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Electrochemistry [email protected] Are you sure I can have that electron? Na Cl I’m positive! Na+ Cl- Useful Links • This presentation: 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 Typical Olympiad Topics • • • • • • Oxidation Numbers Balancing Redox Reactions Galvanic Cell Architecture Standard Cell Potentials Non Standard Conditions Electrolysis • Common Oxyanions • Redox Concepts From 2012 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 H2 Molecule O Atom in O2 Molecule P Atom in P4 Molecule Oxidation Numbers (States) 2) Monatomic Ions – – Oxidation Number = Charge Examples: K+ Oxidation Number = +1 Mg2+ Oxidation Number = +2 Al3+ Oxidation Number = +3 N3- Oxidation Number = -3 S2- Oxidation Number = -2 FeCl2 ON(Fe) = +2 Oxidation Numbers are written with the sign before the number to distinguish them from Actual Charges 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 (O22-) 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 Oxidation Numbers (States) 4) Sum of the Oxidation Number of All Atoms in a Neutral Compound is Zero H2SO3 Oxidation Number of H = +1 (H Bonded to NonMetal) Oxidation Number of O = -2 Sum is Zero 0 = 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 H2AsO4-1 = 2×ON(H) + 1×ON(As) + 4×ON(O) ON(As) = +5 From 2012 From 2014 K = +1 H = +1 O = -2 N = ??? From 2014 Split to ions K+ NH4+ H2O “hydrate” AsO43- 2012 Balancing Redox Reactions by Half-Reactions 2 Zn(s) 2H (aq) Zn (aq) H2 (g) Reduction Half Reaction (Electrons Taken In) 2H+(aq) + 2e- H2(g) Another Example: Ag+(aq) + e- Ag(s) Oxidation Half Reaction (Electrons Given Off) Zn(s) Zn2+(aq) + 2 e- Another Example: C2O42- (aq) 2 CO2(g) + 2 e- Half-Reactions To Balance Electrons, Reductions and Oxidations MUST Occur Simultaneously NO3-(aq) + 4 H+ + 3 e- NO(g) + 2 H2O(l) C2O42- (aq) 2 CO2(g) + 2 e- ID the Reducing Agent in the Unbalanced Reaction: ClO3- + Br- Cl2 + Br2 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 H2O c. Balance H by adding H+ d. Balance residual charge by adding e3. 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 H2O, then cancel out any excess. 2012 From 2014 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. 3 2 Cu(s) NO (aq) Cu (aq) NO(g) Unbalanced eeZn2+ Zn(s) → Zn2+(aq) + 2 e- ClCu2+ Cl- Cu2+(aq) + 2 e- → Cu(s) Voltaic (Galvanic) Cell Electrode Oxidation Electrode Reduction to the cathode to the anode 2008 Local 39. Which occurs at the anode of any voltaic cell? I. A metal electrode dissolves. SO32- + H2O → SO42- + 2 H+ + 2 eII. A substance undergoes oxidation. Fe(s) → Fe2+ + 2 eIII. Positive ions are deposited from the solution. (A) I only (B) II only (C) I and II only (D) I and III only Cell Potential • Voltaic Cell Spontaneous Redox Reaction (Ecell>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) (Ecell) • Volt Difference in Potential Energy per Electrical Charge (1V = 1J/C) (e- charge = 1.60x10-19C) • Potential Difference Between 2 Electrodes Standard Cell Potential T = 25°C Standard State Gas (P = 1atm) Species in Solution (1 M Concentration) E0cell =E0red (cathode) -E0red(anode) • Use Standard Reduction Potentials for the Reduction and Oxidation Half-Reactions • Note: No Multiplying Reduction Potential By Stoichiometry • Voltaic (Galvanic) Cell: Positive Ecell • Electrolytic Cell: Negative Ecell Most easily oxidized 2008 Local 41. What is the standard cell potential for the voltaic cell: Cr | Cr3+ || Pb2+ | Pb ? Pb2+ Cr3+ E0red / V + 2 e- → Pb + 3 e- → Cr (A) 1.09 (B) 0.61 (C) -0.61 (D) -1.09 -0.13 -0.74 Half Reaction Zn2+(aq) + 2e- Zn(s) Cr3+(aq) + e- Cr2+(aq) Tl+(aq) + e- Tl(s) Cu2+(aq) + e- Cu+(aq) Fe3+(aq) + e- Fe2+(aq) E0 (V) -0.763 -0.408 -0.336 +0.161 +0.769 Use the Standard Reduction Potentials to Find the Standard Cell Potential, E0cell, for the Reaction: Zn(s) + 2 Tl+(aq) Zn2+(aq) + 2 Tl(s) Zn(s) Zn2 (aq) 2e 2Tl (aq) 2e Tl(s) 0 cell E =E 0 red (cathode) -E 81Tl: Thallium Ered 0.763V (Anode/Oxidation) Ered 0.336V (Cathode/Reduction) 0 red(anode) -0.336V-(-0.763V) 0.427V Half Reaction Zn2+(aq) + 2e- Zn(s) Cr3+(aq) + e- Cr2+(aq) Tl+(aq) + e- Tl(s) Cu2+(aq) + e- Cu+(aq) Fe3+(aq) + e- Fe2+(aq) Calculate the E0rxn based on the standard reduction potentials above. Which reaction(s) is(are) spontaneous? Cr 2+ (aq) + Fe3+ (aq) Cr3+ (aq) + Fe2+ (aq) Cr 2+ (aq) Cr 3+ (aq) + e- (oxidation, E red = -0.408V) Fe3+ (aq) +e- Fe2+ (aq) (reduction, E red = +0.769V) E 0cell =E 0red (cathode) -E 0red(anode) 0.769V-(-0.408V) 1.177V Cu 2+ (aq) + Fe2+ (aq) Cu+ (aq) + Fe3+ (aq) Fe2+ (aq) Fe3+ (aq) + e- (oxidation, E red = +0.769V) Cu 2+ (aq) + e- Cu + (aq) (reduction, E red = +0.161V) E0cell =E0red (cathode) -E0red(anode) 0.161V-(0.769V) 0.608V E0 (V) -0.763 -0.408 -0.336 +0.161 +0.769 Ecell Non-Standard Conditions G n F E ΔG = ΔG 0 +RTln Q -nFE = -nFE 0 +RTln Q -nFE = -nF -nFE 0 +RTln Q -nF RT E=E ln Q nF 0 0.0592 V E=E log Q n 0 Nernst Equation 2012 An electrochemical cell is constructed with a piece of copper wire in a 1.00 M solution of Cu(NO3)2 and a piece of chromium wire in a 1.00 M solution of Cr(NO3)3. Half Reaction Cr3+(aq) + 3 e- Cr(s) Cu2+(aq) + 2 e- Cu(s) E0 (V) -0.744 +0.340 (1) Write a balanced equation for the spontaneous reaction that occurs in this cell and calculate the potential it produces. (2) Sketch a diagram for this cell. Label the anode. Show the direction of electron flow in the external circuit. Show the direction of movement of nitrate ions. Explain. (3) The cell is allowed to operate until the [Cu2+] = 0.10 M. Find the [Cr3+] (4) Calculate the cell potential at these concentrations. An electrochemical cell is constructed with a piece of copper wire in a 1.00 M solution of Cu(NO3)2 and a piece of chromium wire in a 1.00 M solution of Cr(NO3)3. Half Reaction Cr3+(aq) + 3 e- Cr(s) Cu2+(aq) + 2 e- Cu(s) E0 (V) -0.744 +0.340 (1) Write a balanced equation for the spontaneous reaction that occurs in this cell and calculate the potential it produces. 3Cu 2 (aq) 2Cr(s) 2Cr3 (aq) 3Cu(s) E 1.084V (2) Sketch a diagram for this cell. Label the anode. Show the direction of electron flow in the external circuit. Show the direction of movement of nitrate ions. Explain. An electrochemical cell is constructed with a piece of copper wire in a 1.00 M solution of Cu(NO3)2 and a piece of chromium wire in a 1.00 M solution of Cr(NO3)3. Half Reaction Cr3+(aq) + 3 e- Cr(s) Cu2+(aq) + 2 e- Cu(s) E0 (V) -0.744 +0.340 (3) The cell is allowed to operate until the [Cu2+] = 0.10 M. Find the [Cr3+] Assume Vol 1.00L Change in Moles Cu 2+ =1.00 mol(initial) - 0.10 mol(final) = 0.90 mol 2 mol Cr 3+ 0.90 mol Cu 0.60 mol Cr 3+ 2+ 3 mol Cu Change in Moles Cr 3+ =1.00 mol(initial) + 0.60 mol(produced by oxidation) = 1.60 mol 2+ Concentration Cr 3+ = 1.60M (4) Calculate the cell potential at these concentrations. 2 3+ Cr 0.0592 V 0 E=E log 3 n Cu 2+ 1.60 2 E = 1.084V- 0.0592 V log 1.050V 3 6 0.10 How many moles of electrons must pass through a cell to produce 5.00 kg of Aluminum from Al2O3? Al2O3 + 6 e- 2 Al(s) + 3 O21. Calculate the number of moles of electrons needed. 1 mol Al 6 e 5.00 103 g 26.9815 g 2 Al 556 mol e Using F = 96485 C/mol e-, and A=C/s, 2. How long will this take using a current of 33.5 A? 96485 C 1 A 1 6 556 mol e 1.60 10 s 18.5 days mol e C 33.5 A s Should also know that W = J/s. 2008 Local 42. During the electrolysis of AgNO3, 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. 2012 Local #42 How to memorize the negative ions: ______-ate BO33- CO32- NO3- SiO32- PO43- SO42- ClO3- AsO43- SeO42- BrO3- IO3- 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: NO3- is nitrate, PO43- is phosphate, ClO3- is chlorate Note, not all “-ate”s have a corresponding “-ite”! ______-ide Irregular* Also: H- N3- O2- F- P3- S2- Cl- As3- Se2- Br- I- Rule: Example: 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). N3- is nitride, O2- is oxide, F- is fluoride * Carbide is actually C22-. This does not follow the naming rules above, and you do not need to know this ion. ______-ite NO2- PO33- SO32- ClO2- AsO33- SeO32- BrO2- IO2- 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: NO2- is nitrite, PO33- is phosphite, ClO2- is chlorite Note, not all “-ate”s have a corresponding “-ite”! per-______-ate * O22- ClO4Also: MnO4- BrO4- IO4- 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: ClO4- is perchlorate * Adding one more oxygen to oxide gives O22-. The prefix rule follows with its own suffix for the name peroxide. 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 hydrogen -______-ide OH- HP2- 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. hydrogen ______-ate HCO3- HSiO3- HPO42- HSO4- HAsO42- HSeO4- Rule: Ion consisting of one H+ added to the “-ate” ion with 2- or greater charge. Example: HCO3- is hydrogen carbonate, HPO42- is hydrogen phosphate Important Note: H2PO4- is dihydrogen phosphate Corresponding Acids HF HNO2 H3BO3 H2CO3 HNO3 H2S HCl HClO H3PO3 H2SO3 HClO2 H2SiO3 H3PO4 H2SO4 HClO3 HClO4 H2Se HBr HBrO H3AsO3 H2SeO3 HBrO2 H3AsO4 H2SeO4 HBrO3 HBrO4 Other visual redox stuff: 2 Zn(s) 2H (aq) Zn (aq) H2 (g) - e+ e+ Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn H H Zn+ Zn+ Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Zn Oxidation-Reduction Reactions 2 +1 +2 Zn(s) 2 H (aq) Zn (aq) H2 (g) 0 0 Use Oxidation # to ID Oxidized and Reduced Species Zn is Oxidized (Reducing Agent) to Zn2+ H+ is Reduced (Oxidizing Agent) to H2 Identify Half-Rxn (Ox or Red) +5 +5 VO3- VO2+ +3 +6 CrO2- CrO42+6 Oxidation +6 SO3 SO42+6 Neither: Lewis Acid/Base Neither: Lewis Acid/Base +3 NO3 NO2- Reduction 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 HNO3 → Zn(NO3)2 + 2 NO2 + 2 H2O (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. e- O H ClH+ e- Cu+ H