Chapter 24 Oxidation-reduction Reactions

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1 Chapter 24 Oxidation-reduction Reactions 24-1 Oxidation states First, let s dispense with writing oxidation-reduction all the time. Like many chemistry, let s call these redox reactions In Chapter 10 we looked at the Lewis formula of a couple of compounds to figure out which atoms gained or lost electrons in a redox reaction. Having to write out Lewis formulas for all compounds in a reaction is cumbersome and time consuming, so we will now learn a set of rules that will help us to assign oxidation states to help us locate which atoms gain or lose electrons. As you learn how to assign oxidation state, please keep in mind that the idea of an oxidation state is a simple booking trick that helps us to identify the material getting oxidized or reduced. It does not necessarily equal an actual charge on at atom. The oxidation state for monoatomic ions is easy to do. The oxidation state of the atom is equal to the charge on the ion. The oxidation state in covalent compounds is found by arbitrarily assigning electrons that are actually shared to one atom of a pair. When the electrons are shared between two identical atoms, they are split evenly between the two atoms. When they shared between two different atoms, the electrons are assigned to the more electronegative atom. A good example of that is water. Remember water is a covalent molecule. The covalent bond between the Hydrogen and the oxygen comes from a sharing of one electron from O and one electron from H. While these electrons are shared between the two atoms (They orbit both atoms, not just one or the other), for the oxidation state of this molecule we will assume that oxygen takes the electrons from both of the hydrogens and keeps them to itself. Thus the O has a -2 oxidation state (Because it has 2 more electrons than it needs) while the hydrogens have a 1 oxidation state (because each has 1 less electron than it needs, and so has a 1 charge)

2 Key Concept: Six rules for assigning oxidation states (a little simpler than rule from text) 1. Oxidation state for an atom in an element is 0 (examples Na(s), O 2, O 3, Hg) 2. Oxidation state of a monoatomic ion is the same as it s charge ( Na 1, Cl - 1,) 3. Oxygen is assigned an oxidation state of -2 in covalent compounds. (H2O, 2- CO 2) An exception occurs in peroxides O 2, where the oxidation state is -1 (Prime example H2O 2) 4. In covalent compounds with nonmetals, H is given a 1 NH 3, CH4 5. Fluorine, i.e., always Sum of oxidation state must be zero for an electrically neutral species. For a charged ion the sum of the oxidation state must be equal to the charge of the ion These rules aren t perfect but they should apply for the problems you will get in this class. Example Problems: NaCl Ionic compound, Na and Cl Oxidation state of ionic ions is the same as the charge of the ion Na = 1, Cl=-1 CO 2 O is -2, two of them, for a net of -4, compound is neutral so C must be 4 here SF 6 F is -1, six of them for -6, so S must be 6 NO - 3 O is -2 times 3 = -6, net charge is -1 so N must be 5 CH 4 H is 1, so C must be -4 Clicker questions: Following two examples from the lab KMnO 4 Break into ions; K & MnO 4 K so K=1 - MnO, O is -2, so Mn must be 7 to make net -1 charge 4 H2C2O4 Break into ions H and C O H so H=

3 -2 C2O 4, O= -2, but there are 4 for a total of -8 Net charge on ion is -2 so there must be a net of 6 on the C For 3 on each C Note an interesting case of applying the rule to Fe3O 4 Magnetite (the iron ore that is magnetic) The oxygen gives a -8, so the iron should be 8/3 state! This is not a true oxidation state but occurs because of the way we apply the rules. In this case there are 2 Fe 3 and 1 Fe 2 for a net effect of 8/ Oxidation-Reduction Reactions As we saw in chapter 10 redox reaction involve the transfer of electrons from one atom to another, so they are sometimes called electron transfer reactions Recall from the last chapter that The material that loses electrons, gain charge, is being oxidized, and is considered the reducing agent in the reaction The material that gains electron...loses charge...is being reduced...is considered to be the oxidizing reagent in the reaction. Now that we have oxidation states it is much easier to come up with these species Example: Fe2O 3(s) 3CO(g) 2Fe(s) 3CO 2(g) Oxidation states So Fe goes from 3 to 0 so it is being reduced and Fe O (s) is the oxidizing agent and C goes from 2 to 4 so it is being oxidized and CO(g) is a reducing agent

4 24-3 Half reactions Key Concept: All redox equations can be split into two half reactions, one that represents the oxidation and the other that represents the reduction Example: 2 2 Zn(s) Cu (aq) Zn (aq) Cu(s) has two ½ reactions 2 Zn(s) Zn (aq) 2e oxidation ½ reaction Zn is being oxidized Zn oxidation # Electrons on right-hand side of equation 2 Cu (aq) 2e Cu (s) Reduction ½ reaction Cu is being reduced Cu oxidation # Electrons on Left-hand side of equation Break redox equation into ½ reactions is the key to balancing redox equations 24-4 Balancing Oxidation-Reductions reactions in Acidic and Neutral solutions Balancing redox reactions is trickier than balancing non-redox reaction because you have to balance the electrons in the ½ reactions. So let s learn how to do this. Key Concept: Rules for balancing redox reactions in Neutral or Acidic conditions 1. Write separate equations for reduction and oxidation half reactions 2. For Each half reactions a. First Balance all elements except O and H b. Add H O to balance O 2 c. Add H to balance H d. add electrons to balance net charge 3. If necessary multiply one or both ½ reaction equations by integer so the total number of electrons used in one reaction is equal to the total number of electrons furnished by the other. 4. Add the two ½ reaction equations together and cancel any common terms 5. Double check that all species and charge balance.

5 Let s try this with the reaction VO Sn V Sn Step 1 break into ½ reactions Sn Sn VO V Step 2 a,b,c Balance atoms OK Step 2 d Balance electrons Sn Sn 2e 2 3 VO V H2O 2 3 VO 2H V H O VO 2H e V H O Step 3 Make # of electrons in ½ reaction match Multiply V equation by VO 4H 2e 2V 2H2O Step 4 Add equation together Sn 2VO 4H Sn 2V 2H O Let s try another one: - Ag NO Ag NO Separate Ag Ag 3 Balance all but H and O (Ok as is) Balance O with water Ag Ag 3 NO NO NO NO 2H O Balance H with H Ag Ag Balance charge with electrons Ag Ag e NO 4H NO 2H O NO 4H NO 2H O (4,-1 = 3 on left, 0 on right, need 3- on left - NO 4H 3e NO 2H O Multiply so electrons are even in two half rxns - (Ag Ag e ) x 3 NO 3 4H 3e NO 2H2O - 3Ag 3Ag 3e NO 4H 3e NO 2H O

6 Sum 2 half rxns - 3Ag NO 3 4H 3Ag NO 2H2O Double check that it s all balanced (it is) Let s try one from the lab H C O MnO CO Mn Separate: H C O CO MnO Mn Balance all but H and O H C O 2 CO MnO Mn Balance O with H2O H C O 2 CO MnO Mn 4H O Balance H with H H C O 2 CO 2 H 8H MnO Mn 4H O Balance Charge with e - H C O 2 CO 2 H 2e (Reduction ½ rxn) 8H MnO 5e Mn 4H O (Oxidation ½ rxn) Multiply the oxidation by 5 and the reduction by 2-2 5H C O 10 CO 10 H 10e 16H 2MnO 10e 2Mn 8H O Add and check - 2 5H C O 16H 2MnO 10e 10 CO 10 H 10e 2Mn 8H O Remove common terms 2 5H C O 6H 2MnO 10 CO 2Mn 8H O Balancing Oxidation-Reductions reactions in Basic solutions Occasionally you will get a problem that specifies that the reaction takes place in - a basic solution. What does this mean (ph > 8, OH predominates, H concentration is insignificant.) What would this do to the procedure we just did?? Can t use H as part of the reaction because there isn t any around! How would we balance this? - If you try to add OH directly to balance the equations it is easy to get it all screwed up because you add oxygen and throw the oxygen balance off. McQuarrie has a way around this that works, but I find a little counterintuitive.

7 Let s try this method instead. The trick is to solve it just as you did before, but - - just before the final answer, add OH to both sides of the equation so the OH is equal to any H and they combine to form water. Key Concept: To balance redox equation under basic conditions add one last step New step (insert between step 4 and 5) Count up the number of H in your final equation - Add an equal number of OH to BOTH side of the equation The H and OH become water on one side of the equation - The OH remains on the other side of the equation Double check to see if any of the new waters can be canceled out Example: Ag(s) CN (aq) O (g) (Basic) Ag(CN) ½ reactions? Ag(s) CN (aq) Ag(CN) - - O (g) H O (Notice this is a bit tricky because water was not in the original equation, but remember how we balance oxygen by adding water? Mass balance - - Ag 2CN Ag(CN) O (g) 2H O; 4H O (g) 2H O Charge balance Ag 2CN Ag(CN) e 2 2 4H O 4e 2H O Multiply by 4 4Ag 8CN 4Ag(CN) e 4H O 4e 2H O Sum - 4Ag 8CN 4H O 4Ag(CN) 2H O Now remove H by adding OH 4Ag 8CN 4H O 4 OH 4Ag(CN) 2H O 4OH Ag 8CN O 2 4H2O 4Ag(CN) 2 2H2O 4OH Notice that at this point you have 2 moles of water as a reactant and product, and these two moles could be removed Ag 8CN O 2H O 4Ag(CN) 4OH

8 24-6 Oxidation-Reduction Reactions and Chemical Analysis Skip for now 24-7 Corrosion Now that you have the basics, I think I will skip these details. However if you are curious about how rust forms or how galvanizing iron prevents rust, take a look at section 24-7 It is a pretty quick read.