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A redox reaction is one in which the reactants’ oxidation numbers change.

+8. +8. +6. +6. +4. +4. +2. +2. 0. 0. -2. -2. -4. -4. -6. -6. A redox reaction is one in which the reactants’ oxidation numbers change. What are the oxidation numbers of the metals in the reaction below?. oxidized. +3. +2. +2. +4. 2 FeCl 3 + SnCl 2  2 FeCl 2 + SnCl 4.

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A redox reaction is one in which the reactants’ oxidation numbers change.

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  1. +8 +8 +6 +6 +4 +4 +2 +2 0 0 -2 -2 -4 -4 -6 -6 A redox reaction is one in which the reactants’ oxidation numbers change. What are the oxidation numbers of the metals in the reaction below? oxidized +3 +2 +2 +4 2 FeCl3 + SnCl2 2 FeCl2 + SnCl4 The iron’s charge has become more negative, so it is reduced. The tin’s charge has become more positive, so it is oxidized. reduced Sn Fe

  2. Which reactant is oxidized and which is reduced in the following? +2 +4 +5 0 -1 -1 Synthesis 1) Cl2 + SnCl2 SnCl4 oxidized reduced 0 +1 +2 0 Single-replacement 2) Cu + 2AgNO3 Cu(NO3)2 + 2Ag oxidized Note that a redox reaction can also be another type of reaction. reduced -2 -1 0 Decomposition 3) 2KClO3 2KCl + 3O2 KClO3 oxidized +1+ x + 3(–2) = 0 reduced x = +5

  3. Balancing Redox Reactions +7 +3 +4 (acidic conditions) Rxn: MnO4– + C2O42– Mn2+ + CO2 1. Determine the oxidation number of the redox active species. 2. Split the redox reaction into two half reactions (remember OIL RIG) and add in the #e– being transferred. Red: 2 8 H+ + 5e– + MnO4– Mn2+ + 4 H2O Ox: 5 C2O42– 2 CO2 + 2e– + 16 H+ + 2 MnO4– + 5 C2O42– 2 Mn2+ + 10 CO2 + 8 H2O 3. Balance the oxygens with H2O and the hydrogens with H+. 4. Multiply both half-reactions by a factor that will make the #e– transferred equal. 5. Add the two half-reactions to get the final balanced redox rxn.

  4. Balancing Redox Reactions Under Basic Conditions 1. Balance reaction using acidic conditions. 16 H+ + 2 MnO4– + 5 C2O42– 2 Mn2+ + 10 CO2 + 8 H2O +16 OH– +16 OH– 8 16 H2O 2. Add OH– to both sides to neutralize the H+. 3. Cancel out excess water and rewrite reaction equation. 8 H2O + 2 MnO4– + 5 C2O42– 2 Mn2+ + 10 CO2 + 16 OH–

  5. Voltaic (alt. Galvanic) Electrochemical Cells Ecell = EA + EC ZnZn2+Cu2+Cu Ecell = 0.763V + 0.153 V AnodeCathode 0.92 0.916 V Salt bridge EA = +0.763 V EC = +0.153 V Oxidation takes place at the anode. Reduction takes place at the cathode.

  6. Example: What are the standard cell potentials for the folllowing? Al3+ + 3e–  Al Eo= –1.66 V Ag+ + e–  Ag Eo= +0.80 V Br2 + 2e–  2Br– Eo= +1.06 V 1. AlAl3+Ag+Ag Ox: Al  Al3+ + 3e– +1.66 V +2.46 V Red: Ag+ + e–  Ag +0.80 V 2. AgAg+Br2Br– Ox: Ag  Ag+ + e– –0.80 V +0.26 V Red: Br2 + e–  2 Br– +1.06 V Note: Voltaic cells ALWAYS have positive cell emf’s (voltages).

  7. Some terminology emf– Electromotive force; force causing e– to move SHE– Standard hydrogen electrode; Eo = 0 V by definition. Standard conditions– Solution concentrations = 1 M and gas pressures = 1 atm Oxidizing agent– is reduced during redox rxn Reducing agent– is oxidized during redox rxn Faraday’s constant = 96485 C/mol e– transferred R = 8.314 J/mol·K 1 A = 1 C/s (C  Coulomb = a quantity of charge) 1 V = 1 J/C

  8. Good oxidizing agent = large, positive reduction potential Table of Standard Reduction Potentials  SHE Good reducing agent = large, negative reduction potential

  9. A voltaic cell generates current as a result of a spontaneous redox reaction. The equations relating E, DG and K are: DG = -nFE lnK = -DG/RT = nFE/RT R = 8.314 J/mol·K, T = temperature in Kelvin, F = 96485 C/mol n = # electrons transferred in reaction 20.58 What are DG and K for: 2VO2+ + 4H+ + 2Ag  VO2+ + 2H2O + 2Ag+ n = 2 e– Ecell = EC + EA = 1.00 V + (–0.799 V) = 0.201 V (Remember 1 V = 1 J/C) DG = -nFE = -(2 e–)(96485 C/mol e–)(0.201 V) = -38787 J/mol K = e–DG/RT = e(38787 J/mol)/(8.314 Jmol*298K)= e15.7 = 6294875

  10. Operating under non-standard conditions: Nernst Eq. = 20.50 What is the cell potential for the following if [Ce4+]=2.0M, [Ce3+] = 0.010 M and [Cr3+] = 0.010 M? 3Ce4+ + Cr(s)  3Ce3+ + Cr3+ n = 3e– OX RED Eo = EC + EA = +1.61 V + +0.74 V = 2.35 V = 2.53 V E = 2.35 V – (0.0592/3)log [0.010 M]3[0.010 M] Because [reactants]>>[products] AND the reaction has a large positive Eo, the emf has increased to produce more product. [2.0 M]3

  11. Electrolysis: Decomposition of a compound by passing electricity through it. 20.80 Metallic magnesium can be made by the electrolysis of molten MgCl2. What mass of Mg is formed by passing a current of 5.25 A through molten MgCl2 for 2.50 days? • Calculate total time: • Calculate total charge passed = 5.25 C/s * 216000 s = 1134000 C • Divide by 2 because it takes 2e–/equiv Mg2+= 567000 C • Calculate mol Mg = 567000 C /(96485 C/mol) = 5.877 mol Mg(s) • Calculate grams Mg = 5.877 mol Mg(s) * 24.305 g/mol = 142.8 g Ans: 143 g of Mg will be formed

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