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A. Redox Reactions – Crash course video

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1 A. Redox Reactions – Crash course video
Electrochemistry A. Redox Reactions – Crash course video electrochemistry is the branch of chemistry that studies electron transfer in chemical reactions is a OXIDATION loss of electrons “OIL” 2+ Charge of ion or element (if charge increase, from its reactant, it’s a oxidation reaction eg) Mg(s)  Mg2+(aq) + 2e Electrons are in the product = losing electrons -1 Charge of ion or element (if charge increase, from its reactant, it’s a oxidation reaction 2Cl(aq)  Cl2(g) + 2e Electrons are in the product = losing electrons

2 oxidation and reduction reactions occur together, hence the term redox
is a reduction gain of electrons “RIG” 3+ eg) Fe3+(aq) + 3e  Fe(s) Charge of ion or element (if charge decreases, from its reactant, it’s a reduction reaction Br2(l) + 2e  2Br(aq) Electrons are in the reactant = gaining electrons oxidation and reduction reactions occur together, hence the term redox the reduction and oxidation reactions are called the half reactions “adding” the half reactions together will give you the that takes place during the redox reaction net ionic equation

3 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 multiply one or both (ions not changing) are included! spectator ions NOT the substance that is is called the ( ) (it causes the oxidation by taking e-) reduced oxidizing agent OA the substance that is is called the ( ) (it causes the reduction by giving up e-) oxidized reducing agent RA Video – REDOX reaction

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

5 3 Zn(s) + 2 Au(NO3)3(aq)  2 Au(s) + 3 Zn(NO3)2(aq)
Example 2 Given the following reaction, write the half reactions and the net ionic equation. 3 Zn(s) + 2 Au(NO3)3(aq)  2 Au(s) + 3 Zn(NO3)2(aq) 3+ 1– 2+ 1– ox red NO3- is spectator Ox: 3 [ ] Zn(s)  Zn2+(aq) + 2e- Red: 2 [ ] Au3+(aq) + 3e-  Au(s) Net: 2 Au3+(aq) Zn(s)  2 Au(s) Zn2+(aq) Step 1: identify the charges for each element or ion. Step 2: identify your oxidation (increase in number) and reduction (decrease in number) half reactions. Also identify your spectator ion. Step 3: write your half reactions (charges must be balanced with same number of electrons. Step 4: Ensure you have the same number of electrons on both sides (reactants and products) Step 5: Write the net ionic equation.

6 Practice Questions

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

8 B. Spontaneous Redox Reactions
chemical reactions which occur on their own, without the input of , are called additional energy spontaneous not all reactions are 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)

9 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

10 Try These: For each of the following combinations of substances, state whether the reaction would be spontaneous or non-spontaneous: Cr3+(aq) with Ag(s) I2(s) with K(s) H2O2(l) with Au3+(aq) Sn2+(aq) with Cu(s) Fe2+(aq) with H2O (l) non-spontaneous spontaneous spontaneous non-spontaneous non-spontaneous (both ways)

11 Practice Question

12 Steps are found at the bottom of examples.
C. Predicting Redox Reactions Answer: A 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.

13 Zn2+(aq) + Cr(s)  Zn(s) + Cr2+(aq)
Example 1 Predict the most likely redox reaction when chromium is placed into aqueous zinc sulphate. Cr(s) Zn2+(aq) SO42-(aq) H2O(l) S RA S OA OA with H2O(l) OA/RA SOA (Red): Zn2+(aq) + 2e-  Zn(s) SRA (Ox): Cr(s)  Cr2+(aq) + 2e- spont Net: Zn2+(aq) + Cr(s)  Zn(s) + Cr2+(aq)

14 Cd2+(aq) + 2 Ag(s)  Cd(s) + 2 Ag+(aq)
Example 2 Predict the most likely redox reaction when silver is placed into aqueous cadmium nitrate. Ag(s) Cd2+(aq) NO3-(aq) H2O(l) S RA S OA OA with H+ (aq) OA/RA SOA (Red): Cd2+(aq) + 2e-  Cd(s) SRA (Ox): 2 [ ] Ag(s)  Ag+(aq) + e- nonspont Net: Cd2+(aq) + 2 Ag(s)  Cd(s) Ag+(aq)

15 Example 3 Predict the most likely redox reaction when potassium permanganate is slowly poured into an acidic iron (II) sulphate solution. K+(aq) MnO4-(aq) H+(aq) Fe2+(aq) H2O(l) SO42-(aq) OA S OA with H+ (aq) OA OA/ RA S OA/RA OA with H+ (aq), H2O(l) SOA (Red): MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l) SRA (Ox): 5 [ ] Fe2+(aq)  Fe3+(aq) + e- spont Net: MnO4-(aq) +8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5 Fe3+(aq)

16 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

17 Video: REDOX tables Example 1
Generate a redox table given the following data (useful when all reactions are given: Cu2+(aq) Zn2+(aq) Pb2+(aq) Ag+(aq) Cu(s) Zn(s) Pb(s) Ag(s)  indicates no reaction  indicates a reaction

18 Redox Table SOA Ag+(aq) + e-  Ag(s) Cu2+(aq) + 2e-  Cu(s) Pb2+(aq) + 2e-  Pb(s) Zn2+(aq) + 2e-  Zn(s) SRA Put the oxidizing agents in order from strongest to weakest. Ag+(aq), Cu2+(aq), Pb2+(aq), Zn2+(aq) Put the reducing agents in order from strongest to weakest. Zn(s), Pb(s), Cu(s), Ag(s)

19 Example 2: Generate a redox table given the following data:   Cu(s) + Ag+(aq)  Cu2+(aq) + Ag(s) Zn2+(aq) + Ag(s)  no reaction Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Hg(l) + Ag+(aq)  no reaction Label each as OA or Ra Redox Table SOA Hg2+(aq) + 2e-  Hg(l) Ag+(aq) + e-  Ag(s) Cu2+(aq) + 2e-  Cu(s) Zn2+(aq) + 2e-  Zn(s) SRA

20 Example 2 (continued): Put the oxidizing agents in order from weakest to strongest. Zn2+(aq), Cu2+(aq), Ag+(aq), Hg2+(aq) Put the reducing agents in order from weakest to strongest. Hg(l), Ag(s), Cu(s), Zn(s)

21 Example 3: Generate a redox table given the following data:   2X-(aq) + Y2(g)  spontaneous reaction 2Z-(aq) + Y2 (g)  no reaction 2Z-(aq) + W2 (g)  spontaneous reaction Label each as OA or RA Redox Table SOA W2 (g) + 2e-  2W-(aq) Z2 (g) + 2e-  2Z-(aq) Y2 (g) + 2e-  2Y-(aq) X2 (g) + 2e-  2X-(aq) SRA

22 Example 3 (continued): Put the oxidizing agents in order from strongest to weakest . W2(g), Z2(g), Y2(g), X2(g) Put the reducing agents in order from strongest to weakest . X-(aq), Y-(aq), Z-(aq), W-(aq)

23 Practice Question

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

25 Rules for Assigning Oxidation Numbers:
1. In a pure element, the oxidation number is zero 2. In simple ions, the oxidation number is equal to the 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

26 4. In most compounds containing oxygen, the oxidation number for oxygen is (Exception is the peroxides eg) H2O2, Na2O2 where the oxidation number of oxygen is ) –2 –1 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 PO43- = net charge Zero the charge –3

27 Common Oxidation Numbers
Atom or Ion Oxidation Number Examples All atoms in elements Na is 0 Cl in Cl2 is 0 Hydrogen in all compounds (except: hydrogen in hydrides) +1 -1 H in HCL is +1 H in LiH is -1 Oxygen in all compounds, Except oxygen in peroxides -2 O in H2O is -2 O in H2O2 is -1 All monoatomic ions Charge of Ion Na+ is 1+ S2- is 2-

28 Example What is the oxidation number (state) for the element identified in each of the following substances: a) N in N2O +1 –2 individual oxidation numbers N2 O sum of oxidation numbers +2 –2 = 0 b) N in NO3- +5 –2 N O3– +5 –6 = –1

29 c) C in C2H5OH C2 H5 O H –2 +1 –2 +1 –4 +5 –2 +1 = 0 d) C in C6H12O6 C6 H12 O6 +1 –2 +12 –12 = 0

30 figuring out oxidation numbers can help to identify whether a reaction is a or not
redox reaction for it to be a redox reaction, there has to be an in oxidation number and a in oxidation number seen in the reaction both increase decrease +1 -1 +1 -1 eg) Ag(s) + NaNO3(aq)  Na(s) + AgNO3(aq) Ag increases oxidized redox!!! Na decreases  reduced… +2 -2 +1 -1 +2 -1 +1 -2 PbSO4(aq) KI(aq)  PbI2(s) + K2SO4(aq) nothing changes  NOT a redox reaction!

31 Practice Question

32 Practice question Answer: a

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

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

35 disproportionation occurs when one element
F. Disproportionation disproportionation occurs when one element is both oxidized and reduced in a reaction eg) -1 -2 2 H2O2(aq)  2 H2O(l) + O2(g) +1 -1 Cl2(g) OH-(aq)  ClO-(aq) + Cl-(aq) + H2O(l) Do workbook questions: page 4

36 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 Oxidation Number Method

37 Example 1:Of half Reaction Method
Balance the following half reaction : (+6) (+3) +6 2 +3 2 4 H+(aq) + 3 e– + CrO42-(aq)  CrO2-(aq) + 2 H2O(l) +6 8 = 2 +3 4 = 1 net charge = –1 net charge = –1 (Cr is already balanced)

38 Balance the following half reaction:
Example 2: Balance the following half reaction: (+6) (0) +1 +3 2 6 H+(aq) + 6 e– + 2 HClO2(aq)  Cl2(g) + 4 H2O(l) +1 +3 4 = 0 net charge = 0 net charge = 0

39 Practice Question

40 CrO42-(aq) + SO32-(aq)  CrO2-(aq) + SO42-(aq)
Example 1: Balance the following using oxidation numbers, assuming acidic conditions: Answer: A +6 2 +4 2 +3 2 +6 2 CrO42-(aq) + SO32-(aq)  CrO2-(aq) + SO42-(aq) +6 8 = 2 +4 6 = 2 +3 4 = 1 +6 8 = 2 (+6) (+3) Red 2 [ ] 4 H+(aq) + 3 e– + CrO42-(aq)  CrO2-(aq) + 2 H2O(l) (+4) (+6) Ox 3 [ ] H2O(l) + SO32-(aq)  SO42-(aq) + 2 e– + 2 H+(aq) 8 H+(aq) + 2 CrO42-(aq) + 3 H2O(l) + 3 SO32-(aq)  2 CrO2-(aq) + 4 H2O(l) + 3 SO42-(aq) H+(aq) Net 2 H+(aq) + 2 CrO42-(aq) + 3 SO32-(aq)  2 CrO2-(aq) + H2O(l) + 3 SO42-(aq)

41 Example 1: of Oxidizing Number Method
Balance the following reaction using the oxidation number method. +7 +4 +4 +6 2 2 3 2 3 1 __ H+(aq) + __MnO4-(aq) + __ SO32-(aq)  __ MnO2(aq) + __ SO42-(aq) + __ H2O(l)  = 3  1 atom = 3  2(Co) =6e-  = +2  1 atom = +2  3(Co) =+6e-

42 Balance the following reaction using the oxidation number method.
Example 2: Balance the following reaction using the oxidation number method. +4 -1 +3 +5 6 3 3 1 2 1 6 __ H2O(l) + __N2O4(g) _ Br(aq)  _ NO2 (aq) + _ BrO3(aq) + __ H+(aq)  = 1  2 atoms = 2  3(Co) =6e-  = +6  1 atom = +6 1(Co) =6e-

43 H. Redox Stoichiometry 1. Calculations 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 we can now apply these same calculations to balanced redox equations

44 Example 1 What is the mass of zinc is produced when 100 g of chromium is placed into aqueous zinc sulphate. Cr(s) Zn2+(aq) SO42-(aq) H2O(l) RA OA OA with H2O(l) OA/RA SRA SOA SOA (Red): Zn2+(aq) + 2e-  Zn(s) SRA (Ox): Cr(s)  Cr2+(aq) + 2e- Net: Cr(s) + Zn2+(aq)  Zn(s + Cr2+(aq) )

45 Cr(s) + Zn2+(aq)  Zn(s) + Cr2+(aq)
m = 100 g M = g/mol n = m M = g 52.00 g/mol = … mol 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

46 K+(aq) MnO4-(aq) H+(aq) H2O(l) Fe2+(aq) SO42-(aq) OA OA with H+(aq) OA
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? K+(aq) MnO4-(aq) H+(aq) H2O(l) Fe2+(aq) SO42-(aq) OA OA with H+(aq) OA OA/RA OA/RA OA with H+(aq) SOA SRA OA with H2O(l) SOA (Red): MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l) SRA (Ox): [ ] 5 Fe2+(aq)  Fe3+(aq) + e- Net: MnO4-(aq) +8H+(aq) + 5Fe2+(aq)  Mn2+(aq)+ 4H2O(l) + 5Fe3+(aq)

47 *** Reminder: Use the formulas n=m/M and C = n/v to find quantities
MnO4-(aq)+8H+(aq) + 5Fe2+(aq)  Mn2+(aq)+ 4H2O(l) + 5Fe3+(aq) v = ? c = 1.50 mol/L n = mol x 1/5 = mol v = n C v = mol 1.50 mol/L = L 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

48 Practice Question

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

50 in redox titrations, two common oxidizing agents are used because of their and :
colour strength 1. permanganate ions (MnO4-(aq)) – purple 2. dichromate ions (Cr2O72-(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)

51 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

52 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 primary standard this is done just prior to the titration

53 Example Find the concentration of (standardize) the KMnO4(aq) solution by titrating mL of mol/L acidified tin (II) chloride with the KMnO4(aq). Trial 1 2 3 4 Final Volume (mL) 18.40 35.30 17.30 34.10 Initial Volume (mL) 1.00 0.60 Volume of (mL) Endpoint Colour pink light pink 17.40 16.90 16.70 16.80 KMnO4(aq).

54 endpoint average is calculated by using 3 volumes within 0.20 mL
16.90 mL mL mL 3 = mL Review: Redox reactions

55 determine net ionic redox equation
Analysis: determine net ionic redox equation K+(aq) MnO4-(aq) H+(aq) H2O(l) Sn2+(aq) Cl-(aq) OA OA with H+(aq) OA OA/RA OA/RA RA SOA SRA SOA (Red): 2 [ ] MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l) SRA (Ox): 5 [ ] Sn2+(aq)  Sn4+(aq) e- Net: 2MnO4-(aq)+ 16H+(aq)+ 5Sn2+(aq)  2Mn2+(aq)+ 8H2O(l) + 5Sn4+(aq)

56 use net redox equation to calculate KMnO4(aq) concentration
2MnO4-(aq) +16H+(aq) +5Sn2+(aq) 2Mn2+(aq)+ 8H2O(l) +5Sn4+(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

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

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

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

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

61 electrons flow from the to the through a connecting wire
anode (LEOA) cathode (GERC) 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- move porous cup salt bridge the flow of ions through the solution and e- through the wire maintains overall electrical neutrality Cell song Repeated animation

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

63 Practice Question

64 2. Standard Reduction Potentials
Answer: 4,6,4,1 are the ability of a half cell to reduction potentials attract e- these potentials are measured using a voltmeter each half reaction listed in the Data Booklet has an E value measured in assigned to it 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

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

66 Practice Question

67 Calculate the E net for the reaction of Zn(s) with CuSO4(aq).
Example Calculate the E net for the reaction of Zn(s) with CuSO4(aq). Answer: B Zn(s) Cu2+(aq) SO42-(aq) H2O(l) S RA S OA OA with H2O(l) OA/RA SOA (Red): Cu2+(aq) + 2e-  Cu(s) E = V SRA (Ox): Zn(s)  Zn2+(aq) + 2e- E = V Enet = V Net: Zn(s) + Cu2+(aq)  Cu(s + Zn2+(aq)

68 Practice Question

69 Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s) OR
4. Shorthand Notation Answer: 1.36V Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s) OR Zn(s) / Zn2+(aq) // Cr2O72-(aq) , H+(aq) , Cr3+(aq)/ C(s) ***anode // cathode line separates (/) phases double line represents the or and separates the two (//) porous cup salt bridge half reactions comma separates chemical species in the same phase (,)

70 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

71 Draw and fully label the following electrochemical cell:
Example Draw and fully label the following electrochemical cell: Al(s)/ Al3+(aq) // Ni2+(aq) / Ni(s) e- V Al(s) Ni(s) anode negative terminal cathode positive terminal Ni2+ (electrolyte) anions Al3+ (electrolyte) cations

72 2 Al(s) + 3 Ni2+(aq)  3 Ni(s + 2 Al3+(aq) )
H2O(l) S RA OA RA S OA OA/RA SOA (Red): 3 [ ] Ni2+(aq) + 2e-  Ni(s) E = –0.26 V SRA (Ox): 2 [ ] Al(s)  Al3+(aq) + 3e- E = V Net: 2 Al(s) Ni2+(aq)  3 Ni(s Al3+(aq) ) Enet = V spontaneous: yes

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

74 common batteries of clocks, remote controls, noisy kids toys etc.
a) Dry Cell common batteries of clocks, remote controls, noisy kids toys etc. 1.5 V and 9 V Cathode (Red): 2 MnO2(s) + H2O(l) + 2e-  Mn2O3(aq) + 2 OH-(aq) E= V Anode (Oxid): Zn(s)  Zn2+(aq) + 2e E = V Net: 2 MnO2(s)+ H2O(l) + Zn(s)  Mn2O3(aq) + 2 OH-(aq) + Zn2+(aq) Enet = V the produced causes irreversible side reactions to occur making recharging impossible OH-

75 b) Nickel-Cadmium one type of battery rechargeable
Cat (Red): 2 NiO(OH)(s)+ 2 H2O(l) + 2e-  2 Ni(OH)2(s) +2 OH-(aq) E= V An (Oxid): Cd(s) OH-(aq)  2 Cd(OH)2(s) + 2e- E = V Net: 2 NiO(OH)(s)+ 2 H2O(l) + Cd(s)  2 Ni(OH)2(s)+ 2 Cd(OH)2(s) Enet = V

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

77 Cat (Red): PbO2(s)+ HSO4-(aq) + 3 H+(aq) + 2e-  PbSO4(s)+ 2 H2O(l) E= +1.68 V
An (Oxid): Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2e- E = V Net: Pb(s)+ PbO2(s) + 2 H+(aq) + 2 HSO4-(aq) 2 PbSO4(s)+ 2 H2O(l) Enet = V

78 d) Fuel Cells cells where reactants are continuously supplied the energy from this reaction can be used to run machines one type is the hydrogen-oxygen fuel cell is pumped in at the while is pumped in at the (which both have a lot of surface area) hydrogen gas anode oxygen gas cathode pressure is used to push the H2 through a platinum catalyst which splits the H2 into 2H+ and 2e-

79 the H+ ions and oxygen atoms combine to form water
the 2e- move through an external circuit towards the cathode generating electrical energy the O2 is also pushed through the platinum catalyst forming two oxygen atoms the H+ ions and oxygen atoms combine to form water Cathode (Red): O2(g) + 4 H+(aq) e-  2 H2O(l) E= V +1.23 Anode (Oxid): 2 H2(g)  4 H+(aq) + 4e- E = V 0.00 Net: O2(g) H2(g)  2 H2O(l) Enet = V

80 Hydrogen-oxygen Fuel Cell

81 need a source of hydrogen…reformers are used to convert CH4 or CH3OH into and
unfortunately, is a CO2 greenhouse gas about 24-32% efficient where gas-powered car is about 20% efficient

82 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) electrical nonspontaneous these reactions have a Enet negative commonly used to (eg. gold, silver, bronze, chromium etc), electroplate metals and split compounds into recharge batteries, useful gases (eg. H2, O2, Cl2 etc)

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

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

85 Practice Question

86 some processes are used in industry to produce gases, for example:
Answer: B 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 Hall-Heroult cell Al 2. a for producing …a saturated sodium chloride solution is electrolyzed …chlorine gas is formed and collected at the anodes chlor-alkali plant chlorine gas

87 2 H2O(l)  O2(g) + 4 H+(aq) + 4e-
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. Ni2+(aq) NO3-(aq) H2O(l) OA OA with H+(aq) OA/RA SOA SRA Cathode SOA(Red): 2 [ ] Ni2+(aq) + 2e-  Ni(s) E = V Anode SRA(Ox): 2 H2O(l)  O2(g) H+(aq) + 4e- E = V Net: 2 Ni2+(aq) + 2 H2O(l)  2 Ni(s)+ O2(g) + 4 H+(aq) Enet = V

88 2 H2O(l) +2 e-  H2(g) + 2 OH- (aq)
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. K+(aq) I-(aq) H2O(l) OA RA OA/RA SRA SOA Cathode SOA(Red): 2 H2O(l) +2 e-  H2(g) + 2 OH- (aq) E = V Anode SRA(Ox): 2 I-(aq)  I2(s) + 2e- E = V Net: 2 H2O(l) I-(aq)  H2(g) OH-(aq) + I2(s) Enet = V

89 2 H2O(l)  O2(g) + 4 H+(aq) + 4e-
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. Cu2+(aq) SO42-(aq) H2O(l) OA OA/RA OA/RA SOA SRA Cathode SOA(Red): 2 Cu2+(aq) e-  Cu(s) [ ] E = V Anode SRA(Ox): 2 H2O(l)  O2(g) H+(aq) + 4e- E = V Net: 2 H2O(l) + 2 Cu2+(aq)  2 Cu(s) + O2(g) H+(aq) Enet = V

90 *** 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 H2O to O2 is more difficult …called chloride ions overvoltage

91 2 H2O(l) +2 e-  H2(g) + 2 OH- (aq)
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. Na+(aq) Cl-(aq) H2O(l) OA RA OA/RA SRA SOA Cathode SOA(Red): 2 H2O(l) +2 e-  H2(g) + 2 OH- (aq) E = V Anode SRA(Ox): 2 Cl-(aq)  Cl2(g) + 2e- E = V Net: 2 H2O(l) + 2 Cl-(aq)  H2(g) + Cl2(g) + 2 OH- (aq) Enet = V

92 2. Quantitative Study of Electrolysis
quantitative analysis (stoich) provides information on necessary quantities, current and/or time for electrolytic reactions the unit for charge is the (q) Coulomb (C) one e- carries of charge 1.60 x C this means that one of e- carry of charge mole 9.65 x 104 C is called the (see Data Booklet pg 3) 9.65 x 104 C/mol Faraday constant (F)

93 ne- = q F q = It where: ne- = number of moles of electrons (mol) q = charge in Coulombs (C) F = Faraday constant = 9.65 x 104 C/mol I = current in C/s or Amperes (A) t = time in seconds (s)

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

95 Example 1 An electrochemical cell caused a mol of e- to flow through a wire. Calculate the charge. ne- = q F mol = q 9.65 x 104 C/mol q = 6948 C = 6.95  103 C ne- = mol F = 9.65 x 104 C/mol

96 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. ne- = It F = (0.100 C/s)(3000 s) 9.65 x 104 C/mol = mol I = A (C/s) t = 50.0 min  60 s/min = 3000 s F = 9.65 x 104 C/mol

97 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? Al3+(l) e  Al(l) ne- = It F = (20.0 A)(1.00 h  3600 s/h) 9.65 x 104 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

98 Practice Question

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

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

101 O2(g) + 2H2O(l) + 4e-  4 OH-(aq)
Fe(OH)2(s) rust H2O droplet cathode anode Fe(s) Cathode SOA(Red): O2(g) + 2H2O(l) + 4e-  4 OH-(aq) Anode SRA(Ox): 2 [ ] Fe(s)  Fe2+(aq) + 2e- O2(g) + 2H2O(l) + 2Fe(s)  4 OH-(aq) Fe2+(aq) Net: O2(g) + 2H2O(l) + 2Fe(s)  2 Fe(OH)2(s)

102 Practice Question

103 M. Prevention of Corrosion
Answer: D M. Prevention of Corrosion applying a coating of to protect metal from oxidation paint other metals (eg. Zn, Cr, Sn) can be onto metals that you don’t want to corrode (eg. steel (Fe)) plated 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 stronger reducing agent sacrificial anode Fe Zn coating

104 this method is also called
cathodic 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!

105 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 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 stronger reducing agent e- for reduction

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

107

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

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

110 Electrolytic Cell


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