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Fundamentals K: Oxidation and Reduction

Fundamentals K: Oxidation and Reduction

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Fundamentals K: Oxidation and Reduction

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  1. Fundamentals K: Oxidation and Reduction • To date, we have look at 2 specific types of chemical reaction: • Acid-Base Reactions • Precipitation Reactions There is one more reaction type we need to consider: Oxidation-Reduction or Redox Reactions

  2. Oxidation-Reduction Reactions Let’s look at the reaction of magnesium metal with molecular oxygen: Mg (s) + O2 --> MgO (s) But we know that magnesium oxide is an ionic solid, so its chemical formula is actually: Mg2+ + O2- The elemental magnesium became OXIDIZED or LOST ELECTRONS after reactions with molecular oxygen. The oxygen atoms GAINED ELECTRONS or became REDUCED

  3. Oxidation-Reduction Reactions Let’s look at the reaction of magnesium metal with molecular chlorine in a similar manner: Mg (s) + Cl2 (g) --> MgCl2 (s) But we know that magnesium chloride is an ionic solid, so its chemical formula is actually: Mg2+ + 2Cl- The elemental magnesium became OXIDIZED or LOST ELECTRONS after reactions with molecular chlorine. The chlorine atoms GAINED ELECTRONS or became REDUCED

  4. Oxidation-Reduction Reactions LEOis GERman Loss of Electrons is Oxidation Gain of Electrons is Reduction (Its not politically correct, but it works)

  5. Oxidation-Reduction Reactions Whenever a species is oxidized, another species MUST be reduced Because of this phenomenon, these types of reactions are frequently called Redox reactions

  6. How do we know what is taking place in a Redox Reaction? Oxidation Numbers An oxidation number is a value given to an atom that indicates its redox state and allows us to track the atom during a chemical reaction See Toolbox K.1: Assigning Oxidation Numbers

  7. Oxidation Numbers: Rules Summary The oxidation number of an element in its elemental form is 0 The oxidation number of a monatomic ion is equal to its charge Hydrogen has two possible oxidation numbers: -1 if bonded to a metal and +1 in nonmetal compounds Oxygen’s oxidation number is almost always -2 in most compounds Oxidation increases the oxidation number of an element Reduction decreases the oxidation number of an element

  8. Oxidizing and Reducing Agents The species that causes oxidation is an oxidizing agent. IT IS REDUCED The species that brings about a reduction is a reducing agent. IT IS OXIDIZED

  9. Oxidizing and Reducing Agents Cu2+ is _______ to Cu. It is the _________ agent. Zn is _______ to Zn2+. It is the _________ agent. Zn(s)+ Cu2+ (aq) ------>Zn2+ (aq) + Cu (s)

  10. Balancing Redox Reactions Remember hearing several times during the course of the semester about how a reaction isn’t balanced until the charges are balanced? Here’s where that really becomes important. When balancing the chemical equation for a redox reaction involving ions, the total charge on each side must be balanced.

  11. Examples i) Which species have been oxidized and which species have been reduced in the reaction: 2Cu+ (aq) + I2 (s) ---> 2Cu2+ (aq) + 2I- (aq) To solve: Calculate the oxidation #’s Find the oxidation numbers of sulfur, nitrogen and chlorine in: a) SO32- b) NO2- c) HClO3

  12. Examples In an aqueous solution, Cerium (IV) ions oxidize iodide ions to solid diatomic iodine and are themselves reduced to Cerium (III) ions. Write the net ionic equation for the reaction. To solve: What does the problem tell us? Balance the charges Check your work

  13. Examples A mixture of 5.00g of Cr(NO3)2 and 6.00 g of CuSO4 is dissolved in sufficient water to make 25.00 mL of solution. In the reaction, copper metal is formed and each chromium ion loses one electron. Write the net ionic equation What is the number of electrons transferred in the balanced equation written with the smallest whole number coefficients? What are the molar concentrations of the two anions in the final solution?

  14. Chapter 12: Electrochemistry Expressing Redox Reactions in terms of Half Reactions • Complicated redox reactions require us to think of the reactions in terms of the 2 half reactions occurring: oxidation and reduction • A Half Reaction is the oxidation or reduction part of the reaction considered alone Complete reaction: Zn (s) + 2Ag+ (aq) --> Zn2+ (aq) + 2Ag (s) Oxidation half reaction: Zn (s) --> Zn2+ (aq) + 2e- Reduction half reaction: Ag+ (aq) + 1e- --> Ag (s)

  15. Balancing Redox Reactions Because of differences in charge between reactants and products, it is not always possible to balance redox reactions using simple stoichiometric coefficients. For example, Au3+ (aq) + I- (aq) --> Au (s) + I2 (s) At first glance, it seems that this equation can be balanced by placing a 2 in front of the iodide. Au3+ (aq) + 2I- (aq) --> Au (s) + I2 (s) The stoichiometry looks right, but is the charge?

  16. Balancing Redox Reactions We know that this is a redox reaction, so let’s examine the half reactions: 2I- (aq) --> I2 (s) + 1e- Au3+ (aq) + 3e- --> Au (s) The half reactions give us an indicator of what is going on here. Iodide is losing a single electron, but in order to reduce a single gold ion to elemental gold, we need 3 electrons. We would need 6 total iodide ions in order to reduce a single gold ion, right? 2Au3+ (aq) + 6I- (aq) --> 2Au (s) + 3I2 (s)

  17. Balancing Redox Reactions: The Rules We will deal with 3 types of redox reactions. Straight Reactions: These reactions can be balanced by simply evaluating the oxidation numbers of the reactants and products Redox reactions that occur in acidic conditions: Just as the name implies. Use the rules on the next slide. Redox reactions that occur in basic conditions: Hmm. What do you think this means. Use the rules for acidic reactions WITH ONE ADDITIONAL STEP AT THE END

  18. Balancing Redox Reactions by Oxidation Number Step 1: Try to balance the atoms in the equation by inspection, that is, by the standard technique for balancing non-redox equations. (Many equations for redox reactions can be easily balanced by inspection.) If you successfully balance the atoms, go to Step 2. If you are unable to balance the atoms, go to Step 3. Step 2: Check to be sure that the net charge is the same on both sides of the equation. If it is, you can assume that the equation is correctly balanced. If the charge is not balanced, go to Step 3. Step 3: If you have trouble balancing the atoms and the charge by inspection, determine the oxidation numbers for the atoms in the formula, and use them to decide whether the reaction is a redox reaction. If it is not redox, return to Step 1 and try again. If it is redox, go to Step 4. Step 4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced. Step 5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number (a ratio that makes the number of electrons lost equal to the number of electrons gained). Step 6: Add coefficients to the formulas so as to obtain the correct ratio of the atoms whose oxidation numbers are changing. (These coefficients are usually placed in front of the formulas on the reactant side of the arrow.) Step 7: Balance the rest of the equation by inspection.

  19. Balancing Redox Reactions that occur in Acidic conditions Step 1: Write the skeletons of the oxidation and reduction half-reactions. (The skeleton reactions contain the formulas of the compounds oxidized and reduced, but the atoms and electrons have not yet been balanced.) Step 2: Balance all elements other than H and O. Step 3: Balance the oxygen atoms by adding H2O molecules where needed. Step 4: Balance the hydrogen atoms by adding H+ ions where needed. Step 5: Balance the charge by adding electrons, e-. Step 6: If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number of electrons lost. Step 7: Add the 2 half-reactions as if they were mathematical equations. The electrons will always cancel. If the same formulas are found on opposite sides of the half-reactions, you can cancel them. If the same formulas are found on the same side of both half-reactions, combine them. Step 8: Check to make sure that the atoms and the charges balance.

  20. Balancing Redox Reactions that occur in Basic conditions Steps 1-7: Begin by balancing the equation as if it were in acid solution. If you have H+ ions in your equation at the end of these steps, proceed to Step #8. Otherwise, skip to Step #11. Step 8: Add enough OH- ions to each side to cancel the H+ ions. (Be sure to add the OH- ions to both sides to keep the charge and atoms balanced.) Step 9: Combine the H+ ions and OH- ions that are on the same side of the equation to form water. Step 10: Cancel or combine the H2O molecules. Step 11: Check to make sure that the atoms and the charge balance. If they do balance, you are done. If they do not balance, re-check your work in Steps 1-10.

  21. Balancing Redox Reactions: Examples Balance the following redox equation: HNO3 (aq) + H3AsO3 (aq) --> NO (g) + H3AsO4 (aq) + H2O (l) Step 1: Can we balance it eyeballometrically? Maybe, but it will take a while. The hydrogens are problematic. Step 2: Are we sure it is a redox reaction? Check the oxidation numbers. Step 3: Find a shared number between the two elements such that the gained electron number equals the number of lost electrons. Step 4: Place stoichiometric coefficient in front of the appropriate chemical formula. Step 5: Balance the rest of the equation by inspection.

  22. Balancing Redox Reactions: Examples Balance the following redox equation: Cu(s) + HNO3 (aq) --> Cu(NO3)2 (aq) + NO (g) + H2O (l) Step 1: Can we balance it eyeballometrically? Nicht. Step 2: Are we sure it is a redox reaction? Check the oxidation numbers. Step 3: Find a shared number between the two elements such that the gained electron number equals the number of lost electrons. Step 4: For chemical formulas of compounds involved in redox couples (NOT HNO3), put the coefficients of the common electron count from step 3. Step 5: Balance the rest of the equation by inspection.

  23. Balancing Redox Reactions: Examples Balance the following redox equation: NO2 (g) + H2 (g) --> NH3 (g) + 2H2O (l) Step 1: Can we balance it eyeballometrically? Si. Easy peasy. Step 2: Done!

  24. Balancing Redox Reactions: Acid Conditions Balance the following redox equation: Cr2O72- (aq) + HNO2 (aq) --> Cr3+ (aq) + NO3- (aq) (acidic) Step 1: We need to identify the 2 half reactions involved in the redox reaction Oxidation numbers: Cr: +6 --> +3 Gain of 3 e-, Reduction N: +3 --> +5 Loss of 2 e-, Oxidation Step 2: Write the skeletal half reactions and: i) Balance all atoms other than H and O ii) Balance O by adding H2O to one side iii) Balance H by adding H+ to the other side iv) Balance charge by adding electrons to the side that needs it

  25. Balancing Redox Reactions: Acid Conditions Balance the following redox equation: Cr2O72- (aq) + HNO2 (aq) --> Cr3+ (aq) + NO3- (aq) (acidic) Oxidation Reaction: Cr2O72- (aq) ----> Cr3+ (aq) Cr2O72- (aq) ----> 2Cr3+ + 7H2O Balance Cr and O Cr2O72- (aq) + 14H+ ----> 2Cr3+ + 7H2O Balance H Cr2O72- (aq) + 14H+ + 6e- ----> 2Cr3+ + 7H2O Balance charge Reduction Reaction: HNO2 (aq) ---> NO3- (aq) HNO2 (aq) + H2O ---> NO3- (aq) HNO2 (aq) + H2O ---> NO3- (aq) + 3H+ HNO2 (aq) + H2O ---> NO3- (aq) + 3H+ +2e-

  26. Balancing Redox Reactions: Acid Conditions Balance the following redox equation: Cr2O72- (aq) + HNO2 (aq) --> Cr3+ (aq) + NO3- (aq) (acidic) Step 3: Multiply the half reaction with the lowest number of electrons by the stoichiometric coefficient to make the number of electrons in each half reaction the same. Cr2O72- (aq) + 14H+ + 6e- ----> 2Cr3+ + 7H2O HNO2 (aq) + H2O ---> NO3- (aq) + 3H+ +2e- Multiply by 3 Cr2O72- (aq) + 14H+ + 6e- + 3HNO2 (aq) + 3H2O ---> 2Cr3+ + 7H2O + 3NO3- (aq) + 9H+ +6e-

  27. Balancing Redox Reactions: Acid Conditions Cr2O72- (aq) + 14H+ + 6e- + 3HNO2 (aq) + 3H2O ---> 2Cr3+ + 7H2O + 3NO3- (aq) + 9H+ +6e- Step 4: Cancel the compounds found on both side of the arrow Cr2O72- (aq) + 5H+ + 3HNO2(aq)--> 2Cr3+ + 4H2O + 3NO3-(aq) Step 5: Check work. Done!

  28. Balancing Redox Reactions: Acid Conditions MnO4- (aq) + Br- (aq) ---> MnO2(aq) + BrO3-(aq) Step 1: Half reactions Mn: +7 --> +4 Gain 3e-, Reduction Br: -1 --> +5 Lose 6e-, Oxidation Step 2: Balance half reactions Oxidation: Br- (aq) --> BrO3- 3H2O + Br- (aq) --> BrO3- 3H2O + Br- (aq) --> BrO3- + 6H+ 3H2O + Br- (aq) --> BrO3- + 6H+ + 6e- Reduction: MnO4- --> MnO2 MnO4- --> MnO2 + 2H2O MnO4- + 4H+ --> MnO2 + 2H2O MnO4- + 4H+ + 3e- --> MnO2 + 2H2O

  29. Balancing Redox Reactions: Acid Conditions MnO4- (aq) + Br- (aq) ---> MnO2(aq) + BrO3-(aq) Step 3: We have 6 electrons transferred in the oxidation reaction and 3 electrons in the reduction reactions. Multiply the reduction reaction by 2 and combine with the oxidation reaction. 2MnO4- + Br- + 3H2O + 8H+ + 6e- --> BrO3- + 6H+ + 6e- + 2MnO2 + 4H2O Step 4: Cancel molecules found on both sides of the reaction arrow. 2MnO4- + Br- + 2H+ --> BrO3- + 2MnO2 + H2O Step 5: Done!

  30. Balancing Redox Reactions: Basic Conditions Cr(OH)3(s) + ClO3- (aq) ---> CrO42-(aq) + Cl-(aq) Step 1: Same as acidic conditions. Identify the 2 half reactions Cr: +3 --> +6 Lose 3 electrons, Oxidation Cl: +5 --> -1 Gain 6 electrons, Reduction Step 2: Balance the half reactions Oxidation reaction: Cr(OH)3(s) ---> CrO42- Cr(OH)3(s) + H2O ---> CrO42- Cr(OH)3(s) + H2O ---> CrO42- + 5H+ Cr(OH)3(s) + H2O ---> CrO42- + 5H+ + 3e-

  31. Balancing Redox Reactions: Basic Conditions Cr(OH)3(s) + ClO3- (aq) ---> CrO42-(aq) + Cl-(aq) Step 2 (cont’d): Balance the half reactions Reduction reaction: ClO3- (aq) ---> Cl- ClO3- (aq) ---> Cl- + 3H2O ClO3- (aq) + 6H+ ---> Cl- + 3H2O ClO3- (aq) + 6H+ + 6e- ---> Cl- + 3H2O Step 3: Multiply the oxidation reaction by 2 to get the same # of electrons and combine the equations 2Cr(OH)3(s) + 2H2O + ClO3- (aq) + 6H+ + 6e- --> Cl- + 3H2O + 2CrO42- + 10H+ + 6e-

  32. Balancing Redox Reactions: Basic Conditions 2Cr(OH)3(s) + 2H2O + ClO3- (aq) + 6H+ + 6e- --> Cl- + 3H2O + 2CrO42- + 10H+ + 6e- Step 4: Remove things on both sides of the reaction arrow 2Cr(OH)3(s) + ClO3- (aq) --> Cl- + H2O + 2CrO42- + 4H+ Step 5: Now for the tricky part. Count the number of protons and add the same number of hydroxide ions to BOTH sides. 2Cr(OH)3(s) + ClO3- (aq) + 4 OH- --> Cl- + H2O + 2CrO42- + 4H+ + 4 OH- OR 2Cr(OH)3(s) + ClO3- (aq) + 4 OH- --> Cl- + 5H2O + 2CrO42-