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Electrochemistry Na + Cl - NaCl Are you sure I can have that electron? I’m positive!

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Presentation on theme: "Electrochemistry Na + Cl - NaCl Are you sure I can have that electron? I’m positive!"— Presentation transcript:

1 Electrochemistry szczepas@canisius.edu Na + Cl - NaCl Are you sure I can have that electron? I’m positive!

2 Useful Links This presentation: www.canisius.edu/~szczepas www.canisius.edu/~szczepas Past Years’ Exams + Answers –Google: ACS Chemistry Olympiad –http://www.acs.org/content/acs/en/education/s tudents/highschool/olympiad/pastexams.html

3 Typical Olympiad Topics Oxidation Numbers Balancing Redox Reactions Galvanic Cell Architecture Standard Cell Potentials Non Standard Conditions Electrolysis Common Oxyanions Redox Concepts

4 From 2012

5 Oxidation Numbers (States) Keeping Track of Electrons Gained or Lost 4 Rules for Assigning Oxidation Numbers: In order of Importance! 1)Atom in Elemental Form –Oxidation Number is Always Zero Examples: Fe Atom Ar Atom H Atom in H 2 Molecule O Atom in O 2 Molecule P Atom in P 4 Molecule

6 Oxidation Numbers (States) 2)Monatomic Ions –Oxidation Number = Charge –Examples: K + Oxidation Number = +1 Mg 2+ Oxidation Number = +2 Al 3+ Oxidation Number = +3 N 3- Oxidation Number = -3 S 2- Oxidation Number = -2 FeCl 2 ON(Fe) = +2 Oxidation Numbers are written with the sign before the number to distinguish them from Actual Charges

7 Oxidation Numbers (States) 3)Nonmetals (Usually Negative Oxidation #’s, But Can Be Positive) Fluorine –Oxidation Number is -1 In All Compounds Hydrogen –Oxidation Number is +1 When Bonded to Non Metals –Oxidation Number is -1 When Bonded to Metals Oxygen –Oxidation Number is Usually -2 In Molecular and Ionic Compounds –In Peroxides (O 2 2- ) Oxidation Number is -1 for Each O Atom Other Halogens –Oxidation Number is -1 in Most Binary Compounds –Oxidation Number When Combined with Oxyanions Can Be Positive

8 Oxidation Numbers (States) 4)Sum of the Oxidation Number of All Atoms in a Neutral Compound is Zero H 2 SO 3 Oxidation Number of H = +1 (H Bonded to NonMetal) Oxidation Number of O = -2 Sum is Zero0 = 2×ON(H) + 1×ON(S) + 3×ON(O) 0 = 2× (+1) + 1×ON(S) + 3× (-2) ON(S) = +4 Sum of the Oxidation Numbers in a Polyatomic Ion Equals the Charge of the Ion H 2 AsO 4 - -1 = 2×ON(H) + 1×ON(As) + 4×ON(O) ON(As) = +5

9 From 2012

10 From 2014 K = +1 H = +1 O = -2 N = ???

11 From 2014 Split to ions K + NH 4 + H 2 O “hydrate” AsO 4 3-

12 2012

13 Balancing Redox Reactions by Half-Reactions Reduction Half Reaction (Electrons Taken In) 2H + (aq) + 2e -  H 2 (g) Another Example: Ag + (aq) + e -  Ag(s) Oxidation Half Reaction (Electrons Given Off) Zn(s)  Zn 2+ (aq) + 2 e - Another Example: C 2 O 4 2- (aq)  2 CO 2 (g) + 2 e -

14 To Balance Electrons, Reductions and Oxidations MUST Occur Simultaneously NO 3 - (aq) + 4 H + + 3 e -  NO(g) + 2 H 2 O(l) C 2 O 4 2- (aq)  2 CO 2 (g) + 2 e - ID the Reducing Agent in the Unbalanced Reaction: ClO 3 - + Br -  Cl 2 + Br 2 Half-Reactions

15 Balancing Redox Rxns 1.Divide total reaction into two half reactions. 2.Balance each half a. All elements besides H and O b. Balance O by adding H 2 O c. Balance H by adding H + d. Balance residual charge by adding e - 3.Multiply each half to least common multiple of electrons 4.Add half reactions and cancel 5.Check if balanced The above procedure uses acid or neutral conditions as the default. *To convert from acid to base conditions after steps 1-5, add enough OH - to both sides to neutralize all the H + to H 2 O, then cancel out any excess.

16 2012

17 From 2014

18 A sample of copper metal is dissolved in 6 M nitric acid contained in a round bottom flask. This reaction yields a blue solution and emits a colorless gas which is found to be nitric oxide. Write a balanced equation for this reaction. Unbalanced

19 Zn(s) → Zn 2+ (aq) + 2 e - Zn 2+ e-e- e-e- Cu 2+ Cu 2+ (aq) + 2 e - → Cu(s) Cl -

20 Electrode Oxidation Reduction Voltaic (Galvanic) Cell

21 to the cathode to the anode

22 2008 Local 39. Which occurs at the anode of any voltaic cell? I. A metal electrode dissolves. SO 3 2- + H 2 O → SO 4 2- + 2 H + + 2 e - II. A substance undergoes oxidation. Fe(s) → Fe 2+ + 2 e - III. Positive ions are deposited from the solution. (A)I only (B)II only (C)I and II only (D)I and III only

23 Cell Potential Voltaic Cell Spontaneous Redox Reaction (E cell >0) Used to Perform Electrical Work Similar to a Waterfall (Water Falls from High to Low Potential Energy) Electrons Flow Spontaneously from High to Low Electric Potential Use Cell Potential (Cell EMF) (E cell ) Volt  Difference in Potential Energy per Electrical Charge (1V = 1J/C) (e - charge = 1.60x10 -19 C) Potential Difference Between 2 Electrodes

24 Standard Cell Potential Use Standard Reduction Potentials for the Reduction and Oxidation Half-Reactions Note: No Multiplying Reduction Potential By Stoichiometry Voltaic (Galvanic) Cell: Positive E cell Electrolytic Cell: Negative E cell T = 25°C Standard State  Gas (P = 1atm) Species in Solution (1 M Concentration)

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27 Most easily oxidized

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29 2008 Local 41. What is the standard cell potential for the voltaic cell: Cr | Cr 3+ || Pb 2+ | Pb ? E 0 red / V Pb 2+ + 2 e - → Pb-0.13 Cr 3+ + 3 e - → Cr-0.74 (A)1.09 (B)0.61 (C)-0.61 (D)-1.09

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31 Half ReactionE 0 (V) Zn 2+ (aq) + 2e -  Zn(s)-0.763 Cr 3+ (aq) + e -  Cr 2+ (aq) -0.408 Tl + (aq) + e -  Tl(s) -0.336 Cu 2+ (aq) + e -  Cu + (aq)+0.161 Fe 3+ (aq) + e -  Fe 2+ (aq)+0.769 Use the Standard Reduction Potentials to Find the Standard Cell Potential, E 0 cell, for the Reaction: Zn(s) + 2 Tl + (aq)  Zn 2+ (aq) + 2 Tl(s) 81 Tl: Thallium

32 Calculate the E 0 rxn based on the standard reduction potentials above. Which reaction(s) is(are) spontaneous? Half ReactionE 0 (V) Zn 2+ (aq) + 2e -  Zn(s)-0.763 Cr 3+ (aq) + e -  Cr 2+ (aq) -0.408 Tl + (aq) + e -  Tl(s) -0.336 Cu 2+ (aq) + e -  Cu + (aq)+0.161 Fe 3+ (aq) + e -  Fe 2+ (aq)+0.769

33 E cell Non-Standard Conditions Nernst Equation

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35 2012

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40 How many moles of electrons must pass through a cell to produce 5.00 kg of Aluminum from Al 2 O 3 ? Al 2 O 3 + 6 e -  2 Al(s) + 3 O 2- Using F = 96485 C/mol e -, and A=C/s, 2. How long will this take using a current of 33.5 A? 1. Calculate the number of moles of electrons needed. Should also know that W = J/s.

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42 2008 Local 42. During the electrolysis of AgNO 3, what would happen to the mass of silver metal deposited if the current is doubled and the electrolysis time is decreased to ½ of its initial value? (A)It would stay the same. (B)It would increase to twice its initial value. (C)It would decrease to ¼ of its initial value. (D)It would decrease to ½ of its initial value.

43 2012 Local #42

44 How to memorize the negative ions:

45 ______-ate BO 3 3- CO 3 2- NO 3 - SiO 3 2- PO 4 3- SO 4 2- ClO 3 - AsO 4 3- SeO 4 2- BrO 3 - IO 3 - Rule: Most common ion consisting of one nonmetal atom, 2 or 3 oxygen atoms, and a negative charge. Element name may be truncated, and followed by suffix “ate”. Example:NO 3 - is nitrate, PO 4 3- is phosphate, ClO 3 - is chlorate Note, not all “-ate”s have a corresponding “-ite”!

46 ______-ide Irregular* N 3- O 2- F-F- P 3- S 2- Cl - As 3- Se 2- Br - I-I- Rule: Ion consisting of one nonmetal atom with a negative charge has truncated element name with suffix “ide”. Negative charge is how many steps from right edge of periodic table (noble gas group). Example:N 3- is nitride, O 2- is oxide, F - is fluoride * Carbide is actually C 2 2-. This does not follow the naming rules above, and you do not need to know this ion. Also: H -

47 ______-ite NO 2 - PO 3 3- SO 3 2- ClO 2 - AsO 3 3- SeO 3 2- BrO 2 - IO 2 - Rule: Less common ion consisting of one nonmetal atom, 2 or 3 oxygen atoms, and a negative charge. Element name may be truncated, and followed by suffix “ide”. Same charge, but one less oxygen from the “-ate”s. Example:NO 2 - is nitrite, PO 3 3- is phosphite, ClO 2 - is chlorite Note, not all “-ate”s have a corresponding “-ite”!

48 per-______-ate * O 2 2- ClO 4 - BrO 4 - IO 4 - Rule: Ion consisting of one halogen atom, 4 oxygen atoms, and a negative charge. One more oxygen atom than the “-ate”s. Highest possible nonmetal oxidation state in the halogen group. Example:ClO 4 - is perchlorate * Adding one more oxygen to oxide gives O 2 2-. The prefix rule follows with its own suffix for the name peroxide. Also: MnO 4 -

49 hypo-______-ite ClO - BrO - IO - Rule: Most common ion consisting of one halogen atom, one oxygen atom, and a negative charge. Example:ClO - is hypochlorite

50 hydrogen -______-ide OH - HP 2- HS - Rule: Ion consisting of one H + added to the “-ide” ion with 2- or greater charge. Example:HS - is hydrogen sulfide, OH - is hydroxide (a contraction of hydrogen oxide). Note that there is typically a space between “hydrogen” and the rest of the name.

51 hydrogen ______-ate HCO 3 - HSiO 3 - HPO 4 2- HSO 4 - HAsO 4 2- HSeO 4 - Rule: Ion consisting of one H + added to the “-ate” ion with 2- or greater charge. Example:HCO 3 - is hydrogen carbonate, HPO 4 2- is hydrogen phosphate Important Note: H 2 PO 4 - is dihydrogen phosphate

52 Corresponding Acids HF HNO 2 H 3 BO 3 H 2 CO 3 HNO 3 H2SH2SHCl HClO H 3 PO 3 H 2 SO 3 HClO 2 H 2 SiO 3 H 3 PO 4 H 2 SO 4 HClO 3 HClO 4 H 2 SeHBr HBrO H 3 AsO 3 H 2 SeO 3 HBrO 2 H 3 AsO 4 H 2 SeO 4 HBrO 3 HBrO 4

53 Other visual redox stuff:

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55 Zn Zn + Zn H O H H + Zn + e-e- H O H H + e-e-

56 Zn Zn + Zn Zn + H H

57 Oxidation-Reduction Reactions 0+1 +2 0 Use Oxidation # to ID Oxidized and Reduced Species Zn is Oxidized (Reducing Agent) to Zn 2+ H + is Reduced (Oxidizing Agent) to H 2

58 VO 3 -  VO 2 + CrO 2 -  CrO 4 2- SO 3  SO 4 2- NO 3  NO 2 - Identify Half-Rxn (Ox or Red) Neither: Lewis Acid/Base Oxidation Reduction +5 +3+6 +3

59 2008 Local 38. For a stoichiometric mixture of reactants, which statement best describes the changes that occur when this reaction goes to completion? 0 +5 +2 +5 +4 Zn + 4 HNO 3 → Zn(NO 3 ) 2 + 2 NO 2 + 2 H 2 O (A)All of the zinc is oxidized and some of the nitrogen is reduced. (B)All of the zinc is oxidized and all of the nitrogen is reduced. (C)Some of the zinc is oxidized and all of the nitrogen is reduced. (D)Some of the zinc is oxidized and some of the nitrogen is reduced.

60 Cu e-e- Cu + O H H O HH O H H O H H O H H Cl - H+H+ e-e- e-e-


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