SEATTLE CENTRAL COMMUNITY COLLEGE DIVISION OF SCIENCE AND MATHEMATICS. Oxidation-Reduction

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1 SEATTLE CENTRAL COMMUNITY COLLEGE DIVISION OF SCIENCE AND MATHEMATICS OxidationReduction Oxidation is loss of electrons. (Oxygen is EN enough to grab e away from most elements, so the term originally meant combination with O 2 ). Reduction is gain of electrons. (This term came from the process of reducing ores [usually metal oxides or sulfides] to metals usually with C). Remember: Oxidation increases charge. Reduction reduces charge. Oxidizing Agent the reactant (whole formula) that causes the oxidation of another reactant (thus accepts e [e acceptor, e on left of half reaction] & so itself contains an element that is reduced). Reducing Agent the reactant (whole formula) that causes the reduction of another reactant (thus donates e [e donor, e on right of half reaction] & so itself contains an element that is oxidized). DEFINE A NEW CONCEPT: Oxidation Number (ON) is the charge on an atom if all the electrons in a bond are arbitrarily assigned to the more EN atom in the bond. ON's can be or values. Simple Rules (from the definition) for Oxidation Numbers: A) The sum of all the ON's in a neutral molecule is zero; in a polyatomic ion it is the charge on the ion. B) The ON of an element in an ion by itself is the charge on the ion. C) The ON of any element in its uncombined or element state is zero. (Ne, P 4, F 2, S 8, Fe... all ON = 0 ) D) The ON of F is 1 except in F 2, since F is the most EN element. E) Hydrogen has ON = 1 except in metal hydrides or bonded to itself. (in LiH, H has ON = 1; in H 2, ON is zero). F) Oxygen has an ON =2 except when it is bonded to itself or F. (In OF 2, ON O =2; in O2 & O3, ON O =0; in peroxides, ON O =1; in superoxides, ONO=½) 1

2 A. Oxidation Number Method Balancing OxidationReduction Equations 1. Write the unbalance equation with correct formulas for reactants and products. 2. Assign ON and find change ( ) in ON for each kind of atom. (2½. If an element oxidized or reduced has a subscript, balance it and change the ON by multiplying by the # of that atom.) 3. Balance the atoms oxidized and reduced so that the ON for whatever is oxidized equals in size the ON for whatever is reduced. 4. Balance the other atoms. (If the reaction is in water, at the end use H 2 O to balance O's and then H to balance H's). 5. Check to see that all charges and all atoms balance. 6. (If problem is stated to be in basic solution, add the # of OH ions to each side equal to the # of H ions on one side of the equation; react OH H H 2 O on the side with the H ions, and then cancel H 2 O's if they appear on both sides of equation.) B. The Halfreaction method 1. Write separate half reactions for whatever is oxidized and for whatever is reduced: oxidation: A reduced A oxidized n e reduction: B oxidized m e B reduced 2. Balance each half reaction separately for all the atoms: a. Balance the atom being oxidized or reduced. b. Balance all other atoms & ions(na, K, SO 4...) except O & H atoms. c. Balance O atoms by putting H 2 O on opposite side. d. Balance H atoms by putting H on opposite side. 3. Add e to one side or the other of each halfreaction to balance the charge. 4. Multiply the half reactions by factors to make them have the same # of e. 5. Add the two half reactions canceling substances that are on both sides (including the e which must cancel completely). 6. (If basic, use rule 6 in ON method above.) 2

3 HOW TO BALANCE OXIDATIONREDUCTION EQUATIONS Oxidationreduction reactions (often called redox) are those in which a change in oxidation number occurs. The oxidizing agent gains electrons, hence, decreases in oxidation number. On the other hand, the reducing agent gives up electrons and suffers an oxidation number increase. The following are methods that can be used to balance oxidationreduction equations. Such methods can be used to balance this type of equation only. These are very common reactions, however, so that the following can be used very often in chemistry. Oxidationreduction reactions occur in both acid and alkaline solution. For the acid solution, two methods can be used: the halfreaction or the whole reaction method. For the basic reaction, similar but slightly modified methods will be used. ACID SOLUTION (halfreaction method) For redox (oxidationreduction) reactions in acid solution, first write the net ionic equation, unbalanced. As an example of this type of reaction, let us take the reaction of potassium permanganate and tin (II) chloride, forming their usual acid products: MnO 4 Sn H Mn Sn H 2 O Determine the change in oxidation state for each atom reacting. In the above reaction, manganese goes from 7 in MnO." to 2 in Mn, a net change of 5. Similarly, tin goes from 2 to 4, a net change of 2. From atomic theory, in order for an atom to undergo a change in oxidation number, it must gain or lose electrons. Balancing redox equations uses this fact by balancing the electrons gained by the oxidizing agent with those lost by the reducing agent. To use the halfreaction method, write each reactant and product in a separate, or halfreaction, balancing each with the proper number of electrons. For the above reaction, this would be accomplished as follows; (Oxidizing agent) MnO 4 5e Mn (Reducing agent) Sn Sn 2e Notice, however, that the first reaction is not yet balanced, since there are four oxygen atoms on the left side and none on the right. These oxygen atoms pick up the hydrogen ions from the acid to form water molecules. Thus, we rewrite the equation with the water added: MnO 4 5e 8H Mn 4H 2 O Sn Sn 2e 3

4 Note that the numbers of H ions added were just enough to balance the hydrogen atoms in the water molecules formed. Since only whole numbers of electrons can be exchanged (i.e., there is no such thing as 1/2 e), we find in order to obtain the resultant combination of these two equations we must have two of the Mn reactions and five of the Sn reactions, so that each will involve ten electrons. Multiplying each equation by the appropriate number and adding, we have: 2MnO 4 10e 16H 2Mn 8H 2 O 5Sn 5Sn 10e 2MnO 4 5Sn 16H 2Mn 5Sn 8H 2 O The result is a balanced, net ionic equation. To check its balance, check the net charges on each side. They should be equal. If they are not, the equation is NOT BALANCED. In the above equation, we have a net charge of 24 on each side. ALWAYS CHECK THE CHARGES! It takes little time and avoids many mistakes. All of this may seem very complicated, but really it is not. It is merely a systematic method. Let us summarize as follows: 1. Write the net ionic equation with the proper products. 2. Determine the change in oxidation number of the oxide and reduce agents. 3. Write the halfreactions, balancing the oxidation state changes with electrons. Also, at this time, balance any oxygen atoms with the appropriate number of water molecules and H ions. 4. Multiply each equation by the smallest number you can use to produce the same number of electrons in each of the reactions and add the two, canceling the electrons from the result. 5. Check the charges in the result. Both sides will be equal, if the equation is balanced properly. 4

5 WHOLE REACTION METHOD (acid solution) This method is really the same as the one given above, with the electrons balanced for the oxidizing agent and the reducing agent simultaneously. To use this method, write the net ionic reaction, and by some means indicate the change in oxidation state for each element involved. One way to indicate this is as follows: 5 MnO 4 Sn H Mn Sn H 2 O 2 After this has been done, note as above the number that must be used as the lowest common factor, and multiply each reactant and its product by the appropriate number: 5 X 2 = 10 2MnO4 5Sn H 2Mn 5Sn H2O 2 X 5 = 10 Then simply balance the charge by adding H to the positive chargedeficient side and add H^O as needed to balance the oxygen atoms, and the result is a balanced net equation. 2MnO 4 5Sn 16H 2Mn 5Sn 8H 2 O AGAIN, BE SURE TO CHECK THE IONIC CHARGES. BASIC SOLUTION (halfreaction method) Here again, the method is very similar to the acid halfreaction method. First, the ionic reaction is written. Take for an example: Alkaline CrO 2 H 2 O 2 CrO 4 H 2 O Solution Then, as before, the halfreactions are written: CrO 2 CrO 4 3e H 2 O 2 2e H 2 O Here, however, the halfreactions are balanced by balancing the charges with OH ions, supplied by the base. For the first reaction (CrO2, etc.), the left side has a total of 1 and the right side 5. Therefore, we must add four OH ions to the left side so that both sides have 5: CrO 2 4OH CrO 4 3e 2H 2 O 5

6 Note that water molecules were added to balance the hydrogen atoms in the resulting equation. Now the same is done for the other equation: RIGHT SIDE = 0; LEFT SIDE = 2. Therefore, we must add two OH ions to the right side so that both sides have 2. H 2 O 2 2e H 2 0 H 2 O 2OH Again, one water molecule has been added to the left side to balance the hydrogen atoms. Now, as before, we find the proper numbers to multiply each equation (to make the same number of electrons in both), and add: 2CrO 2 8OH 2Cr0 4 6e 4H 2 O (Mult. by 2) 3H 2 O 2 6e 3H 2 O 3H 2 O 6OH (Mult. by 3) 2CrO 2 3H2 O 2 80H 3H 2 O 2CrO 4 7H 2 O 6OH The extra OH and H 2 O that occur on both sides of the equation after adding must be cancelled, leaving a net ionic equation: 2CrO 2 3H 2 O 2 2OH 2CrO 4 4H 2 O This equation is checked for balance by checking the number of oxygen atoms on each side of the equation. They should be the same. In this equation both sides have a total of 12 oxygen atoms. Always check this. To summarize, for a reaction in basic solution: 1. Write the net ionic reaction with the proper products, 2. Determine the change of oxidation state of the oxide and reduce agents. 3. Write the halfreactions, balancing the changes in oxidation states with electrons. Complete the balance of the halfreaction by balancing the charges with OH ions, and the excess hydrogen atoms with water molecules. 4. Multiply each reaction by the smallest number you can use to produce the same number of electrons in both. 5. Add the resulting equations, canceling the electrons and the excess OH or H 2 0 that may result. 6. Check the number of oxygen atoms in the result. They will be the same on both sides of the reaction if the equation is balanced properly. Whole reaction method (ACID solution) The method exactly parallels the whole reaction method for acid solution except that balance is accomplished with OH and H 2 O. 6

7 EQUATION BALANCING PRACTICE Complete and balance each of the following equations. Net ionic equations are required. The equations are not balanced if the numbers of atoms of each element are not equal on each side. Similarly, the number of net charges on each side must be equal. Substances that are essentially entirely ionized must be shown in ionic form. Conversely, unionized compounds must be shown, as they exist in the reaction medium. Ionic charges must be shown. In acid solution H and H 2 O may be added for balancing purposes; in basic solution you may need OH and H 2 O. H 2 O may be needed for balance elsewhere. I. REACTIONS WITH NO ELECTRON TRANSFER a. NH 4 OH NH4 OH b. HCO 3 H2 O CO 3 H 3 O c. H 2 S H 2 O H 3 O HS d. HS H 2 O H 3 O S e. Ag NH 4 OH Ag(NH 3 ) 2 f. H OH g. H NH 4 OH NH 4 h. Pb 2 H 2 S PbS (acidic) i. CaCO 3 H Ca 2 CO 2 J. H 2 PO 4 HPO 4 (basic) k. OAc H 2 O OH HOAc 1. Zn(OH) 2 OH Zn(OH) 4 m. Zn(OH) 3 H Zn 2 n. Ag Cl AgCl o. Ba 2 S0 4 p. Hg 2 S 2 BaSO 4 HgS q. Hg 2 Cl Hg 2 Cl 2 r. Pb 2 Cl PbC1 2 s. AgCl NH 4 OH Ag(NH 3 ) 2 Cl t. Ag(NH 3 ) 2 Cl H AgCl NH 4 u. Pb 2 CrO 4 PbCrO 4 7

8 v. Hg 2 H 2 S HgS (acidic) w. Bi 3 Cl BiOCl (basic) x. SbCl 4 SbOCl Cl (basic) y. H 3 AsO 4 H 2 S As 2 S 5 z. H 3 AsO 3 H 2 S As 2 S 3 aa. AsCl 3 H 2 O H 3 AsO 3 Cl (acidic) bb. AsCl 5 H 2 O H 3 AsO 4 Cl (acidic) cc. (NH 4 ) 2 O As 2 O 3 (NH 4 ) 3 AsO 3 dd. PbO N 2 O 5 Pb(NO 3 ) 2 ee. H 2 O SO 3 H SO4 2 ff. Cu 2 NH 4 OH Cu(OH) 2 NH 4 gg. Cu(OH) 2 NH 4 OH 2 Cu(NH 3 ) 4 (basic) hh. Sn(OH) 2 H Sn 2 (acidic) ii. 2 Cd(NH 3 ) 4 CN Cd(CN)4 2 NH3 jj. H 3 AsS 4 H 2 S As 2 S 5 kk. 3 AsS 4 H As 2 S 5 H 2 S ll. Sb 2 S 3 H Sb 3 H 2 S mm. H 3 AsO 4 Ag Ag 3 AsO 4 (acidic) nn. Co 2 OH Co(OH) 2 2 oo. Co(OH) 2 NH 4 OH Co(NH 3 ) 6 2 pp. Ni(NH 3 ) 6 S 2 NiS NH 3 (basic) qq. Fe 3 S 2 Fe 2 S 3 rr. Fe 2 S 3 H 2 O Fe(OH) 3 HS (basic) ss. Fe 2 Fe(CN) 6 3 Fe 3 (Fe(CN) 6 ) 2 tt. Co 2 Cl CoCl 4 3 uu. Co(NH 3 ) 6 H Co 3 NH 4 vv. Al 3 NH 4 OH Al(OH) 3 NH 4 ww. Al(OH) 3 OH AlO 2 xx. Al(OH) 3 H Al 3 8

9 yy. Zn(OH) 2 NH 4 OH Zn(NH 3 ) 4 2 zz. CrO 4 H Cr 2 O 7 (basic) ab. NH 4 ac. CO 3 OH NH 3 H 2 O HCO 3 (basic) ad. Mg 2 HPO 4 NH 4 OH MgNH 4 PO 4 ae. CO 3 H CO 2 af. AgCl CO 3 Ag 2 CO 3 Cl ag. SO 3 H H 2 O SO 2 ah. Mg 2 NH 4 AsO 4 3 MgNH 4 AsO 4 ai. MgNH 4 AsO 4 Ag AgAsO 4 Mg 2 NH 4 3 aj. MoO 4 PO 4 NH 4 H (NH 4 ) 3 PO 4 12 MoO 3 ak. B 4 O 7 H H 2 O H 3 BO 3 al. NaHCO 3 Na 2 CO 3 CO 2 H 2 O am. NO (g) N 2 O 2(g) an. N 2 O 4(g) NO 2(g) ao. O 2 O 3 ap. SO 2(g) H 2 O HSO 3 (acidic) aq. ZnO OH H 2 O Zn(OH) 4 ar. CaO H 2 O Ca(OH) 2 as. Ag 2 O NH 3 H 2 O Ag(NH 3 ) 2 at. NH 3 CO 2 H 2 O HCO 3 NH 4 au. SiO 2 HF SiF 6 (basic) (acidic) 9

10 II. OXIDATIONREDUCTION REACTIONS a. SO 2(g) NO 2(g) NO (g) SO 3(g) b. SO 2(g) O 2(g) SO 3(g) c. CO (g) H 2 O (g) CO 2(g) H 2(g) d. NH 3(g) O 2(g) H 2 O (g) NO (g) e. N 2(g) H 2(g) NH 3(g) f. Mg (s) H g. H 3 AsO 3 Cr 2 O 7 H 3 AsO 4 Cr 3 (acid) h. Sn 2 Fe 3 Sn 4 Fe 2 (acid) i. Cr 2 O 7 H 2 S Cr 3 S (s) (acid) j. Fe 2 MnO4 Mn 2 k. FeS NO 3 Fe 3 SO 4 NO (g) (acid) l. As 2 S 3 NO 3 NO (g) SO 4 H 3 AsO 4 (acid) m. Mn 2 BiO 3 NO (g) SO 4 Cl (acid) n. H 2 SO 3 ClO 3 SO 4 Cl (acid) o. Cr 2 O 7 Sn 2 Sn 4 Cr 3 p. Sb 2 S 3 NO 3 NO (g) S Sb 2 O 5 (acid) q. S 2 O 3 MnO 4 Mn 2 SO 4 (acid) r. FeS SO 4 SO 2(g) Fe 3 (acid) s. Sn(OH) 4 MnO 4 MnO 2(s) Sn(OH) 6 (base) t. Cr(OH) 4 H 2 O 2 CrO 4 H 2 O (base) u. S 2 O 4 CrO 4 Cr(OH) 4 SO 4 (base) v. CN Cu 2 Cu(CN) 2 CNO (base) w. Hg 2 Cl 2 NH 4 OH Hg HgNH 2 Cl Cl x. H 3 AsO 4 H 2 S H 3 AsO 3 S y. Sn 2 H 2 O 2 Sn 4 (acid) z. H 3 AsO 3 H 2 O 2 H 3 AsO 4 (acid) 10

11 aa. NO 3 bb. NO 3 Cl NO (g) Cl 2(g) (acid) H 2 S NO (g) S (s) (acid) cc. Fe 3 H 2 S Fe 2 S (acid) dd. Cr 2 O 7 H 2 S Cr 3 S (acid) ee. Cr 2 O 7 Cl Cr 3 Cl 2(g) (acid) ff. CuS (s) NO 3 gg. HgS NO 3 Cu 2 S NO (g) (acid) Cl HgCl 2 NO (g) S (s) (acid) hh. Hg 2 Sn 2 Cl Hg 2 Cl 2 Sn 4 (acid) ii. Hg 2 Cl 2 Sn 2 Hg Sn 4 Cl jj. Hg 2 Sn 2 Hg Sn 4 kk. Sn(OH) 4 O 2(g) Sn(OH) 6 ll. Sn(OH) 4 Sn Sn(OH) 6 Cu (base) mm. Cu 2 S 2 O 4 nn. Cu S 2 O 4 SO 3 SO 3 Cu (base) Cu (acid) oo. S 2 H 2 S (g) S (acid) pp. Sn Sb 3 Sn 2 Sb qq. Sn H Sn 2 H 2(g) rr. S 2 O 3 H 2 O SO 4 H 2 S (g) ss. Co(OH) 2 O 2 Co(OH) 3 tt. H 2 O 2 Mn(OH) 2 MnO 2 uu. Co(OH) 2 H 2 O 2 Co(OH) 3 vv. MnO 2 Cl Mn 2 Cl 2(g) (acid) ww. MnO 4 xx. MnO 4 Cl Mn 2 Cl 2(g) (acid) H 2 S Mn 2 S (acid) yy. CrO 5 Cr 3 O 2(g) (acid) zz. Ag NO 3 Cl AgCl NO (g) (acid) ab. Pb NO 3 Cl PbCl 4 NO (g) (acid) 11

12 ac. Cu NO 3 Cu 2 NO 2(g) (acid) ad. MnO 2 Cl Mn 2 Cl 2(g) (acid) ae. I NO 3 NO (g) I 2(g) (acid) af. NO 2 ag. MnO 4 I NO (g) I 2(g) (acid) Br Mn 2 Br 2(g) (acid) ah. Cl 2 2I Cl I 2 ai. Br S 2 O 8 SO 4 Br 2 aj. Fe 2 NO 3 Fe 3 NO (g) (acid) ak. S (g) F 2(g) SF 6(g) al. Se (s) O 2(g) SeO 2(s) am. Rb H 2(g) RbH an. In Cl 2(g) InCl 3(s) ao. PH 3(g) N 2 O (g) H 3 PO 4(l) N 2(g) ap. SF 6(l) H 2 S (g) HF (l) S (s) aq. PbO 2(s) Mn 2 Pb 2 MnO 4 (acid) ar. Bi 3 SnO 2 Bi (s) SnO 3 (base) as. ClO 3 Fe 2 Cl Fe 3 (acid) 12