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Week 6.1 oxidation & reduction

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  • 1. Chapter 3 Oxidation and Reduction Oxidation-Reduction Concepts Electrochemical Cells Voltaic Cell Electrolytic Cell Prepared by: Mrs Faraziehan Senusi PA-A11-7C Corrosion & Prevention Reference: Chemistry: the Molecular Nature of Matter and Change, 6th ed, 2011, Martin S. Silberberg, McGraw-Hill
  • 2. Oxidation-Reduction • ‘Oxidation’ : combination of a substance with oxygen  results in an increase in the oxidation number of an element in that substance  Example:  formation of rust, Fe2O3, iron(III) oxide 4Fe (s) +3O2 (g)  2Fe2O3 (s)  Combustion reactions C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) ‘loss of electron by an element increase in oxidation number’
  • 3. Oxidation-Reduction • ‘Reduction’ : removal of oxygen from a compound  results in an decrease in the oxidation number of an element in that substance  Example:  tungsten used in light bulb filaments can be prepared by reduction of tungsten(VI) oxide with hydrogen at 1200°C WO3 (s) +3H2 (g) W(s) +3H2O (g) ‘gain of electron by an element decrease in oxidation number’
  • 4. Oxidation-Reduction • Oxidation-Reduction Concepts  Oxidation – Loss of Electrons (LEO)  Reduction – Gain of Electrons (GER)  Oxidizing Agent – Species that causes another species to be oxidized (lose electrons) Oxidizing agent is reduced (gains e-)  Reducing Agent – Species that cause another species to be reduced (gain electrons) Reducing agent is oxidized (loses e-) • Oxidation (e- loss) always accompanies Reduction (e- gain) • Total number of electrons gained by the atoms/ions of the oxidizing agent always equals the total number of electrons lost by the reducing agent
  • 5. Oxidation-Reduction A summary of redox terminology :
  • 6. Oxidation-Reduction Oxidation Number – Also called ‘oxidation state’, signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely. – The oxidation number in a binary ionic compound equals the ionic charge – The oxidation number for each element in a covalent compound (or polyatomic ion) are assigned according to the relative attraction of an atom for electrons
  • 7. Oxidation-Reduction (ions composed of only one atom)
  • 8. Oxidation-Reduction Some guideline to assign oxidation numbers : ON is -1 when it combines with elements that are less electronegative that itself)
  • 9. Lets try!! Oxidation-Reduction Assign oxidation numbers to all the elements in the following compounds and ion: (a) Li2O (b) HNO3 (c) Cr2O7 2– Answer: By rule 2, Li has ON = +1 (Li+) and Oxygen has ON = -2 (O2-) Formula for nitric acid, yields a H+ ion and NO3– ion in solution. By rule 4, H has ON = +1. So, nitrate ion must have ON = -1. Oxygen has ON = -2, [N(x) O3 (2-)]– So, x + 3(-2) = -1 and x = +5 (c) By rule 6, sum of ON in dichromate ion, Cr2O7 2– = -2. Oxygen ON = -2, [Cr2(y) O7 (2-)]2– So, 2(y) + 7(-2) = -2 and y = +6 (a) (b)
  • 10. Oxidation-Reduction Problem Assign oxidation numbers to all the elements in the following compounds and ion: a) Sulfur trioxide, SO3 b) BrO2c) Manganate ion, MnO42d) 4Fe + 3O2  2Fe2O3 e) WO3 (s) +3H2 (g) W(s) +3H2O (g) David P. White Prentice Hall © 2003
  • 11. Redox Reactions • Balancing Redox Reactions  Oxidation Number Method  Half-Reaction Method • The balancing process must insure that: The number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent
  • 12. Redox Reactions Oxidation Number Method  Assign oxidation numbers to all elements in the reaction  From changes in oxidation number of given elements, identify oxidized and reduced species  For each element that undergoes a change of oxidation number, compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number change (Draw tie-lines between these atoms)  Multiply one or both these number by appropriate factors to make the electrons lost equal to the electrons gained  Use factors as coefficients in reaction equation
  • 13. Practice Problem Balance equation with Oxidation Number method: Cu(s) + HNO3 (aq)  Cu(NO3 )2 + NO2 (g) + H2O(l) Balance the equation Cu(s) + 4HNO3 (aq)  Cu(NO3 )2 + 2NO2 (g) + 2H2O(l)
  • 14. Redox Reactions • Half-Reaction Method – Applicable to Acid or Base solutions – Separate the oxidation & reduction steps, reflects their actual physical separation in electrochemical cells – Does not usually require Oxidation Numbers (ON) • Procedure  Divide the overall reaction into:  Oxidation Half-Reaction  Reduction Half-Reaction  Balance each half-reaction for atoms & charge  Atoms balanced in order: atoms other than O and H, then O, then H  Charge balanced by adding e Multiply one or both reactions by some integer to make electrons gained equal to electrons lost  Recombine to given balanced redox equation
  • 15. Half-Reaction Method in a “Acidic” solution # When occur in acidic solution, H2O molecules and H+ are available for balancing • Redox Half-Reaction Method – Example Cr2O 7 2- (aq) + I - (aq)  Cr +3 (aq) + I 2 (s)  Divide steps into Half-Reactions
  • 16. Half-Reaction Method in a “Acidic” solution  Balance Atoms & Charges for Cr2O72- / Cr3+ * Balance atoms other than O and H * Add 7 Water molecules to balance Oxygen * Add 14 H+ ions on left to balance 14 H on right Add 6 electrons (e-) on left to balance reaction charges  Balance Atoms & Charges for I-/ I2 No need to add H2O or H+ Add 2 electrons (e-) on right to balance reaction charges
  • 17. Half-Reaction Method in a “Acidic” solution  Multiply each half-reaction, if necessary, by an integer to balance electrons lost/gained  2 e- lost in oxidation reaction and 6 e- gained in reduction   Multiply oxidation half-reaction by 3  Add 2 half-reactions together # check the atoms and charges balances
  • 18. Half-Reaction Method in a “Basic” solution # When occur in basic solution, add OH– ion to both sides of the equation for every H+ ion present  Mn 7  O4 2  - (aq) +   3 2 2C2 O 4  Mn 4 O 2 (s) 2  4 + C O3 Half-Reactions  Multiply each reaction by appropriate integer  2- (aq)
  • 19. Half-Reaction Method in a “Basic” solution  Add reactions  Add OH- to neutralize H+ , balance H2O, and form “basic” solution 2MnO 4(aq)+ 3 C 2O 4 2- + 4OH(aq) 2MnO 2 (s) + 6CO 3 2- + 2H 2O (aq) (aq) (l) # check the atoms and charges balances