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A. Introduction to Electrochemistry

Electrochemistry. A. Introduction to Electrochemistry. electrochemistry is the branch of chemistry that studies. electron transfer in chemical reactions. electron transfer occurs in. living systems. eg). photosynthesis, cellular respiration. also occurs in. non-living systems. eg).

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A. Introduction to Electrochemistry

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  1. Electrochemistry A. Introduction to Electrochemistry • electrochemistry is the branch of chemistry that studies electron transfer in chemical reactions • electron transfer occurs in living systems eg) photosynthesis, cellular respiration

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

  3. is a oxidation loss of electrons “LEO” eg) Mg(s) Mg2+(aq) + 2e 2Cl(aq) Cl2(g) + 2e • is a reduction gain of electrons “GER” eg) Fe3+(aq) + 3e Fe(s) Br2(l) + 2e 2Br(aq)

  4. B. Redox Reactions • 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 • the e lost in the oxidation half reaction the e gained in the reduction half reaction must equal

  5. you may have to of the half reactions to balance the e multiplyone or both • (ions not changing) are included! spectator ions NOT

  6. Example 1 Given the following reaction, write the half reactions and the net ionic equation. Na(s) + LiCl(aq) Li(s) + NaCl(aq) 0 1+ 1– 0 1– 1+ ox red Cl- is spectator Ox: Na(s) Na+(aq) + 1e- Li+(aq) + 1e- Li(s) Red: Net: Li+(aq) + Na(s) Li(s) + Na+(aq)

  7. Example 2 Given the following reaction, write the half reactions and the net ionic equation. Zn(s) + Au(NO3)3(aq) Au(s) + Zn(NO3)2(aq) 0 3+ 1– 0 1– 2+ ox red NO3- is spectator [ ] Ox: 3 Zn(s) Zn2+(aq) + 2e- [ ] Au3+(aq) + 3e- Au(s) Red: 2 Net: 2 Au3+(aq) + 3 Zn(s) 2 Au(s) + 3 Zn2+(aq)

  8. C. Spontaneous Redox Reactions • chemical reactions which occur on their own, without the input of , are called additional energy spontaneous • not all reactions are spontaneous reduced • the substance that is is called the ( ) (it causes the oxidation by taking e-) oxidizing agent OA oxidized • the substance that is is called the ( ) (it causes the reduction by giving up e-) reducing agent RA

  9. OA + spontaneous RA reaction RA + non- spontaneous OA reaction • in the table of redox half reactions (see pg 9 in Data Booklet), the is at the top left and the is at the bottom right strongest oxidizing agent(SOA) strongest reducing agent(SRA) • 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. D. 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: 1. List all species present as reactants • dissociate and soluble ionic compounds acids • do not dissociate molecular compounds • include ions if it is H+(aq) acidic • always include H2O(l)

  12. 2. Identify each as or (***some can be both so memorize them… , , , ) OA RA Fe2+ Cr2+ Sn2+ H2O(l) 3. Identify the and using the table. SOA SRA 4. Write out the for the SOA and SRA. half reactions 5. Determine the net ionic reaction 6. Determine spontaneity

  13. Example 1 Predict the most likely redox reaction when chromium is placed into aqueous zinc sulphate. Cr(s) H2O(l) Zn2+(aq) SO42-(aq) 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. Example 2 Predict the most likely redox reaction when silver is placed into aqueous cadmium nitrate. Ag(s) H2O(l) Cd2+(aq) NO3-(aq) S RA S OA OA with H+(aq) OA/RA SOA (Red): Cd2+(aq) + 2e- Cd(s) SRA (Ox): [ ] Ag(s) Ag+(aq) + e- 2 nonspont Net: Cd2+(aq) + 2 Ag(s) Cd(s) + 2 Ag+(aq)

  15. Example 3 Predict the most likely redox reaction when potassium permanganate is slowly poured into an acidic iron (II) sulphate solution. MnO4-(aq) H+(aq) Fe2+(aq) SO42-(aq) K+(aq) H2O(l) S OA OA/ RA S OA with H+(aq), H2O(l) OA OA with H+(aq OA/RA 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. E. 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 9 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. Cu2+(aq) Zn2+(aq) Pb2+(aq) Ag+(aq) Cu(s)     Zn(s)     Pb(s)     Ag(s)     Example 1 Generate a redox table given the following data:  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

  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 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 3: Generate a redox table given the following data:   2X-(aq) + Y2 spontaneous reaction 2Z-(aq) + Y2 no reaction 2Z-(aq) + W2  spontaneous reaction Redox Table SOA W2 + 2e- 2W-(aq) Z2 + 2e- 2Z-(aq) Y2 + 2e- 2Y-(aq) X2 + 2e- 2X-(aq) SRA 1. pg 572 Lab 13.A, 2. pg 573 # 11-14 3. pg 5 in workbook

  21. F. Oxidation Numbers (States) • an or is the charge an atom to have when found in a or charged oxidation number state 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

  22. 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

  23. 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 0 –3

  24. 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

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

  26. G. Balancing Redox Reactions • sometimes most reactants and products are known but the complete reaction is not given…called a reaction skeleton • we can use the oxidation numbers to balance the equations either in or conditions: acidic basic

  27. 1. In an Acidic Solution 1. Assign oxidation numbers (ON). 2. Balance the that changes in oxidation number. element 3. Add to balance the change in oxidation number. e– total (ON  subscript  coefficient) 4. Balance O using H2O(l). 5. Balance H using H+(aq). 6. Check that the half-reaction is balanced with respect to net charge.

  28. Example 1: Balance the following half reaction assuming acidic conditions: (+6) (+3) +3 2 +6 2 + 2 H2O(l) 4 H+(aq) + 3 e– + CrO42-(aq)  CrO2-(aq) +6 8 = 2 +3 4 = 1 net charge = –1 net charge = –1 (Cr is already balanced)

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

  30. 1. In a Basic Solution 1. Assign oxidation numbers. 2. Balance the that changes in oxidation number. element 3. Add to balance the change in oxidation number. e– total (ON  subscript  coefficient) 4. Balance equation charge OH–(aq). 5. Balance H using H2O(l). 6. Check that the half-reaction is balanced with respect to oxygen.

  31. Example 1: Balance the following half reaction assuming basic conditions: (+4) (+6) +6 2 +4 2 + 2 e– + H2O(l) 2 OH–(aq) + SO32-(aq) SO42-(aq) +4 6 = 2 +6 8 = 2 # oxygen = 2 + 3 = 5 # oxygen = 4 + 1 = 5 (S is already balanced)

  32. 3. Putting it All Together • now we are going to combine coming up with our own half reactions with figuring out the net redox reaction

  33. Steps 1. Assign oxidation numbers. 2. Separate the partial net equation into two (omit any , , or ). half reactions H2O(l) H+(aq) OH-(aq) 3. Balance each half-reaction. 4. of the equations so e lost = e gained. Multiplyone or both 5. Add the equations to produce a balanced net ionic equation 6. Check to see if all elements and charges are balanced. Simplify.

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

  35. Example 2: Balance the following using oxidation numbers, assuming basic conditions: +3 0 +6 +4 2 2 TeO32-(aq) + Cr3+(aq) Te(s) + Cr2O72- (aq) +4 6 = 2 +12 14 = 2 (+4) (0) Red [ ] 3 H2O(l) + 4 e– + TeO32-(aq) Te(s) + 6 OH–(aq) 3 (+6) (+12) 2 [ ] 14 OH–(aq) 2 Cr3+ (aq) Cr2O72-(aq) + 6 e– + 7 H2O(l) Ox + 9 H2O(l) + 3 TeO32-(aq) + 28 OH-(aq) + 4 Cr3+ (aq) 3 Te(s) + 18 OH-(aq) + 2 Cr2O72-(aq) + 14 H2O(l) Net 3 TeO32-(aq) + 10 OH-(aq) + 4 Cr3+ (aq) 3 Te(s) + 2 Cr2O72-(aq) + 5 H2O(l)

  36. 4. Disproportionation • disproportionation occurs when one element is both oxidized and reduced in a reaction eg) -1 -2 0 2 H2O2(aq) 2 H2O(l) + O2(g) 0 +1 -1 Cl2(g) + 2 OH-(aq) ClO-(aq) + Cl-(aq) + H2O(l)

  37. 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

  38. Example 1 What is the mass of zinc is produced when 100 g of chromium is placed into aqueous zinc sulphate. Cr(s) H2O(l) Zn2+(aq) SO42-(aq) 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))

  39. Cr(s) + Zn2+(aq)Zn(s) + Cr2+(aq) m = 100 g M = 52.00 g/mol n = m M = 100 g 52.00 g/mol = 1.923… mol m = ? M = 65.39 g/mol n = 1.923… mol x 1/1 = 1.923… mol m = nM = (1.923…mol)(65.39 g/mol) = 125.75 g = 126 g

  40. 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): Fe2+(aq) Fe3+(aq) + e- 5 Net: MnO4-(aq) +8H+(aq) + 5Fe2+(aq) Mn2+(aq)+ 4H2O(l) + 5Fe3+(aq)

  41. MnO4-(aq) +8H+(aq) + 5Fe2+(aq) Mn2+(aq)+ 4H2O(l) + 5 Fe3+(aq) v = ? c = 1.50 mol/L n = 1.125 mol x 1/5 = 0.225 mol v = n C v = 0.225 mol 1.50 mol/L = 0.150 L v = 0.500 L c = 2.25 mol/L n = cv = (2.25 mol/L)(.500L) = 1.125 mol

  42. 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

  43. 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 eg) MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) purple colourless • as long as the sample (RA) in the flask is reacting with the the sample will be permanganate ions (dichromate ions) colourless (orange)

  44. 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 • permanganate solution as it oxidizes distilled water decomposes

  45. 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

  46. Trial 1 2 3 4 Final Volume (mL) 18.40 35.30 17.30 34.10 Initial Volume (mL) 1.00 18.40 0.60 17.30 Volume of (mL) Endpoint Colour pink light pink light pink light pink Example Find the concentration of (standardize) the KMnO4(aq) solution by titrating 10.00 mL of 0.500 mol/L acidified tin (II) chloride with the KMnO4(aq). 17.40 16.90 16.70 16.80 titrant

  47. endpoint average is calculated by using 3 volumes within 0.20 mL 16.90 mL + 16.70 mL + 16.80 mL 3 Endpoint average = = 16.80 mL

  48. 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): MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) 2 [ ] [ ] SRA (Ox): Sn2+(aq) Sn4+(aq) + 2 e- 5 Net: 2MnO4-(aq) + 16H+(aq) + 5Sn2+(aq) 2Mn2+(aq)+ 8H2O(l) + 5Sn4+(aq)

  49. use net redox equation to calculate KMnO4(aq) concentration 2MnO4-(aq) + 16H+(aq) + 5Sn2+(aq) 2Mn2+(aq)+ 8H2O(l) +5Sn4+(aq) v = 0. 01000 L c = 0.500 mol/L n = cv = (0.500 mol/L)(0.01000 L) = 0.00500 mol v = 0.01680 L C = ? n = 0.00500 mol x 2/5 = 0.00200 mol C = n v C = 0.00200 mol 0.01680 L = 0.119 mol/L

  50. I. Electrochemical Cells 1. Voltaic/Galvanic Cells • 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 conducting wire • the transfer of e- can occur through a separating the two substances in containers called half cells

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