HEIN17_437-467v3-hr 8/28/06 11:26 AM Page 437 CHAPTER 17 Oxidation Reduction The Golden Gate Bridge in San Francisco requires frequent painting to prevent oxidation of the metal in the sea air. Chapter Outline 17.1 17.2 17.3 Oxidation Number Oxidation Reduction Balancing Oxidation Reduction Equations 17.4 17.5 17.6 Balancing Ionic Redox Equations Activity Series of Metals Electrolytic and Voltaic Cells 437
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 438 438 CHAPTER 17 OXIDATION REDUCTION The variety of oxidation reduction reactions that affect us every day is amazing. Our society runs on batteries in our calculators, laptop computers, cars, toys, radios, televisions, and more. We paint iron railings and galvanize nails to combat corrosion. We electroplate jewelry and computer chips with very thin coatings of gold or silver. We bleach our clothes using chemical reactions that involve electron transfer. We test for glucose in urine or alcohol in the breath with reactions that show vivid color changes. Plants turn energy into chemical compounds through a series of reactions called photosynthesis. These reactions all involve the transfer of electrons between substances in a chemical process called oxidation reduction. 17.1 Oxidation Number Table 17.1 Oxidation Numbers for Common Ions Ion Oxidation number H 1 Na 1 K 1 Li 1 Ag 1 Cu 2 2 Ca 2 2 Ba 2 2 Fe 2 2 Mg 2 2 Zn 2 2 Al 3 3 Fe 3 3 Cl - -1 Br - -1 F - -1 I - -1 S 2- -2 O 2- -2 The oxidation number of an atom (sometimes called its oxidation state) represents the number of electrons lost, gained, or unequally shared by an atom. Oxidation numbers can be zero, positive, or negative. An oxidation number of zero means the atom has the same number of electrons assigned to it as there are in the free neutral atom. A positive oxidation number means the atom has fewer electrons assigned to it than in the neutral atom, and a negative oxidation number means the atom has more electrons assigned to it than in the neutral atom. The oxidation number of an atom that has lost or gained electrons to form an ion is the same as the positive or negative charge of the ion. (See Table 17.1.) In the ionic compound NaCl, the oxidation numbers are clearly 1 for the Na ion and -1 for the Cl - ion. The Na ion has one less electron than the neutral Na atom, and the Cl - ion has one more electron than the neutral Cl atom. In MgCl 2, two electrons have transferred from the Mg atom to the two Cl atoms; the oxidation number of Mg is 2. In covalently bonded substances, where electrons are shared between two atoms, oxidation numbers are assigned by an arbitrary system based on relative electronegativities. For symmetrical covalent molecules such as H 2 and Cl 2, each atom is assigned an oxidation number of zero because the bonding pair of electrons is shared equally between two like atoms, neither of which is more electronegative than the other: H H When the covalent bond is between two unlike atoms, the bonding electrons are shared unequally because the more electronegative element has a greater attraction for them. In this case the oxidation numbers are determined by assigning both electrons to the more electronegative element. Thus in compounds with covalent bonds such as NH 3 and H 2 O, H N H H Cl Cl shared pairs of electrons H O H the pairs of electrons are unequally shared between the atoms and are attracted toward the more electronegative elements, N and O. This causes the N and O atoms to be relatively negative with respect to the H atoms. At the same
HEIN17_437-467v2.1-hr 8/25/06 8:36 AM Page 439 17.1 OXIDATION NUMBER 439 time, it causes the H atoms to be relatively positive with respect to the N and O atoms. In H 2 O, both pairs of shared electrons are assigned to the O atom, giving it two electrons more than the neutral O atom, and each H atom is assigned one electron less than the neutral H atom. Therefore, the oxidation number of the O atom is -2, and the oxidation number of each H atom is 1. In NH 3, the three pairs of shared electrons are assigned to the N atom, giving it three electrons more than the neutral N atom, and each H atom has one electron less than the neutral atom. Therefore, the oxidation number of the N atom is -3, and the oxidation number of each H atom is 1. Assigning correct oxidation numbers to elements is essential for balancing oxidation reduction equations. The oxidation number or oxidation state of an element is an integer value assigned to each element in a compound or ion that allows us to keep track of electrons associated with each atom. Oxidation numbers have a variety of uses in chemistry from writing formulas to predicting properties of compounds and assisting in the balancing of oxidation reduction reactions in which electrons are transferred. As a starting point, the oxidation number of an uncombined element, regardless of whether it is monatomic or diatomic, is zero. Rules for assigning oxidation numbers are summarized in Table 17.2. Many elements have multiple oxidation numbers; for example, nitrogen 1 2 0 1 H O H H N H H 1 3 1 1 Oxidation number Oxidation state O N 0 N 2 N 2 O NO N 2 O 3 NO 2 N 2 O 5 NO 3 - Oxidation number 0 1 2 3 4 5 5 Use the following steps to find the oxidation number for an element within a compound: Step 1 Step 2 Step 3 Write the oxidation number of each known atom below the atom in the formula. Multiply each oxidation number by the number of atoms of that element in the compound. Write an expression indicating the sum of all the oxidation numbers in the compound. Remember: The sum of the oxidation numbers in a compound must equal zero. The oxidation number for Cu metal is 0 while the oxidation number for Cu 2 ions in the crystal is 2. Table 17.2 Rules for Assigning Oxidation Number 1. All elements in their free state (uncombined with other elements) have an oxidation number of zero (e.g., Na, Cu, Mg, H 2, O 2, Cl 2, N 2 ). 2. H is 1, except in metal hydrides, where it is -1 (e.g., NaH, CaH 2 ). 3. O is -2, except in peroxides, where it is -1, and in OF 2, where it is 2. 4. The metallic element in an ionic compound has a positive oxidation number. 5. In covalent compounds the negative oxidation number is assigned to the most electronegative atom. 6. The algebraic sum of the oxidation numbers of the elements in a compound is zero. 7. The algebraic sum of the oxidation numbers of the elements in a polyatomic ion is equal to the charge of the ion.
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 440 440 CHAPTER 17 OXIDATION REDUCTION Example 17.1 Determine the oxidation number for carbon in carbon dioxide: SOLUTION Step 1 Step 2 Step 3 Step 4 CO 2 2 ( 2)2 C (-4) = 0 C = 4 (oxidation number for carbon) Example 17.2 Determine the oxidation number for sulfur in sulfuric acid: SOLUTION Step 1 Step 2 Step 3 Step 4 H 2 SO 4 ±1 2 2(1) = 2 4(-2) = -8 2 S(-8) = 0 S = 6 (oxidation number for sulfur) Practice 17.1 Determine the oxidation number of (a) S in Na 2 SO 4, (b) As in K 3 AsO 4, and (c) C in CaCO 3. Oxidation numbers in a polyatomic ion (ions containing more than one atom) are determined in a similar fashion, except that in a polyatomic ion the sum of the oxidation numbers must equal the charge on the ion instead of zero. Example 17.3 SOLUTION Determine the oxidation number for manganese in the permanganate ion MnO 4 - : Step 1 Step 2 Step 3 MnO 4 2 ( 2)4 Mn (-8) = -1 (the charge on the ion) Mn = 7 (oxidation number for manganese) Example 17.4 SOLUTION Determine the oxidation number for carbon in the oxalate ion C 2 O 2-4 : Step 1 Step 2 Step 3 C 2 O 2 4 2 ( 2)4 2C (-8) = -2 (the charge on the ion) 2C = 6 C = 3 (oxidation number for C)
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 441 17.2 OXIDATION REDUCTION 441 Practice 17.2 Determine the oxidation numbers of (a) N in NH (b) Cr in Cr 2 O 2-4, 7, and (c) P in PO 3-4. Determine the oxidation number of each element in (a) KNO and (b) SO 2-3 4. (a) Potassium is a Group 1A metal; therefore, it has an oxidation number of 1. The oxidation number of each O atom is -2 (Table 17.2, Rule 3). Using these values and the fact that the sum of the oxidation numbers of all the atoms in a compound is zero, we can determine the oxidation number of N: KNO 3 Example 17.5 SOLUTION 1 N 3( 2) 0 N 6 1 5 The oxidation numbers are K, 1; N, 5; O, -2. (b) Because SO 2-4 is an ion, the sum of oxidation numbers of the S and the O atoms must be -2, the charge of the ion. The oxidation number of each O atom is -2 (Table 17.2, Rule 3). Then SO 4 2 S 4( 2) 2, S 2 8 6 S 8 2 The oxidation numbers are S, 6; O, - 2. Practice 17.3 Determine the oxidation number of each element in these species: (a) BeCl (b) (c) H (d) NH 4 (e) BrO3-2 HClO 2 O 2 17.2 Oxidation Reduction Oxidation reduction, also known as redox, is a chemical process in which the oxidation number of an element is changed. The process may involve the complete transfer of electrons to form ionic bonds or only a partial transfer or shift of electrons to form covalent bonds. Oxidation occurs whenever the oxidation number of an element increases as a result of losing electrons. Conversely, reduction occurs whenever the oxidation number of an element decreases as a result of gaining electrons. For example, a change in oxidation number from 2 to 3 or from -1 to 0 is oxidation; a change from 5 to 2 or from -2 to -4 is reduction (see Figure 17.1). Oxidation and reduction occur simultaneously in a chemical reaction; one cannot take place without the other. Oxidation (loss of electrons) Oxidation number 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 Reduction (gain of electrons) oxidation-reduction redox oxidation reduction Figure 17.1 Oxidation and reduction. Oxidation results in an increase in the oxidation number, and reduction results in a decrease in the oxidation number.
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 442 442 CHAPTER 17 OXIDATION REDUCTION Many combination, decomposition, and single-displacement reactions involve oxidation reduction. Let s examine the combustion of hydrogen and oxygen from this point of view: 2 H 2 O 2 2 H 2 O oxidizing agent reducing agent Both reactants, hydrogen and oxygen, are elements in the free state and have an oxidation number of zero. In the product (water), hydrogen has been oxidized to 1 and oxygen reduced to -2. The substance that causes an increase in the oxidation state of another substance is called an oxidizing agent. The substance that causes a decrease in the oxidation state of another substance is called a reducing agent. In this reaction the oxidizing agent is free oxygen, and the reducing agent is free hydrogen. In the reaction Zn(s) H 2 SO 4 (aq) ZnSO 4 (aq) H 2 (g) metallic zinc is oxidized, and hydrogen ions are reduced. Zinc is the reducing agent, and hydrogen ions, the oxidizing agent. Electrons are transferred from the zinc metal to the hydrogen ions. The electron transfer is more clearly expressed as Zn 0 2 H SO 2-4 Zn 2 SO 2-4 H 0 2 Oxidation: Reduction: Increase in oxidation number Loss of electrons Decrease in oxidation number Gain of electrons In the reaction between Zn and H 2 SO 4 the Zn is oxidized while hydrogen is reduced. The superscript 0 shows that the oxidation number is 0 for elements in their uncombined state. The oxidizing agent is reduced and gains electrons. The reducing agent is oxidized and loses electrons. The transfer of electrons is characteristic of all redox reactions. 17.3 Balancing Oxidation Reduction Equations Many simple redox equations can be balanced readily by inspection, or by trial and error: Na Cl 2 NaCl (unbalanced) 2 Na Cl 2 2 NaCl (balanced) Balancing this equation is certainly not complicated. But as we study more complex reactions and equations such as P HNO 3 H 2 O NO H 3 PO 4 (unbalanced) 3 P 5 HNO 3 2 H 2 O 5 NO 3 H 3 PO 4 (balanced) the trial-and-error method of balancing equations takes an unnecessarily long time. One systematic method for balancing oxidation reduction equations is based on the transfer of electrons between the oxidizing and reducing agents. Consider the first equation again: Na 0 Cl 0 2 Na Cl - (unbalanced)
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 443 17.3 BALANCING OXIDATION REDUCTION EQUATIONS 443 In this reaction, sodium metal loses one electron per atom when it changes to a sodium ion. At the same time chlorine gains one electron per atom. Because chlorine is diatomic, two electrons per molecule are needed to form a chloride ion from each atom. These electrons are furnished by two sodium atoms. Stepwise, the reaction may be written as two half-reactions, the oxidation halfreaction and the reduction half-reaction: Cl 0 2 Cl 0 2 2 Na 0 2 e - 2 Na 0 2 Na 2 e - 2 Cl - 2 Na Cl - oxidation half-reaction reduction half-reaction When the two half-reactions, each containing the same number of electrons, are added together algebraically, the electrons cancel out. In this reaction there are no excess electrons; the two electrons lost by the two sodium atoms are utilized by chlorine. In all redox reactions the loss of electrons by the reducing agent must equal the gain of electrons by the oxidizing agent. Here sodium is oxidized and chlorine is reduced. Chlorine is the oxidizing agent; sodium is the reducing agent. In the following examples, we use the change-in-oxidation-number method, a system for balancing more complicated redox equations. Balance the equation Sn HNO 3 SnO 2 NO 2 H 2 O (unbalanced) Example 17.6 Step 1 Assign oxidation numbers to each element to identify the elements being oxidized and those being reduced. Write the oxidation numbers below each element to avoid confusing them with ionic charge: Sn HNO 3 SnO 2 NO 2 H 2 O SOLUTION Step 2 0 1 5 2 4 2 4 2 1 2 Note that the oxidation numbers of Sn and N have changed. Now write two new equations (half-reactions), using only the elements that change in oxidation number. Then add electrons to bring the equations into electrical balance. One equation represents the oxidation step; the other represents the reduction step. Remember: Oxidation produces electrons; reduction uses electrons. Sn 0 Sn 4 4 e - Sn 0 loses 4 electrons N 5 1 e - N 4 N 5 gains 1 electron (oxidation) (reduction) Step 3 Multiply the two equations by the smallest whole numbers that will make the electrons lost by oxidation equal to the number of electrons gained by reduction. In this reaction the oxidation step is multiplied by 1 and the reduction step by 4. The equations become Sn 0 Sn 4 4 e - Sn 0 loses 4 electrons (oxidation)
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 444 444 CHAPTER 17 OXIDATION REDUCTION In balancing the final elements, we must not change the ratio of the elements that were oxidized and reduced. Step 4 4 N 5 4 e - 4 N 4 4 N 5 gains 4 electrons (reduction) We have now established the ratio of the oxidizing to the reducing agent as being four atoms of N to one atom of Sn. Transfer the coefficient in front of each substance in the balanced oxidation reduction equations to the corresponding substance in the original equation. We need to use 1 Sn, 1 SnO 2, 4 HNO 3, and 4 NO 2 : Sn 4 HNO 3 SnO 2 4 NO 2 H 2 O (unbalanced) Step 5 In the usual manner, balance the remaining elements that are not oxidized or reduced to give the final balanced equation: Sn 4 HNO 3 SnO 2 4 NO 2 2 H 2 O (balanced) Finally, check to ensure that both sides of the equation have the same number of atoms of each element. The final balanced equation contains 1 atom of Sn, 4 atoms of N, 4 atoms of H, and 12 atoms of O on each side. Because each new equation presents a slightly different problem and because proficiency in balancing equations requires practice, let s work through two more examples. Example 17.7 Balance the equation I 2 Cl 2 H 2 O HIO 3 HCl (unbalanced) SOLUTION Step 1 Step 2 Step 3 Step 4 Assign oxidation numbers: I 2 Cl 2 H 2 O HIO 3 HCl 0 0 1 2 1 5 2 1 1 The oxidation numbers of I 2 and Cl 2 have changed, I 2 from 0 to 5, and Cl 2 from 0 to -1. 2 e Write the oxidation and reduction steps. Balance the number of atoms and then balance the electrical charge using electrons: I 2 2 I 5 10 e - (oxidation) ( 10 e - are needed to balance I 2 loses 10 electrons the 10 charge) Cl 2 2 e - 2 Cl - (reduction) ( - are needed to balance Cl 2 gains 2 electrons the -2 charge) Adjust loss and gain of electrons so that they are equal. Multiply the oxidation step by 1 and the reduction step by 5: I 2 2 I 5 10 e - (oxidation) loses 10 electrons I 2 5 Cl 2 10 e - 10 Cl - 5 Cl 2 gain 10 electrons (reduction) Transfer the coefficients from the balanced redox equations into the original equation. We need to use 1 I 2, 2 HIO 3, 5 Cl 2, and 10 HCl: I 2 5 Cl 2 H 2 O 2 HIO 3 10 HCl (unbalanced)
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 445 17.3 BALANCING OXIDATION REDUCTION EQUATIONS 445 Step 5 Balance the remaining elements, H and O: I 2 5 Cl 2 6 H 2 O 2 HIO 3 10 HCl (balanced) Check: The final balanced equation contains 2 atoms of I, 10 atoms of Cl, 12 atoms of H, and 6 atoms of O on each side. Balance the equation K 2 Cr 2 O 7 FeCl 2 HCl CrCl 3 KCl FeCl 3 H 2 O (unbalanced) Step 1 Assign oxidation numbers ( Cr and Fe have changed): K 2 Cr 2 O 7 FeCl 2 HCl CrCl 3 KCl FeCl 3 H 2 O Example 17.8 SOLUTION Step 2 Step 3 Step 4 Step 5 1 6 2 2 1 1 1 3 1 1 1 3 1 1 2 Write the oxidation and reduction steps. Balance the number of atoms and then balance the electrical charge using electrons: Fe 2 Fe 3 1 e - (oxidation) Fe 2 loses 1 electron 2 Cr 6 6 e - 2 Cr 3 (reduction) 2 Cr 6 gain 6 electrons Balance the loss and gain of electrons. Multiply the oxidation step by 6 and the reduction step by 1 to equalize the transfer of electrons. 6 Fe 2 6 Fe 3 6 e - (oxidation) 6 Fe 2 lose 6 electrons 2 Cr 6 6 e - 2 Cr 3 (reduction) 2 Cr 6 gain 6 electrons Transfer the coefficients from the balanced redox equations into the original equation. (Note that one formula unit of K 2 Cr 2 O 7 contains two Cr atoms.) We need to use 1 K 2 Cr 2 O 7, 2 CrCl 3, 6 FeCl 2, and 6 FeCl 3 : K 2 Cr 2 O 7 6 FeCl 2 HCl Balance the remaining elements in the order K, Cl, H, O: K 2 Cr 2 O 7 6 FeCl 2 14 HCl 2 CrCl 3 KCl 6 FeCl 3 H 2 O (unbalanced) 2 CrCl 3 2 KCl 6 FeCl 3 7 H 2 O (balanced) Check: The final balanced equation contains 2 K atoms, 2 Cr atoms, 7 O atoms, 6 Fe atoms, 26 Cl atoms, and 14 H atoms on each side. Practice 17.4 Balance these equations using the change-in-oxidation-number method: (a) HNO 3 S NO 2 H 2 SO 4 H 2 O (b) CrCl 3 MnO 2 H 2 O MnCl 2 H 2 CrO 4 (c) KMnO 4 HCl H 2 S KCl MnCl 2 S H 2 O
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 446 446 CHAPTER 17 OXIDATION REDUCTION 17.4 Balancing Ionic Redox Equations The main difference between balancing ionic redox equations and molecular redox equations is in how we handle the ions. In the ionic redox equations, besides having the same number of atoms of each element on both sides of the final equation, we must also have equal net charges. In assigning oxidation numbers, we must therefore remember to consider the ionic charge. Several methods are used to balance ionic redox equations, including, with slight modification, the oxidation-number method just shown for molecular equations. But the most popular method is probably the ion electron method. The ion electron method uses ionic charges and electrons to balance ionic redox equations. Oxidation numbers are not formally used, but it is necessary to determine what is being oxidized and what is being reduced. The method is as follows: Step 1 Write the two half-reactions that contain the elements being oxidized and reduced using the entire formula of the ion or molecule. Step 2 Balance the elements other than oxygen and hydrogen. Step 3 Balance oxygen and hydrogen. Acidic solution: For reactions in acidic solution, use H and H 2 O to balance oxygen and hydrogen. For each oxygen needed, use one H Then add H 2 O. as needed to balance the hydrogen atoms. Basic solution: For reactions in alkaline solutions, first balance as though the reaction were in an acidic solution, using Steps 1 3. Then add as many OH - ions to each side of the equation as there are H ions in the equation. Now combine the H and OH - ions into water (for example, 4 H and 4 OH - give 4 H 2 O). Rewrite the equation, canceling equal numbers of water molecules that appear on opposite sides of the equation. Step 4 Add electrons (e - ) to each half-reaction to bring them into electrical balance. Step 5 Since the loss and gain of electrons must be equal, multiply each half-reaction by the appropriate number to make the number of electrons the same in each half-reaction. Step 6 Add the two half-reactions together, canceling electrons and any other identical substances that appear on opposite sides of the equation. Example 17.9 SOLUTION Balance this equation using the ion electron method: MnO 4 - S 2- Mn 2 S 0 Step 1 Step 2 (acidic solution) Write two half-reactions, one containing the element being oxidized and the other the element being reduced (use the entire molecule or ion): S 2- S 0 (oxidation) MnO4 - Mn 2 (reduction) Balance elements other than oxygen and hydrogen (accomplished in Step 1: 1 S and 1 Mn on each side).
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 447 CHEMISTRY IN ACTION Sensitive Sunglasses xidation reduction reactions are the basis for many interesting applications. Consider photochromic glass, which is used for lenses in light-sensitive glasses. These lenses, manufactured by the Corning Glass Company, can change from transmitting 85% of light to transmitting only 22% of light when exposed to bright sunlight. Photochromic glass is composed of linked tetrahedrons of silicon and oxygen atoms jumbled in a disorderly array, with crystals of silver chloride caught between the silica tetrahedrons. When the glass is clear, the visible light passes right through the molecules. The glass absorbs ultraviolet light, however, and this energy triggers an oxidation reduction reaction between Ag and Cl-: O Ag Cl- UV light " Ag 0 Cl0 To prevent the reaction from reversing itself immediately, a few ions of Cu are incorporated into the silver chloride crystal. These Cu ions react with the newly formed chlorine atoms: Cu Cl0 Cu2 ClThe silver atoms move to the surface of the crystal and form small colloidal clusters of silver metal. This metallic silver absorbs visible light, making the lens appear dark (colored). As the glass is removed from the light, the Cu2 ions slowly move to the surface of the crystal, where they interact with the silver metal: Cu2 Ag 0 Cu Ag The glass clears as the silver ions rejoin chloride ions in the crystals. An oxidation reduction reaction causes these photochromic glasses to change from light to dark in bright sunlight. Step 3 Balance O and H. Remember the solution is acidic. The oxidation requires neither O nor H, but the reduction equation needs 4 H 2O on the right and 8 H on the left: S2- S0 8 H MnO 4- Mn2 4 H 2O Step 4 Balance each half-reaction electrically with electrons: S2- S0 2 enet charge = -2 on each side 5 e- 8 H MnO 4- Mn2 4 H 2O net charge = 2 on each side Step 5 Equalize loss and gain of electrons. In this case, multiply the oxidation equation by 5 and the reduction equation by 2: 5 S2-5 S0 10 e10 e- 16 H 2 MnO 4-2 Mn2 8 H 2O Step 6 Add the two half-reactions together, canceling the 10 e- from each side, to obtain the balanced equation: 5 S2-5 S0 10 e 10 e 16 H 2 MnO 4-2 Mn2 8 H 2O 16 H 2 MnO 4-5 S2-2 Mn2 5 S0 8 H 2O (balanced) Check: Both sides of the equation have a charge of 4 and contain the same number of atoms of each element. 447
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 448 448 CHAPTER 17 OXIDATION REDUCTION Example 17.10 SOLUTION Balance this equation: CrO 2-4 Fe(OH) 2 Cr(OH) 3 Fe(OH) 3 Step 1 Write the two half-reactions: Fe(OH) 2 Fe(OH) 3 (oxidation) CrO 2-4 Cr(OH) 3 (reduction) (basic solution) Step 2 Balance elements other than H and O (accomplished in Step 1). Step 3 Remember the solution is basic. Balance O and H as though the solution were acidic. Use H and H 2 O. To balance O and H in the oxidation equation, add 1 H 2 O on the left and 1 H on the right side: Fe(OH) 2 H 2 O Fe(OH) 3 H Add 1 OH - to each side: Fe(OH) 2 H 2 O OH - Fe(OH) 3 H OH - Combine H and OH - as H 2 O and rewrite, canceling H 2 O on each side: Fe(OH) 2 H 2 O OH - Fe(OH) 3 H 2 O Fe(OH) 2 OH - Fe(OH) 3 (oxidation) To balance O and H in the reduction equation, add 1 H 2 O on the right and 5 H on the left: CrO 4 2-5 H Cr(OH) 3 H 2 O Add 5 OH - to each side: CrO 2-4 5 H 5 OH - Cr(OH) 3 H 2 O 5 OH - Combine 5 H 5 OH - 5 H 2 O: CrO 2-4 5 H 2 O Cr(OH) 3 H 2 O 5 OH - Rewrite, canceling 1 H 2 O from each side: CrO 2-4 4 H 2 O Cr(OH) 3 5 OH - (reduction) Step 4 Step 5 Balance each half-reaction electrically with electrons: Fe(OH) 2 OH - Fe(OH) 3 e - CrO 2-4 4 H 2 O 3 e - Cr(OH) 3 5 OH - (balanced oxidation equation) (balanced reduction equation) Equalize the loss and gain of electrons. Multiply the oxidation reaction by 3: 3 Fe(OH) 2 3 OH - 3 Fe(OH) 3 3 e - CrO 2-4 4 H 2 O 3 e - Cr(OH) 3 5 OH -
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 449 17.4 BALANCING IONIC REDOX EQUATIONS 449 Step 6 Add the two half-reactions together, canceling the 3 e - and 3 OH - from each side of the equation: 3 Fe(OH) 2 3 OH - 3 Fe(OH) 3 3 e - CrO 2-4 4 H 2 O 3 e - Cr(OH) 3 5 OH - CrO 2-4 3 Fe(OH) 2 4 H 2 O Cr(OH) 3 3 Fe(OH) 3 2 OH - Check: Each side of the equation has a charge of -2 and contains the same number of atoms of each element. (balanced) Practice 17.5 Balance these equations using the ion electron method: (a) I - NO2 - I 2 NO (acidic solution) (b) Cl 2 IO - 3 IO - 4 Cl - (basic solution) (c) AuCl4 - Sn 2 Sn 4 AuCl Cl - Ionic equations can also be balanced using the change-in-oxidation-number method shown in Example 17.6. To illustrate this method, let s use the equation from Example 17.10. Balance this equation using the change-in-oxidation-number method: CrO 2-4 Fe(OH) 2 Cr(OH) 3 Fe(OH) 3 (basic solution) Steps 1 and 2 Assign oxidation numbers and balance the charges with electrons: Cr 6 3 e - Cr 3 (reduction) Cr 6 gains 3 e - Fe 2 Fe 3 e - (oxidation) Fe 2 loses 1 e - Step 3 Equalize the loss and gain of electrons, and then multiply the oxidation step by 3: Cr 6 3 e - Cr 3 (reduction) Cr 6 gains 3 e - 3 Fe 2 3 Fe 3 3 e - (oxidation) 3 Fe 2 loses 3 e - Step 4 Transfer coefficients back to the original equation: CrO 2-4 3 Fe(OH) 2 Cr(OH) 3 3 Fe(OH) 3 Step 5 Balance electrically. Because the solution is basic, use OH - to balance charges. The charge on the left side is -2 and on the right side is 0. Add 2 OH - ions to the right side of the equation: CrO 2-4 3 Fe(OH) 2 Cr(OH) 3 3 Fe(OH) 3 2 OH - Adding 4 H 2 O to the left side balances the equation: CrO 2-4 3 Fe(OH) 2 4 H 2 O Cr(OH) 3 3 Fe(OH) 3 2 OH - (balanced) Check: Each side of the equation has a charge of -2 and contains the same number of atoms of each element. Example 17.11 SOLUTION
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 450 450 CHAPTER 17 OX I DAT I O N R E D U C T I O N Practice 17.6 Balance these equations using the change-in-oxidation-number method: (a) Zn Zn(OH)2(basic solution) 4 H2 (b) H 2O2 Sn2 Sn4 (acidic solution) (c) Cu Cu2 Cu2O (basic solution) 17.5 Activity Series of Metals Knowledge of the relative chemical reactivities of the elements helps us predict the course of many chemical reactions. For example, calcium reacts with cold water to produce hydrogen, and magnesium reacts with steam to produce hydrogen. Therefore, calcium is considered a more reactive metal than magnesium: Figure 17.2 A coil of copper placed in a silver nitrate solution forms silver crystals on the wire. The pale blue of the solution indicates the presence of copper ions. Table 17.3 Activity Series of Metals Ease of oxidation K Ba Ca Na Mg Al Zn Cr Fe Ni Sn Pb H2 Cu As Ag Hg Au K Ba2 Ca2 Na Mg2 Al3 Zn2 Cr3 Fe2 Ni2 Sn2 Pb2 2 H± Cu2 As3 Ag Hg2 Au3 e2 e2 ee2 e3 e2 e3 e2 e2 e2 e2 e2 e 2 e3 ee2 e3 e- activity series of metals Ca(s) 2 H 2O(l) Ca(OH)2(aq) H 2(g) Mg(s) H 2O(g) MgO(s) H 2(g) steam The difference in their activity is attributed to the fact that calcium loses its two valence electrons more easily than magnesium and is therefore more reactive and/or more readily oxidized than magnesium. When a coil of copper is placed in a solution of silver nitrate (AgNO3), free silver begins to plate out on the copper. (See Figure 17.2.) After the reaction has continued for some time, we can observe a blue color in the solution, indicating the presence of copper(ii) ions. The equations are Cu0(s) 2 AgNO3(aq) 2 Ag 0(s) Cu(NO3)2(aq) Cu0(s) 2 Ag (aq) 2 Ag 0(s) Cu2(aq) Cu0(s) Cu2(aq) 2 eag (aq) e- Ag 0(s) (net ionic equation) (oxidation of Cu0 ) (reduction of Ag ) If a coil of silver is placed in a solution of copper(ii) nitrate, Cu(NO3)2, no reaction is visible. Ag 0(s) Cu(NO3)2(aq) no reaction In the reaction between Cu and AgNO3, electrons are transferred from Cu0 atoms to Ag ions in solution. Copper has a greater tendency than silver to lose electrons, so an electrochemical force is exerted upon silver ions to accept electrons from copper atoms. When an Ag ion accepts an electron, it is reduced to an Ag 0 atom and is no longer soluble in solution. At the same time, Cu0 is oxidized and goes into solution as Cu2 ions. From this reaction, we can conclude that copper is more reactive than silver. Metals such as sodium, magnesium, zinc, and iron that react with solutions of acids to liberate hydrogen are more reactive than hydrogen. Metals such as copper, silver, and mercury that do not react with solutions of acids to liberate hydrogen are less reactive than hydrogen. By studying a series of reactions such as these, we can list metals according to their chemical activity, placing the most active at the top and the least active at the bottom. This list is called the activity series of metals. Table 17.3 lists some of the common metals in the series. The arrangement corresponds to the ease with which the elements
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 451 17.5 ACTIVITY SERIES OF METALS 451 are oxidized or lose electrons, with the most easily oxidizable element listed first. More extensive tables are available in chemistry reference books. The general principles governing the arrangement and use of the activity series are as follows: 1. The reactivity of the metals listed decreases from top to bottom. 2. A free metal can displace the ion of a second metal from solution, provided that the free metal is above the second metal in the activity series. 3. Free metals above hydrogen react with nonoxidizing acids in solution to liberate hydrogen gas. 4. Free metals below hydrogen do not liberate hydrogen from acids. 5. Conditions such as temperature and concentration may affect the relative position of some of these elements. Here are two examples using the activity series of metals. Will zinc metal react with dilute sulfuric acid? From Table 17.3, we see that zinc is above hydrogen; therefore, zinc atoms will lose electrons more readily than hydrogen atoms. Hence zinc atoms will reduce hydrogen ions from the acid to form hydrogen gas and zinc ions. In fact, these reagents are commonly used for the laboratory preparation of hydrogen. The equation is Zn(s) H 2 SO 4 (aq) ZnSO 4 (aq) H 2 (g) Zn(s) 2 H (aq) Zn 2 (aq) H 2 (g) (net ionic equation) Will a reaction occur when copper metal is placed in an iron(ii) sulfate solution? No, copper lies below iron in the series, loses electrons less easily than iron, and therefore will not displace iron(ii) ions from solution. In fact, the reverse is true. When an iron nail is dipped into a copper(ii) sulfate solution, it becomes coated with free copper. The equations are Cu(s) FeSO 4 (aq) no reaction Fe(s) CuSO 4 (aq) FeSO 4 (aq) Cu(s) From Table 17.3, we may abstract the following pair in their relative position to each other: Fe Fe 2 2 e - Cu Cu 2 2 e - According to Principle 2 on the use of the activity series, we can predict that free iron will react with copper(ii) ions in solution to form free copper metal and iron(ii) ions in solution: Fe(s) Cu 2 (aq) Fe 2 (aq) Cu(s) (net ionic equation) Example 17.12 SOLUTION Example 17.13 SOLUTION Practice 17.7 Indicate whether these reactions will occur: (a) Sodium metal is placed in dilute hydrochloric acid. (b) A piece of lead is placed in magnesium nitrate solution. (c) Mercury is placed in a solution of silver nitrate.
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 452 452 CHAPTER 17 OXIDATION REDUCTION 17.6 Electrolytic and Voltaic Cells electrolysis electrolytic cell cathode anode The process in which electrical energy is used to bring about chemical change is known as electrolysis. An electrolytic cell uses electrical energy to produce a chemical reaction. The use of electrical energy has many applications in industry for example, in the production of sodium, sodium hydroxide, chlorine, fluorine, magnesium, aluminum, and pure hydrogen and oxygen, and in the purification and electroplating of metals. What happens when an electric current is passed through a solution? Let s consider a hydrochloric acid solution in a simple electrolytic cell, as shown in Figure 17.3. The cell consists of a source of direct current (a battery) connected to two electrodes that are immersed in a solution of hydrochloric acid. The negative electrode is called the cathode because cations are attracted to it. The positive electrode is called the anode because anions are attracted to it. The cathode is attached to the negative pole and the anode to the positive pole of the battery. The battery supplies electrons to the cathode. When the electric circuit is completed, positive hydronium ions (H 3 O ) migrate to the cathode, where they pick up electrons and evolve as hydrogen gas. At the same time the negative chloride ions (Cl - ) migrate to the anode, where they lose electrons and evolve as chlorine gas. Reaction at the cathode: H 3 O 1 e - H 0 H 2 O (reduction) H 0 H 0 H 2 Reaction at the anode: Cl - Cl 0 1 e - (oxidation) Cl 0 Cl 0 Cl 2 2 HCl(aq) electrolysis 9999: H 2 (g) Cl 2 (g) (net reaction) Note that oxidation reduction has taken place. Chloride ions lost electrons (were oxidized) at the anode, and hydronium ions gained electrons (were reduced) at the cathode. e Voltage source e Anode () Figure 17.3 During the electrolysis of a hydrochloric acid solution, positive hydronium ions are attracted to the cathode, where they gain electrons and form hydrogen gas. Chloride ions migrate to the anode, where they lose electrons and form chlorine gas. The equation for this process is 2 HCl(aq) H 2 (g) Cl 2 (g). Cathode ( ) Cl H 3 O
HEIN17_437-467v2.1-hr 1 7.6 8/24/06 4:09 PM Page 453 453 E L E C T R O LY T I C A N D VO LTA I C C E L L S Oxidation always occurs at the anode and reduction at the cathode. When concentrated sodium chloride solutions (brines) are electrolyzed, the products are sodium hydroxide, hydrogen, and chlorine. The overall reaction is electrolysis 2 Na(aq) 2 Cl-(aq) 2 H 2O(l) 9999: 2 Na(aq) 2 OH -(aq) H 2(g) Cl 2(g) The net ionic equation is 2 Cl-(aq) 2 H 2O(l) 2 OH -(aq) H 2(g) Cl 2(g) During electrolysis, Na ions move toward the cathode and Cl- ions move toward the anode. The anode reaction is similar to that of hydrochloric acid; the chlorine is liberated: 2 Cl-(aq) Cl 2(g) 2 eeven though Na ions are attracted by the cathode, the facts show that hydrogen is liberated there. No evidence of metallic sodium is found, but the area around the cathode tests alkaline from the accumulated OH - ions. The reaction at the cathode is 2 H 2O(l) 2 e- H 2(g) 2 OH -(aq) If electrolysis is allowed to continue until all the chloride is reacted, the solution remaining will contain only sodium hydroxide, which on evaporation yields solid NaOH. Large amounts of sodium hydroxide and chlorine are made by this process. When molten sodium chloride (without water) is subjected to electrolysis, metallic sodium and chlorine gas are formed: electrolysis 2 Na(l) 2 Cl-(l) 9999: 2 Na(s) Cl 2(g) An important electrochemical application is the electroplating of metals. Electroplating is the art of covering a surface or an object with a thin adherent electrodeposited metal coating. Electroplating is done for protection of the surface of the base metal or for a purely decorative effect. The layer deposited is surprisingly thin, varying from as little as 5 * 10-5 cm to 2 * 10-3 cm, depending on the metal and the intended use. The object to be plated is set up as the cathode and is immersed in a solution containing ions of the plating metal. When an electric current passes through the solution, metal ions that migrate to the cathode are reduced, depositing on the object as the free metal. In most cases, the metal deposited on the object is replaced in the solution by using an anode of the same metal. The following equations show the chemical changes in the electroplating of nickel: Reaction at the cathode: Ni2(aq) 2 e- Ni(s) Reaction at the anode: Ni(s) Ni2(aq) 2 e- (Ni plated out on an object) (Ni replenished in solution) Metals commonly used in commercial electroplating are copper, nickel, zinc, lead, cadmium, chromium, tin, gold, and silver. Jewelry and eyeglasses are electroplated with rhodium to prevent tarnishing.
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 454 454 CHAPTER 17 OXIDATION REDUCTION voltalc cell In the electrolytic cell shown in Figure 17.3, electrical energy from the voltage source is used to bring about nonspontaneous redox reactions. The hydrogen and chlorine produced have more potential energy than was present in the hydrochloric acid before electrolysis. Conversely, some spontaneous redox reactions can be made to supply useful amounts of electrical energy. When a piece of zinc is put in a copper(ii) sulfate solution, the zinc quickly becomes coated with metallic copper. We expect this coating to happen because zinc is above copper in the activity series; copper(ii) ions are therefore reduced by zinc atoms: Zn 0 (s) Cu 2 (aq) Zn 2 (aq) Cu 0 (s) This reaction is clearly a spontaneous redox reaction, but simply dipping a zinc rod into a copper(ii) sulfate solution will not produce useful electric current. However, when we carry out this reaction in the cell shown in Figure 17.4, an electric current is produced. The cell consists of a piece of zinc immersed in a zinc sulfate solution and connected by a wire through a voltmeter to a piece of copper immersed in copper(ii) sulfate solution. The two solutions are connected by a salt bridge. Such a cell produces an electric current and a potential of about 1.1 volts when both solutions are 1.0 M in concentration. A cell that produces electric current from a spontaneous chemical reaction is called a voltaic cell. A voltaic cell is also known as a galvanic cell. The driving force responsible for the electric current in the zinc copper cell originates in the great tendency of zinc atoms to lose electrons relative to the tendency of copper(ii) ions to gain electrons. In the cell shown in Figure 17.4, zinc atoms lose electrons and are converted to zinc ions at the zinc electrode surface; the electrons flow through the wire (external circuit) to the copper electrode. Here copper(ii) ions pick up electrons and are reduced to copper atoms, which plate out on the copper electrode. Sulfate ions flow from the CuSO 4 solution via the salt bridge into the ZnSO 4 solution (internal circuit) to complete the circuit. Flow of electrons e Digital voltmeter 1.1 SO 4 2 Salt bridge K 2 SO 4 K Zinc (anode) Copper (cathode) Zn 2 Cu 2 SO 2 4 SO 2 4 (1M ZnSO 4 solution) (1M CuSO 4 solution) Zn Zn 2 2 e Cu 2 2 e Figure 17.4 Zinc copper voltaic cell. The cell has a potential of 1.1 volts when ZnSO 4 and CuSO 4 solutions are 1.0 M. The salt bridge provides electrical contact between the two half-cells. Cu
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 455 17.6 ELECTROLYTIC AND VOLTAIC CELLS 455 The equations for the reactions of this cell are anode Zn 0 (s) Zn 2 (aq) 2 e - (oxidation) cathode Cu 2 (aq) 2 e - Cu 0 (s) (reduction) net ionic overall Zn 0 (s) Cu 2 (aq) Zn 2 (aq) Cu 0 (s) Zn(s) CuSO 4 (aq) ZnSO 4 (aq) Cu(s) The redox reaction, the movement of electrons in the metallic or external part of the circuit, and the movement of ions in the solution or internal part of the circuit of the zinc copper cell are very similar to the actions that occur in the electrolytic cell of Figure 17.3. The only important difference is that the reactions of the zinc copper cell are spontaneous. This spontaneity is the crucial difference between all voltaic and electrolytic cells. Voltaic cells use chemical reactions to produce electrical energy. Electrolytic cells use electrical energy to produce chemical reactions. Calculators, watches, radios, cell phones, and portable CD players are powered by small efficient voltaic cells called dry cell batteries. These batteries are called dry cells because they do not contain a liquid electrolyte (like the voltaic cells discussed earlier). Dry cell batteries are found in several different versions. The acid-type dry cell battery contains a zinc inner case that functions as the anode. A carbon (graphite) rod runs through the center and is in contact with the zinc case at one end and a moist paste of solid MnO 2, NH 4 Cl, and carbon that functions as the cathode. (See Figure 17.5a.) The cell produces about 1.5 volts. The alkaline-type dry cell battery is the same as the acid type except the NH 4 Cl is replaced by either KOH or NaOH. These dry cells typically last longer because the zinc anode corrodes more slowly in basic conditions. A third type of dry cell is the zinc mercury cell shown in Figure 17.5b. The reactions occurring in this cell are anode Zn 0 2 OH - ZnO H 2 O 2 e - (oxidation) cathode HgO H 2 O 2 e - Hg 0 2 OH - (reduction) net ionic Zn 0 Hg 2 Zn 2 Hg 0 overall Zn 0 HgO ZnO Hg 0 Cathode (graphite rod) Anode (zinc inner case) (a) Paste of MnO 2, NH 4 Cl, and carbon Insulation Steel cathode HgO in basic medium (KOH) e Flow of electrons (b) Zinc anode Figure 17.5 (a) A common acid-type dry cell. (b) Diagram of an alkaline zinc mercury cell.
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 456 456 CHAPTER 17 OXIDATION REDUCTION Pb-Sb alloy grids Spongy Pb H 2 SO 4 solution (electrolyte) PbO 2 Figure 17.6 Cross-sectional diagram of a lead storage battery cell. Perforated separator To offset the relatively high initial cost, this cell (1) provides current at a very steady potential of about 1.5 volts; (2) has an exceptionally long service life that is, high energy output to weight ratio; (3) is completely self-contained; and (4) can be stored for relatively long periods of time when not in use. An automobile storage battery is an energy reservoir. The charged battery acts as a voltaic cell and through chemical reactions furnishes electrical energy to operate the starter, lights, radio, and so on. When the engine is running, a generator or alternator produces and forces an electric current through the battery and, by electrolytic chemical action, restores it to the charged condition. The cell unit consists of a lead plate filled with spongy lead and a lead(iv) oxide plate, both immersed in dilute sulfuric acid solution, which serves as the electrolyte (see Figure 17.6). When the cell is discharging, or acting as a voltaic cell, these reactions occur: Pb plate (anode): Pb 0 Pb 2 2 e - (oxidation) PbO plate (cathode): PbO 2 4 H 2 e - Pb 2 2 2 H 2 O (reduction) Net ionic redox reaction: Pb 0 PbO 2 4 H 2 Pb 2 2 H 2 O Precipitation reaction on plates: Pb 2 (aq) SO 2-4 (aq) PbSO 4 (s) Because lead(ii) sulfate is insoluble, the Pb 2 ions combine with SO 2-4 ions to form a coating of PbSO 4 on each plate. The overall chemical reaction of the cell is Pb(s) PbO 2 (s) 2 H 2 SO 4 (aq) 9999: discharge 2 PbSO 4 (s) 2 H 2 O(l) cycle The cell can be recharged by reversing the chemical reaction. This reversal is accomplished by forcing an electric current through the cell in the opposite direction. Lead sulfate and water are reconverted to lead, lead(iv) oxide, and sulfuric acid: 2 PbSO 4 (s) 2 H 2 O(l) charge 999: Pb(s) PbO 2 (s) 2 H 2 SO 4 (aq) cycle The electrolyte in a lead storage battery is a 38% by mass sulfuric acid solution having a density of 1.29 g>ml. As the battery is discharged, sulfuric acid is removed, thereby decreasing the density of the electrolyte solution. The state of
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 457 CHEMISTRY IN ACTION Superbattery Uses Hungry Iron Ions Scientists are constantly trying to make longer-lasting environmentally friendly batteries to fuel our many electronic devices. Now a new type of alkaline batteries called super-iron batteries have been developed by chemists at Technion-Israel Institute of Technology in Israel. A traditional cell of this type is shown on p. 455. In the new battery, the heavy manganese dioxide cathode is replaced with super iron. (See accompanying diagram.) This special type of iron compound contains iron(vi) in compounds such as K 2 FeO 4 or BaFeO 4. Iron typically has an oxida- tion state of 2 or 3. In super iron, each iron atom is missing six electrons instead of the usual 2 or 3. This allows the battery to store 1505 J more energy than other alkaline batteries. When the battery is used (or the cell discharges), the following reaction occurs: 2 MFeO 4 3 Zn Fe 2 O 3 ZnO 2 MZnO 2 (M = K 2 or Ba) The iron compounds used in this battery are much less expensive than the current MnO 2 compounds and the products are more environmentally friendly ( Fe 2 O 3 is a form of rust). Manganese dioxide molecules in conventional batteries can only each accept 1 electron while iron(vi) compounds can absorb 3 electrons each. The super-iron compounds are highly conductive, which means the superiron battery will work well in our high-drain-rate electronic items. The accompanying graph shows a comparison between a conventional battery and a super-iron battery (both AAA). The conventional AAA battery lasts less than half as long as a super-iron AAA battery. Just think how many more CDs you could play in that time without changing the batteries! Super-iron battery Electrolyte (aqueous KOH) Super-iron cathode Separator Case, cathode current collector Cover Zinc paste anode Anode current collector AAA battery discharge potential, volts 1.5 1.0 Super-iron battery High-power alkaline battery 0.5 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 AAA battery 0.7-watt discharge time, hours charge or discharge of the battery can be estimated by measuring the density (or specific gravity) of the electrolyte solution with a hydrometer. When the density has dropped to about 1.05 g>ml, the battery needs recharging. In a commerical battery, each cell consists of a series of cell units of alternating lead lead(iv) oxide plates separated and supported by wood, glass wool, or fiberglass. The energy storage capacity of a single cell is limited, and its electrical potential is only about 2 volts. Therefore a bank of six cells is connected in series to provide the 12-volt output of the usual automobile battery. Practice 17.8 Consider the reaction Sn 2 (aq) Cu(s) Sn(s) Cu 2 (aq) (a) Which metal is oxidized and which is reduced? (b) Write the reaction occurring at the anode and at the cathode. 457
HEIN17_437-467v2.1-hr 8/24/06 4:09 PM Page 458 458 CHAPTER 17 OXIDATION REDUCTION Chapter 17 Review 17.1 Oxidation Number KEY TERM Oxidation number or oxidation state To assign an oxidation number: 1. All elements in their free state (uncombined with other elements) have an oxidation number of zero (e.g., Na, Cu, Mg, H 2, O 2, Cl 2, N 2 ). 2. H is 1, except in metal hydrides, where it is -1 (e.g., NaH, CaH 2 ). 3. O is -2, except in peroxides, where it is -1, and in OF 2, where it is 2. 4. The metallic element in an ionic compound has a positive oxidation number. 5. In covalent compounds the negative oxidation number is assigned to the most electronegative atom. 6. The algebraic sum of the oxidation numbers of the elements in a compound is zero. 7. The algebraic sum of the oxidation numbers of the elements in a polyatomic ion is equal to the charge of the ion. 17.2 Oxidation Reduction KEY TERMS Oxidation reduction or redox Oxidation Reduction Oxidizing agent Reducing agent Oxidation reduction is a chemical process in which electrons are transferred from one atom to another to change the oxidation number of the atom. When the oxidation number increases, oxidation occurs, resulting in the loss of electrons. When the oxidation number decreases, reduction occurs, resulting in the gain of electrons. 17.3 Balancing Oxidation Reduction Equations Trial and error or inspection methods. Can be balanced by writing and balancing the halfreactions for the overall reaction. Can be balanced by the change-in-oxidation-number method. 17.4 Balancing Ionic Redox Equations To balance equations that are ionic, charge must also be balanced (in addition to atoms and ions). The ion electron method for balancing equations is: Step 1 Write the two half-reactions that contain the elements being oxidized and reduced using the entire formula of the ion or molecule. Step 2 Balance the elements other than oxygen and hydrogen. Step 3 Balance oxygen and hydrogen. Acidic solution: For reactions in acidic solution, use H and H 2 O to balance oxygen and hydrogen. For each oxygen needed, use one H 2 O. Then add H as needed to balance the hydrogen atoms. Basic solution: For reactions in alkaline solutions, first balance as though the reaction were in an acidic solution, using Steps 1 3. Then add as many OH - ions to each side of the equation as there are H ions in the equation. Now combine the H and OH - ions into water (for example, 4 H and 4 OH - give 4 H 2 O). Rewrite the equation, canceling equal numbers of water molecules that appear on opposite sides of the equation. Step 4 Add electrons (e - ) to each half-reaction to bring them into electrical balance. Step 5 Since the loss and gain of electrons must be equal, multiply each half-reaction by the appropriate number to make the number of electrons the same in each half-reaction. Step 6 Add the two half-reactions together, canceling electrons and any other identical substances that appear on opposite sides of the equation. 17.5 Activity Series of Metals KEY TERM Activity series of metals The activity series lists metals from most to least reactive. A free metal can displace anything lower on the activity series. 17.6 Electrolytic and Voltaic Cells KEY TERMS Electrolysis Electrolytic cell Cathode Anode Voltaic cell Electrolysis is the process of using electricity to bring about chemical change. A typical electrolytic cell is shown next: