen, the oxidation number for hydrogen is. (Exception is the metal hydrides eg) LiH where the oxidation number of hydrogen is ). +1 –1–1

27 5. The sum of oxidation numbers of all atoms in a substance must equal the of the substance. ( for compounds and of the polyatomic ion) eg) sum of MgO = sum of PO 4 3- = 4. In most compounds containing oxygen, the oxidation number for oxygen is. (Exception is the peroxides eg) H 2 O 2, Na 2 O 2 where the oxidation number of oxygen is ) –1–1 –2–2 net charge Zerothe charge 0–3–3

28 Common Oxidation Numbers Atom or IonOxidation NumberExamples All atoms in elements0Na is 0 Cl in Cl 2 is 0 Hydrogen in all compounds (except: hydrogen in hydrides) +1 H in HCL is +1 H in LiH is -1 Oxygen in all compounds, Except oxygen in peroxides -2 O in H 2 O is -2 O in H 2 O 2 is -1 All monoatomic ionsCharge of IonNa + is 1+ S 2- is 2-

29 Example What is the oxidation number (state) for the element identified in each of the following substances: a) N in N 2 O N2N2 O individual oxidation numbers sum of oxidation numbers –2 = b) N in NO 3 - NO3–O3– –2 –6 = –1+5

30 c) C in C 2 H 5 OH –2+1 = 0–4 –2 d) C in C 6 H 12 O 6 –2 –12 = 00 0 C6C6 H 12 O6O6 C2C2 H5H5 OH +1 –

31  figuring out oxidation numbers can help to identify whether a reaction is a or not  for it to be a redox reaction, there has to be an in oxidation number and a in oxidation number seen in the reaction Ag(s) + NaNO 3 (aq)  Na(s) + AgNO 3 (aq) PbSO 4 (aq) + 2 KI(aq)  PbI 2 (s) + K 2 SO 4 (aq) eg) redox reaction Ag increases  oxidized Na decreases  reduced… redox!!! nothing changes  NOT a redox reaction! both increasedecrease

32 Practice Question

33 Practice question Answer: a

34  electron transfer occurs in eg) living systems photosynthesis, cellular respiration Answer: C

35  also occurs in eg) non-living systems combustion, bleaching, metallurgy

36 F. Disproportionation  disproportionation occurs when one element is both oxidized and reduced in a reaction eg) 2 H 2 O 2(aq)  2 H 2 O (l) + O 2(g) -20 Cl 2(g) + 2 OH - (aq)  ClO - (aq) + Cl - (aq) + H 2 O (l) 0+1 Do workbook questions: page 4

37 G. Balancing Redox Reactions  sometimes most reactants and products are known but the complete reaction is not given…called a reaction skeleton  There are two different ways you can balance redox reactions: either can be used so find one way that works best for you…  Half Reaction Method Half Reaction Method  Oxidation Number Method Oxidation Number Method

38 Example 1:Of half Reaction Method Balance the following half reaction : +6 22 +3 22 44=  88=  2 (Cr is already balanced) +3 e – + 2 H 2 O(l) +4 H + (aq) CrO 4 2- (aq)  CrO 2 - (aq) net charge = –1 (+6)(+3)

39 Example 2: Balance the following half reaction: +1 22 0 44 = 0 +6 e – +4 H 2 O(l) +6 H + (aq) HClO 2 (aq)  Cl 2 (g) net charge = 0 (+6)(0)

40 Practice Question

41 Example 1: Balance the following using oxidation numbers, assuming acidic conditions: +4 22 88=  2 +3 e – +2 H 2 O (l) +4 H + (aq) CrO 4 2- (aq)  CrO 2 - (aq) (+6)(+3) CrO 4 2- (aq) + SO 3 2- (aq)  CrO 2 - (aq) + SO 4 2- (aq) 22 66=  2 22 22 88 44=  1=  e – + H 2 O (l) +2 H + (aq) SO 3 2- (aq)  SO 4 2- (aq) (+4)(+6) Red Ox [ ] H + (aq) + 2 CrO 4 2- (aq) + 3 H 2 O (l) + 3 SO 3 2- (aq)  2 CrO 2 - (aq) + 4 H 2 O (l) + 3 SO 4 2- (aq) + 6 H + (aq) 2 H + (aq) + 2 CrO 4 2- (aq) + 3 SO 3 2- (aq)  2 CrO 2 - (aq) + H 2 O (l) + 3 SO 4 2- (aq) Net Answer: A

42 Example 1: of Oxidizing Number Method Balance the following reaction using the oxidation number method. __ H + (aq) + __MnO 4 - (aq) + __ SO 3 2- (aq)  __ MnO 2 (aq) + __ SO 4 2- (aq) + __ H 2 O(l)  =  3  1 atom =  = +2  1 atom = 33  2(Co) =  6e- +2  3(Co) =+6e-

43 Example 2: Balance the following reaction using the oxidation number method. __ H 2 O(l) + __N 2 O 4 (g) + _ Br  (aq)  _ NO 2  (aq) + _ BrO 3  (aq) + __ H + (aq)  =  1  2 atoms =  = +6  1 atom = 22  3(Co) =  6e- +6  1(Co) =  6e- 6

44 H. Redox Stoichiometry  stoichiometry can be used to predict or analyze a quantity of a chemical involved in a chemical reaction  in the past we have used balanced chemical equations to do stoich calculations 1. Calculations  we can now apply these same calculations to balanced redox equations

45 Example 1 What is the mass of zinc is produced when 100 g of chromium is placed into aqueous zinc sulphate. SOA (Red): SRA (Ox): Net: Zn 2+ (aq) + 2e -  Zn (s) Cr (s)  Cr 2+ (aq) + 2e - Cr (s) + Zn 2+ (aq)  Zn (s + Cr 2+ (aq) ) Cr (s) Zn 2+ (aq) SO 4 2- (aq) H 2 O (l) RAOAOA with H 2 O (l) OA/RA SRASOA

46 m = 100 g M = g/mol n = m M = 100 g g/mol = 1.923… mol Cr (s) + Zn 2+ (aq)  Zn (s) + Cr 2+ (aq) m = ? M = g/mol n = 1.923… mol x 1/1 = 1.923… mol m = nM = (1.923…mol)(65.39 g/mol) = g = 126 g

47 Example 2 What volume of 1.50 mol/L potassium permanganate is needed to react with 500 mL of 2.25 mol/L acidic iron (II) sulphate solution? SOA (Red): SRA (Ox): Net: MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O (l) Fe 2+ (aq)  Fe 3+ (aq) + e - MnO 4 - (aq) +8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 4H 2 O (l) + 5Fe 3+ (aq) K + (aq) MnO 4 - (aq) H + (aq) H 2 O (l) Fe 2+ (aq) SO 4 2- (aq) OA with H + (aq) OA/RA SRASOA OA OA/RA OA with H + (aq) OA with H 2 O (l) [ ] 5

48 v = ? c = 1.50 mol/L n = mol x 1/5 = mol v = n C v = mol 1.50 mol/L = L MnO 4 - (aq) +8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 4H 2 O (l) + 5Fe 3+ (aq) v = L c = 2.25 mol/L n = cv = (2.25 mol/L)(.500L) = mol *** Reminder: Use the formulas n=m/M and C = n/v to find quantities

49 Practice Question

50  a titration is a lab process used to determine the of a substance needed to react completely with another substance  this volume can then be used to calculate an unknown using stoichiometry 2. Titrations  one reagent ( - ) is slowly added to another ( - ) until an abrupt change ( ) occurs, usually in colour volume concentration titrantOA sample RA endpoint Answer:

51  in redox titrations, two common oxidizing agents are used because of their and : colourstrength permanganate ions (MnO 4 - (aq) ) – purple dichromate ions (Cr 2 O 7 2- (aq) ) – orange  as long as the sample (RA) in the flask is reacting with the the sample will be permanganate ions (dichromate ions) colourless (green)

52  when the reaction is complete, any unreacted permanganate ions will turn the sample (pink) (with dichromate, sample goes from orange to green) purple  the volume of titrant (OA) needed to reach the endpoint is called the equivalence point  the of the titrant must be accurately known concentration

53  the concentration of the permanganate solution must be calculated against a (a solution of known concentration) before it can be used in a titration itself  this is done just prior to the titration primary standard

54 Example Find the concentration of (standardize) the KMnO 4 (aq) solution by titrating mL of mol/L acidified tin (II) chloride with the KMnO 4 (aq). Trial1234 Final Volume (mL) Initial Volume (mL) Volume of (mL) Endpoint Colour pinklight pink KMnO4(aq)

55  endpoint average is calculated by using 3 volumes within 0.20 mL Endpoint average = mL mL mL 3 = mL Review: Redox reactionsRedox reactions

56 SOA (Red): SRA (Ox): Net: MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O (l) Sn 2+ (aq)  Sn 4+ (aq) + 2 e - 2MnO 4 - (aq) + 16H + (aq) + 5Sn 2+ (aq)  2Mn 2+ (aq) + 8H 2 O (l) + 5Sn 4+ (aq) K + (aq) MnO 4 - (aq) H + (aq) H 2 O (l) Sn 2+ (aq) Cl - (aq) OA with H + (aq) OA/RA SRASOA OA OA/RARA [ ] 5  determine net ionic redox equation Analysis: [ ]2

57  use net redox equation to calculate KMnO 4(aq) concentration 2MnO 4 - (aq) +16H + (aq) +5Sn 2+ (aq)  2Mn 2+ (aq) + 8H 2 O (l) +5Sn 4+ (aq) v = L C = ? n = mol x 2/5 = mol C = n v C = mol L = mol/L v = L c = mol/L n = cv = (0.500 mol/L)( L) = mol

58 I. Electrochemical Cells 1. Voltaic Cells - animationanimation  are devices that convert energy into energy  in redox reactions, e - are transferred from the to the  the transfer of e - can occur through a separating the two substances in containers called electric cellschemical electrical oxidized substancereduced substance conducting wire half cells

59  a is an arrangement where are joined so that the and can move between them  are made of good conducting materials so e - can flow…can be the of the solution or inert such as  the is a solution that contains ions and will transmit ions (charged particles) voltaic cell two half cellsions e-e- electrodes metal carbon electrolyte

60  the electrode where occurs is called the  if the anode is a metal, it mass as the cell operates  the anode is labelled as since it is the electrode where the electrons originate oxidation anode loses  the move to the since this electrode electrons (leaving a net charge in the electrode) negative anionsanode losespositive

61  the electrode where occurs is called the  if the cathode is a metal, it mass as the cell operates reduction cathode gains  the cathode is labelled as since the anode is labelled negative  the move to the since this electrode electrons (leaving a net charge in the electrode) positive cations cathode negative accepts

62  electrons flow from the to the through a connecting wire  ions must be able to to their attracting electrode (either through the or a ) otherwise a buildup of charge will occur opposing the movement of e -  the flow of ions through the solution and e - through the wire maintains overall anode (LEOA) cathode (GERC) move porous cup salt bridge electrical neutrality Cell song Repeated animation

63 Summary of Voltaic Cells Anode (An-ox)Cathode (Red Cat) -+ Is the SRAIs the SOA Is being OxidizedIs being Reduced “That Cat is a Son Of A Rascal”

64 Practice Question

65 2. Standard Reduction Potentials  are the ability of a half cell to  these potentials are measured using a  each half reaction listed in the Data Booklet has an E  value measured in assigned to it reduction potentials attract e - voltmeter volts  all values in the table are arbitrarily assigned based on a standard  the half reaction has been set as the standard and has an E  value of hydrogen cell 0.00 V Answer: 4,6,4,1

66 3. Predicting Voltage of a Voltaic Cell  the standard cell potential is determined by the for the two half reactions  the on the E  value for the half reaction must be  if you multiply an equation to balance e -, you multiply the E  value (voltage is independent of number of e - transferred) (E  net ) adding signoxidation reversed DO NOT  a E  net is a reaction positivespontaneous  a E  net is a reaction negativenonspontaneous E  values

67 Practice Question

68 Example Calculate the E  net for the reaction of Zn (s) with CuSO 4(aq). SOA (Red): SRA (Ox): Net: Cu 2+ (aq) + 2e -  Cu (s) Zn (s)  Zn 2+ (aq) + 2e - Zn (s) + Cu 2+ (aq)  Cu (s + Zn 2+ (aq) Zn (s) Cu 2+ (aq) SO 4 2- (aq) H 2 O (l) RAOAOA with H 2 O (l) OA/RASS E  = V E  = V E  net = V Answer: B

69 Practice Question

70 4. Shorthand Notation  line separates  double line represents the or and separates the two ORZn (s) / Zn 2+ (aq) // Cu 2+ (aq) / Cu (s) Zn (s) / Zn 2+ (aq) // Cr 2 O 7 2- (aq), H + (aq), Cr 3+ (aq) / C (s) (/)(/)phases (//) porous cup salt bridgehalf reactions  comma separates(,)(,) chemical species in the same phase ***anode // cathode Answer: 1.36V

71 5. Drawing Cells  when drawing a cell from the shorthand notation, you have to be able to label the cathode, anode, positive terminal, negative terminal, electrolytes, direction of e  flow, and directions of cation and anion flow  you also have to show and label the reduction half reaction, oxidation half reaction and net reaction including E  values, E  net and spontaneity

72 Example Draw and fully label the following electrochemical cell: Al (s) / Al 3+ (aq) // Ni 2+ (aq) / Ni (s) Ni (s) Al (s) Ni 2+ (electrolyte) Al 3+ (electrolyte) e-e- V anions cations cathode positive terminal anode negative terminal

73 SOA (Red): SRA (Ox): Net: Ni 2+ (aq) + 2e -  Ni (s) Al (s)  Al 3+ (aq) + 3e - 2 Al (s) + 3 Ni 2+ (aq)  3 Ni (s + 2 Al 3+ (aq) ) Al (s) Al 3+ (aq) Ni 2+ (aq) H 2 O (l) RAOA OA/RASS Ni (s) RA [ ] 3 2 E  = –0.26 V E  = V E  net = V spontaneous:yes

74 J. Commercial Cells  are made by connecting two or more voltaic cells in  the of the battery is the of the batteries series (one after the other) voltage individual cells sum  there are many types of batteries:

75 a) Dry Cell  common batteries of clocks, remote controls, noisy kids toys etc. Cathode (Red):2 MnO 2(s) + H 2 O (l) + 2e -  Mn 2 O 3(aq) + 2 OH - (aq) E  = V Anode (Oxid): Zn (s)  Zn 2+ (aq) + 2e - E  = V Net: 2 MnO 2(s) + H 2 O (l) + Zn (s)  Mn 2 O 3(aq) + 2 OH - (aq) + Zn 2+ (aq) E  net = V  the produced causes irreversible side reactions to occur making recharging impossible OH V and 9 V

76 b) Nickel-Cadmium  one type of battery Cat (Red): 2 NiO(OH) (s) + 2 H 2 O (l) + 2e -  2 Ni(OH) 2(s) +2 OH - (aq) E  = V An (Oxid): Cd (s) + 2 OH - (aq)  2 Cd(OH) 2(s) + 2e - E  = V Net: 2 NiO(OH) (s) + 2 H 2 O (l) + Cd (s)  2 Ni(OH) 2(s) + 2 Cd(OH) 2(s) E  net = V rechargeable

77 c) Lead Storage Battery  where serves as the anode, and serves as the cathode  both electrodes dip into an electrolyte solution of  are connected in series typical car batterylead lead coated with lead dioxide sulfuric acid six cells

78 Cat (Red): PbO 2(s) + HSO 4 - (aq) + 3 H + (aq) + 2e -  PbSO 4(s) + 2 H 2 O (l) E  = V An (Oxid):Pb (s) + HSO 4 - (aq)  PbSO 4(s) + H + (aq) + 2e - E  = V Net: Pb (s) + PbO 2(s) + 2 H + (aq) + 2 HSO 4 - (aq)  2 PbSO 4(s) + 2 H 2 O (l) E  net = V

79 d) Fuel Cells  cells where reactants are  the energy from this reaction can be used to  one type is the  is pumped in at the while is pumped in at the (which both have a lot of surface area) continuously supplied run machines hydrogen-oxygen fuel cell hydrogen gasanode oxygen gas cathode  pressure is used to push the H 2 through a platinum catalyst which splits the H 2 into 2H + and 2e -

80 Cathode (Red):O 2(g) + 4 H + (aq) + 4e -  2 H 2 O (l) E  = V Anode (Oxid):2 H 2(g)  4 H + (aq) + 4e - E  = V Net: O 2(g) + 2 H 2(g)  2 H 2 O (l) E  net = V  the 2e - move through an external circuit towards the cathode generating electrical energy  the O 2 is also pushed through the platinum catalyst forming two oxygen atoms  the H + ions and oxygen atoms combine to form water

81 Hydrogen-oxygen Fuel Cell

82  need a source of hydrogen…reformers are used to convert CH 4 or CH 3 OH into and  unfortunately, is a  about 24-32% efficient where gas-powered car is about 20% efficient H2H2 CO 2 greenhouse gas

83 K. Electrolytic Cells 1. The Basics  in an electrolytic cell, energy is used to force a chemical reaction to occur (opposite of a voltaic cell)  commonly used to (eg. gold, silver, bronze, chromium etc),  these reactions have a E  net nonspontaneous negative electroplate metals (eg. H 2, O 2, Cl 2 etc) recharge batteries, useful gases and split compounds into electrical

84  the electrolytic cell is hooked up to a (instead of load or external circuit) so the flow of e - is  the of the electrolytic cell is connected to the of the battery and therefore is battery or power supply “pushed” by an outside force cathode anode negative  the of the electrolytic cell is connected to the of the battery and therefore is anode cathode Positive Electrolysis song

85 Voltaic CellsElectrolytic Cells  chemical to electrical energy  electrical to chemical energy  usually contains porous cup or salt bridge  does not (usually) contain a porous cup or salt bridge  e – flow from anode to cathode  oxidation at anode  reduction at cathode  cations migrate to cathode  anions migrate to anode  E  net is positive (spont)  E  net is negative (nonspont)  has a voltmeter or external load  has a power supply  cathode + anode –  cathode – anode +

86 Practice Question

87  some processes are used in industry to produce gases, for example: 1.the for producing …aluminum oxide is electrolyzed using carbon electrodes …liquid aluminum is collected 2.a for producing …a saturated sodium chloride solution is electrolyzed …chlorine gas is formed and collected at the anodes ages/FG18_18.JPG&imgrefurl=http://wps.prenhall.com/wps/media/objects/602/616516/Chapter_18.html&h=756&w=1600&sz=183& tbnid=4cFJrFlK4noQXM:&tbnh=70&tbnw=150&hl=en&start=11&prev=/images%3Fq%3Dhall- heroult%2Bcell%26svnum%3D10%26hl%3Den%26lr%3Dprenhall Hall-Heroult cellAl chlor-alkali plant chlorine gas Answer: B

88 Example 1 An electric current is passed through a solution of nickel (II) nitrate using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. Cathode SOA(Red) : Anode SRA(Ox): Net: Ni 2+ (aq) + 2e -  Ni (s) 2 H 2 O (l)  O 2(g) + 4 H + (aq) + 4e - 2 Ni 2+ (aq) + 2 H 2 O (l)  2 Ni (s) + O 2(g) + 4 H + (aq) Ni 2+ (aq) NO 3 - (aq) H 2 O (l) OAOA with H + (aq) OA/RA SRASOA E  = V E  = V E  net = V [ ]2

89 Example 2 An electric current is passed through a solution of potassium iodide using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. Cathode SOA(Red) : Anode SRA(Ox): Net: 2 H 2 O (l) +2 e -  H 2(g) + 2 OH - (aq) 2 I - (aq)  I 2(s) + 2e - 2 H 2 O (l) + 2 I - (aq)  H 2(g) + 2 OH - (aq) + I 2(s) K + (aq) I - (aq) H 2 O (l) OARAOA/RA SRASOA E  = V E  = V E  net = V

90 Example 3 An electric current is passed through a solution of copper(II) sulphate using a carbon electrode and a metal electrode. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. Cathode SOA(Red) : Anode SRA(Ox): Net: Cu 2+ (aq) + 2 e -  Cu (s) 2 H 2 O (l) + 2 Cu 2+ (aq)  2 Cu (s) + O 2(g) + 4 H + (aq) Cu 2+ (aq) SO 4 2- (aq) H 2 O (l) OAOA/RA SRASOA E  = V E  net = V 2 H 2 O (l)  O 2(g) + 4 H + (aq) + 4e - E  = V [ ] 2

91 *** Note: is an exception to our rules… chlorine (ions) when water and chlorine are competing as reducing agents, water is the stronger RA but is chosen because the transfer of e - from H 2 O to O 2 is more difficult …called overvoltage chloride ions

92 Example 4 An electric current is passed through a solution of sodium chloride. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. Cathode SOA(Red) : Anode SRA(Ox): Net: 2 H 2 O (l) + 2 Cl - (aq)  H 2(g) + Cl 2(g) + 2 OH - (aq) Na + (aq) Cl - (aq) H 2 O (l) OA RAOA/RA SRASOA E  net = V 2 Cl - (aq)  Cl 2(g) + 2e - E  = V 2 H 2 O (l) +2 e -  H 2(g) + 2 OH - (aq) E  = V

93 2. Quantitative Study of Electrolysis  quantitative analysis (stoich) provides information on necessary quantities, current and/or time for electrolytic reactions  one e - carries of charge  the unit for charge is the  this means that one of e - carry of charge  is called the (see Data Booklet pg 3) (q)Coulomb (C) 1.60 x C mole9.65 x 10 4 C 9.65 x 10 4 C/mol Faraday constant (F)

94 n e- = q F where:n e- = number of moles of electrons (mol) q = charge in Coulombs (C) F = Faraday constant = 9.65 x 10 4 C/mol q = It I = current in C/s or Amperes (A) t = time in seconds (s)

95 n e- = It F  the above equations can be combined into one equation:  we can use these equations in stoichiometric calculations for current, time, mass, moles of a substance and moles e -

96 Example 1 An electrochemical cell caused a mol of e - to flow through a wire. Calculate the charge. n e- = mol F = 9.65 x 10 4 C/mol n e- = q F mol = q 9.65 x 10 4 C/mol q = 6948 C = 6.95  10 3 C

97 Example 2 Determine the number of moles of electrons supplied by a dry cell supplying a current of A to a radio for 50.0 minutes. I = A (C/s) t = 50.0 min  60 s/min = 3000 s F = 9.65 x 10 4 C/mol n e- = It F = (0.100 C/s)(3000 s) 9.65 x 10 4 C/mol = mol

98 Example 3 If a 20.0 A current flows through an electrolytic cell containing molten aluminum oxide for 1.00 hours, what mass of Al (l) will be deposited at the cathode? n e- = It F = (20.0 A)(1.00 h  3600 s/h) 9.65 x 10 4 C/mol = 0.746…mol n = 0.746…mol  1/3 = 0.248…mol M = g/mol m = nM = (0.248…mol)(26.98g/mol) = 6.71 g Al 3+ (l) + 3 e -  Al (l)

99 Practice Question

100 L. Rust and Corrosion  the metal is oxidized causing the loss of  corrosion can be viewed as the process of returning metals to their natural state (ore) structural integrity Answer: D

101  commonly, the oxide coating will scale off leaving new metal exposed an susceptible to corrosion  salt will by acting as aspeed up the oxidation salt bridge  most metals develop a thin oxide coating which then protects their internal atoms against further oxidation

102 Fe (s) H 2 O droplet O 2(g) anode cathode Fe(OH) 2(s) rust Cathode SOA(Red) : Anode SRA(Ox): Net: O 2(g) + 2H 2 O (l) + 4e -  4 OH - (aq) O 2(g) + 2H 2 O (l) + 2Fe (s)  4 OH - (aq) + 2 Fe 2+ (aq) Fe (s)  Fe 2+ (aq) + 2e - [ ] 2 O 2(g) + 2H 2 O (l) + 2Fe (s)  2 Fe(OH) 2(s)

103 Practice Question

104 M. Prevention of Corrosion  other metals (eg. Zn, Cr, Sn) can be onto metals that you don’t want to corrode (eg. steel (Fe))  applying a coating of to protect metal from oxidation  this coating is of a metal that is a than the metal that is to be protected…the coating metal will react instead and is called the paint plated stronger reducing agent sacrificial anode Fe Zn coating Answer: D

105  this method is also calledcathodic protection  has been in use before the science of electrochemistry was developed  Sir Humphrey Davy first used cathodic protection on British naval ships in 1824!

106  because the attached metal is a than the iron in the steel, the more active metal supplies the and therefore the steel (iron) becomes the cathode and is protected  can be used to protect any metal but steel (iron) is most commonly protected  we use steel (iron) for many structures such as buried fuel tanks, septic tanks, pipelines, hulls of ships, bridge supports etc  to protect these structures by cathodic protection, an is connected by a to the structure active metal (eg. Mg, Zn, Al) wire stronger reducing agent e - for reduction

107 Fe (tank) more active metal eg) Mg, Zn, Al Video – sacrificial anodesacrificial anode

108

109  stainless steel contains chromium and nickel, changing steels reduction potential to one characteristic of (basically unreactive)  another protection method is alloying pure metals, which changes their reduction potential noble metals like gold

110  is the process of by metal ions in solution  an object can be plated by making it the in an containing ions of the plating metal depositing the neutral metal on the cathode reducing cathode electrolytic cell electroplating Video - electroplating

111 Electrolytic Cell iscoveryeducati on.com/index.c fm?guidAssetI d=D5A677D5- 947C-46E9- BEFF C12D46 &blnFromSear ch=1&productc ode=US


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