Chapter 14 Oxidation-Reduction Reactions

Transcription

1 Chapter 14 Oxidation-Reduction Reactions 14.1 (a) electrolytic cell; (b) cathode; (c) oxidation; (d) half-reaction; (e) electrochemistry; (f) electrode; (g) oxidizing agent; (h) oxidation-reduction reaction; (i) corrosion; (j) oxidation number 14.2 (a) voltaic (galvanic)cell; (b) anode; (c) reduction; (d) half-cell; (e) battery; (f) salt bridge; (g) reducing agent; (h) electrolysis; (i) spontaneous 14.3 An oxidation-reduction reaction (redox reaction) is a reaction where electrons are transferred from one reactant to another Oxidation is always coupled with reduction because the electrons released in the oxidation process must be absorbed by another substance, resulting in its reduction A substance is oxidized when its oxidation number increases. For monatomic ions, the charge becomes more positive as a result of losing electrons (i.e. Zn Zn e ; the charge on Zn changes from zero to 2+) A substance is reduced when its oxidation number decreases. For monatomic ions, the charge becomes more negative as a result of gaining electrons (i.e. Cl 2 + 2e 2Cl ; the charge on Cl changes from zero to 2 ) (a) Mg(s) + Cu(NO 3 ) 2 (aq) Cu(s) + Mg(NO 3 ) 2 (aq) (b) Magnesium is in its elemental form as a reactant, so we know that its charge is zero. However, as a product, the charge on magnesium is 2+ (because the charge of each nitrate ion is 1 ). The charge on magnesium increases by two when it transfers two electrons to copper (Mg Mg e ) Mg( s ) + Cu (NO 3)2( a q) Cu ( s ) + Mg(NO 3)2( a q) (c) As a reactant, the charge on copper is 2+ (because the charge of each nitrate ion is 1 ). As a product, copper is in its elemental form so its charge is zero. The charge on copper decreases by 2 when it gains 2 electrons from magnesium (Cu e Cu) (a) Ni(s) + Pb(NO 3 ) 2 (aq) Pb(s) + Ni(NO 3 ) 2 (aq) (b) As a reactant, nickel is in its elemental form so we know that its charge is zero. As a product, the charge on nickel is 2+ (because the charge of each nitrate ion is 1 ). The charge on nickel increases by 2 when it transfers 2 electrons to lead (Ni Ni e ) Ni( s ) + Pb (NO 3)2( a q) Pb( s ) + Ni (NO 3)2( a q) (c) As a reactant, lead has a charge of 2+ (because the charge of each nitrate ion is 1 ). As a product, the charge on lead is zero because it is in its elemental form. The charge on lead decreases by two when it gains two electrons from nickel (Pb e Pb) To begin, we determine the charge on each reactant and product. The species whose charge increases is oxidized, and the species whose charge decreases is reduced. The species that contains the element being reduced is the oxidizing agent, and the species that contains the element being oxidized is the reducing agent. 14 1

2 Mg( s ) + SnSO 4( a q) MgSO 4( a q) + Sn ( s ) (a) The oxidation number of magnesium increases (0 2+), so Mg is oxidized in this reaction. (b) The oxidation number of tin decreases (2+ 0), so Sn 2+ is reduced in this reaction. (c) SnO 4 is the oxidizing agent because it accepts electrons from Mg. (d) Mg is the reducing agent because it gives two electrons to Sn 2+. (e) Sulfate is a spectator ion. Tin ions (red) make contact with the surface of the magnesium metal. Magnesium atoms (gray) transfer two electrons to the tin ions. The tin ions deposit as tin metal. Mg 2 + Mg 2 e Sn To begin, we determine the charge on each reactant and product. The species whose charge increases is oxidized, and the species whose charge decreases is reduced. The species that contains the element being reduced is the oxidizing agent, and the species that contains the element being oxidized is the reducing agent Mn(s) + CdSO 4 (aq) MnSO 4 (aq) + Cd(s) (a) The oxidation number of manganese increases (0 2+), so Mn is oxidized in this reaction. (b) The oxidation number of cadmium decreases (2+ 0), so Cd 2+ is reduced in this reaction. (c) Cd 2+ is the oxidizing agent because it accepts electrons from Mn. (d) Mn is the reducing agent because it gives two electrons to Cd 2+. (e) Sulfate is a spectator ion. Cadmium ions (red) make contact with the surface of the manganese metal. Manganese atoms (grey) transfer two electrons to the cadmium ions. The cadmium ions deposit as cadmium metal. 14 2

3 M n 2 + M n 2 e C d An oxidation number is part of a bookkeeping method for tracking where electrons go during a chemical reaction. We assign oxidation numbers to atoms in covalent compounds or polyatomic ions as if they were ions, giving them charges established by a set of rules. Then we can determine which substance contains an atom(s) that loses electrons, and which contains an atom(s) that gains electrons. Substances with atoms that lose electrons are oxidized, and the oxidation number of the oxidized atom(s) increases. Substances with atoms that gain electrons are reduced, and the oxidation number of the oxidized atom(s) decreases The oxidation number of the atom that is oxidized increases. The oxidation number of the atom that is reduced decreases The oxidation number of atoms in their elemental form is zero. The charges of monatomic ions are their oxidation numbers. (a) Cr, 0; (b) Br 2, 0; (c) Cr 3+, 3+; (d) Br, The oxidation number of atoms in their elemental form is zero. The charges of monatomic ions are their oxidation numbers. (a) Ca, 0; (b) S 8, 0; (c) Ca 2+, 2+; (d) S 2, In compounds and ions, we generally assign to the most electronegative element in a substance the oxidation number we expect from its position in the periodic table. For example, oxygen is more electronegative than sulfur, so the oxidation number of oxygen is 2 (because an oxygen atom needs two more electrons to fill its valence shell, Table 14.1 Rule 6). We determine the oxidation number of sulfur from the chemical formula and net charge (if any) on the species. For SO 3 we can write: Net charge = + SO 3 1 S 3 O SO 3 does not have a charge, so its net charge is zero. From Table 14.1 (Rule 6), assign oxygen an oxidation number of oxygen of 2. We assign an oxidation number to sulfur that gives SO 3 a net charge of zero. Net charge = 0 Total negative oxidation numbers = 3 ( 2) =

4 total positive 0 = + 6 S ( ) The oxidation number of sulfur in SO 3 is 6+. S = 6+, O = In compounds and ions, we generally assign to the most electronegative element in a substance the oxidation number we expect from its position in the periodic table. For example, oxygen is more electronegative than nitrogen, so the oxidation number of oxygen is 2 (because an oxygen atom needs two more electrons to fill its valence shell, Table 14.1 Rule 6). We determine the oxidation number of nitrogen from the chemical formula of the species and its net charge (if any). For NO 2 we can write: Net charge = + NO 2 1 N 2 O NO 2 does not have a charge, so its net charge is zero. From Table 14.1 (Rule 6), we assign oxygen an oxidation number of oxygen of 2. We assign an oxidation number to nitrogen that gives NO 2 a net charge of zero. Net charge = 0 Total negative oxidation numbers = 2 ( 2) = 4. total positive 0 = + 4 N ( ) The oxidation number of nitrogen in NO 2 is 4+. N = 4+, O = In compounds and ions, we assign to the most electronegative element in a substance the oxidation number we expect from its position in the periodic table. For example, oxygen is more electronegative than phosphorus, so the oxidation number we assign to oxygen is 2 (because an oxygen atom needs two more electrons to fill its valence shell, Table 14.1 Rule 6). We assign an oxidation number to phosphorus based on the chemical formula and net charge (if any) of the substance. (a) P 4 O 10 Net charge = + PO N 10 O Net charge = 0 Total negative oxidation numbers = 10 ( 2) = 20 total positive 4 P 0 = (20) 14 4

5 P = 20 4 = 5 Each phosphorus atom in P 4 O 10 has an oxidation number of 5+. (b) P 4 O 6 Net charge = + PO N 6 O Net charge = 0 Total negative oxidation numbers = 6 ( 2) = 12 total positive 4 P 0 = (12) P = 12 4 = 3 Each phosphorus atom in P 4 O 6 has an oxidation number of 3+. (c) P 4 O 8 Net charge = + PO N 8 O Net charge = 0 Total negative oxidation numbers = 8 ( 2) = 16 total positive 4 P 0 = (16) P = 16 4 = 4 Each phosphorus atom in P 4 O 8 has an oxidation number of In compounds and ions, the most electronegative element in a substance generally takes the oxidation number we expect from its position in the periodic table. For example, oxygen is more electronegative than chlorine, so the oxidation number of oxygen is 2 (because an oxygen atom needs two more electrons to fill its valence shell, Rule 6). We assign oxidation numbers to chlorine based on the chemical formulas and net charges of the substances. (a) Cl 2 O 14 5

6 Net charge = + Cl oxidation numbers oxidation numbers 2O 2 Cl 1 O Net charge = 0 Total negative oxidation numbers = 1 ( 2) = 2 total positive 2 Cl 0 = (2) Cl = 2 2 = 1 Each chorine atom in Cl 2 O has an oxidation number of 1+. (b) ClO 3 Net charge = + ClO 3 1 Cl 3 O Net charge = 0 Total negative oxidation numbers = 3 ( 2) = 6 total positive Cl 0 = (6) Cl = 6 Each chlorine atom in ClO 3 has an oxidation number of 6+. (c) Cl 2 O 3 Net charge = + Cl oxidation numbers oxidation numbers 2O 3 2 Cl 3 O Net charge = 0 Total negative oxidation numbers = 3 ( 2) = 6 total positive 2 Cl 0 = (6) 14 6

7 Cl = 6 2 = 3 Each chlorine atom in Cl 2 O 3 has an oxidation number of 3+. (d) ClO 2 Net charge = + ClO 2 1 Cl 2 O Net charge = 0 Total negative oxidation numbers = 2 ( 2) = 4 total positive Cl 0 = (4) Cl = 4 Each chlorine atom in ClO 2 has an oxidation number of 4+. (e) Cl 2 O 5 Net charge = + Cl oxidation numbers oxidation numbers 2O 5 2 Cl 5 O Net charge = 0 Total negative oxidation numbers = 5 ( 2) = 10 This give us: total positive 2 Cl 0 = (10) Cl = 10 2 = 5 Each chlorine atom in Cl 2 O 5 has an oxidation number of In compounds and ions, the most electronegative element in a substance generally takes the oxidation number we expect from its position in the periodic table. When a substance contains hydrogen, we must remember (Table 14.1 Rule 4) that the oxidation number of hydrogen is 1+, unless it is combined with a metal (where it is 1 ). (a) AlPO 4 Aluminum s oxidation number is 3+ (predicted from the periodic table). Net charge = + AlPO 4 1 Al + 1 P 4 O 14 7

8 Net charge = 0 Total negative oxidation numbers = 4 ( 2) = 8 total positive 3+ P 0 = (8) 0 = 3 + P 8 P = 5 The oxidation number of P in AlPO 4 is 5+. (b) PF 5 Fluorine s oxidation number is 1 (predicted from Table 14.1 Rule 3 and the periodic table). Net charge = + PF 5 P 5 F Net charge = 0 Total negative oxidation numbers = 5 ( 1) = 5 This gives: total positive P 0 = (5) P = 5 The oxidation number of P in PF 5 is 5+. (c) H 3 PO 4 The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic table). Because we analyzed the phosphate ion in part (a), we can predict that the oxidation number of phosphorus is 5+. Net charge = + H oxidation numbers oxidation numbers 3PO 4 3 H + 1 P 4 O Net charge = 0 Total negative oxidation numbers = 4 ( 2) = 8 total positive 3+ P 0 = (8) 0 = 3 + P 8 P = 5 The oxidation number of P in H 3 PO 4 is

9 (d) H 3 PO 2 The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic table). Net charge = + H oxidation numbers oxidation numbers 3PO 2 3 H + 1 P 2 O Net charge = 0 Total negative oxidation numbers = 2 ( 2) = 4 This gives: total positive 3+ P 0 = (4) 0 = 3 + P 4 P = 1 The oxidation number of P in H 3 PO 2 is 1+. (e) PH 3 The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic table). In this example, hydrogen has a positive oxidation number, so phosphorus has a negative oxidation number. Net charge = + PH 3 3 H P Net charge = 0 Total positive oxidation numbers = 3 (+1) = 3 total positive 3 0 = oxidation numbers ( P) 0 = 3 + P P = 3 The oxidation number of P in PH 3 is 3. (f) H 3 PO 3 The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic table). total positive total negative Net charge = oxidation numbers + oxidation numbers H3PO3 3 H + 1 P 3 O Net charge = 0 Total negative oxidation numbers = 3 ( 2) =

10 total positive 3+ P 0 = (6) 0 = 3 + P 6 P = 3 The oxidation number of P in H 3 PO 3 is In compounds and ions, the most electronegative element in a substance generally takes the oxidation number we expect from its position in the periodic table. (a) NaI NaI is an ionic compound, so sodium s oxidation number is 1+, based on its position in the periodic table. Because the overall charge on NaI is zero, iodine must have a 1 oxidation number. The oxidation number of I in NaI is 1. (b) I 2 Any element in its pure form (i.e. O 3, P 4, S 8, Fe, Na) has an oxidation number of zero. The oxidation number of I in I 2 is 0. (c) IO 2 Net charge = + IO 2 1 I 2 O Net charge = 0 Total negative oxidation numbers = 2 ( 2) = 4 total positive I 0 = (4) I = 4 The oxidation number of I in IO 2 is 4+. (d) I 2 O 7 Net charge = + IO I 7 O Net charge = 0 Total negative oxidation numbers = 7 ( 2) =

11 total positive 2 I 0 = (14) I = 14 2 I = 7 The oxidation number of I in I 2 O 7 is 7+. (e) KIO 4 Potassium s oxidation number is 1+ (predicted from Table 14.1 Rule 4 and the periodic table). Net charge = + KIO 4 1 K + 1 I 4 O Net charge = 0 Total negative oxidation numbers = 4 ( 2) = 8 This give us: total positive 1+ I 0 = (8) 0 = 1 + I 8 I = 7 The oxidation number of I in KIO 4 is 7+. (f) Ca(IO) 2 Calcium s oxidation number is 2+ (predicted from Table 14.1 Rule 4 and the periodic table). Net charge = + Ca(IO) 2 1 Ca + 2 I 2 O Net charge = 0 Total negative oxidation numbers = 2 ( 2) = 4 total positive I 0 = (4) 0 = I 4 I =

12 14.21 SO 4 2 I = 1+ The oxidation number of I in Ca(IO) 2 is 1+. Net charge = + 2 oxidation numbers oxidation numbers SO 4 S 4 O Net charge = 2 Total negative oxidation numbers = 4 ( 2) = 8 total positive S 2 = (8) S = NO 3 The oxidation numbers of sulfur and oxygen in SO 4 2 are 6+ and 2, respectively. Net charge = + oxidation numbers oxidation numbers NO 3 N 3 O Net charge = 1 Total negative oxidation numbers = 3 ( 2) = 6 total positive N 1 = (6) N = 5 The oxidation numbers of nitrogen and oxygen in NO 3 are 5+ and 2, respectively (a) BrF 7 Fluorine s oxidation number is 1 (predicted from Table 14.1 Rule 3 and the periodic table). Net charge = + BrF 7 1 Br 7 F Net charge = 0 Total negative oxidation numbers = 7 ( 1) =

13 total positive Br 0 = (7) Br = 7 The oxidation number of Br in BrF 7 is 7+. (b) BrO 3 Net charge = + oxidation numbers oxidation numbers BrO 3 1 Br 3 O Net charge = 1 Total negative oxidation numbers = 3 ( 2) = 6 total positive Br 1 = (6) Br = 5 The oxidation number of Br in BrO 3 is 5+. (c) For monatomic ions, the oxidation number and charge are the same. Br has an oxidation number of 1. (d) BrCl 3 Chloride has an oxidation number of 1 (predicted from Table 14.1 Rule 5c and the periodic table). Net charge = + BrCl 3 1 Br 3 Cl Net charge = 0 Total negative oxidation numbers = 3 ( 2) = 3 total positive I 0 = (3) Br = 3 The oxidation number of Br in BrCl 3 is

14 (e) BrOCl 3 We predict the oxidation numbers of oxygen (2 ) and chlorine (1 ) from Table 14.1 Rules 4 and 6, and the periodic table. Net charge = + BrOCl 3 1 Br 1 O + 3 Cl Net charge = 0 Total negative oxidation numbers = 1 (2 ) + 3 ( 1) = 5 total positive Br 0 = (5) Br = 5 The oxidation number of Br in BrOCl 3 is (a) CrCl 2 The oxidation number of chlorine is 1 (periodic table). This means that the oxidation number of chromium in CrCl 2 must be 2+. (b) CrSO 4 The sulfate ion has a 2 charge. This means that the oxidation number of chromium in CrSO 4 is 2+. (c) Cr(NO 3 ) 3 Each nitrate ion has a 1 charge. Because there are three nitrate ions in the chemical formula, the oxidation number of chromium in Cr(NO 3 ) 3 is 3+. (d) Cr 2 (SO 4 ) 3 Each sulfate ion has a 2 charge. We can write: Net charge = + Cr oxidation numbers oxidation numbers 2(SO 4) 3 2 Cr 2 3 ( SO 4 ) Total negative oxidation numbers = 3 ( 2) = 6 total positive 2 Cr 0 = (6) Cr = 6 2 = 3 The oxidation number of chromium in Cr 2 (SO 4 ) 3 is (a) ClO 2 Net charge = + ClO 2 1 Cl 2 O Net charge = 0 Total negative oxidation numbers = 2 ( 2) =

15 total positive Cl 0 = (4) Cl = 4 The oxidation numbers of oxygen and chlorine in ClO 2 are 2 and 4+, respectively. (b) CaF 2 Because fluorine is the most electronegative element, it s oxidation number is 1. Because there are two fluoride ions and only one calcium ion, calcium s oxidation number is 2+. (c) H 2 TeO 3 We predict the oxidation numbers of hydrogen (1+) and oxygen (2 ) from Table 14.1 Rules 4 and 6, and the periodic table. Net charge = + H oxidation numbers oxidation numbers 2TeO 3 2 H + 1 Te 3 O Net charge = 0 Total negative oxidation numbers = 3 ( 2) = 6 total positive 2 + Te 0 = (6) Te = 4 The oxidation numbers of H, O, and Te in H 2 TeO 3 are 1+, 2, and 4+, respectively. (d) NaH In compounds with metals, the oxidation number of hydrogen is 1 (Table 14.1 Rule 4). The oxidation number of sodium is (a) Na 2 O 2 The oxidation number of sodium ion is always 1+ in compounds. Because each sodium ion has a 1+ oxidation number, each oxygen ion must have a 1 oxidation number (there is a 1:1 Na:O ratio). (b) Fe(NO 3 ) 3 The charge on each nitrate ion is 1. Because there are three nitrate ions in the chemical formula, the total negative charge is 3. The oxidation number of iron is 3+. The oxidation number of oxygen is 2 (Table 14.1 Rule 6 and the periodic table) so, in NO 3, the oxidation number of nitrogen is 5+. Because we know the charge of a nitrate ion, we do not need to include the iron in the calculation. Net charge = + oxidation numbers oxidation numbers NO 3 N 3 O Net charge = 1 Total negative oxidation numbers = 3 ( 2) =

16 total positive N 1 = (6) N = 5 (c) Sc 2 O 3 The oxidation number of oxygen is 2 (Table 14.1 Rule 6 and the periodic table). The oxidation number of scandium is 3+. Net charge = + Sc oxidation numbers oxidation numbers 2O 3 2 Sc 3 O Net charge = 0 Total negative oxidation numbers = 3 ( 2) = 6 total positive 2 Sc 0 = (6) Sc = 6 2 = 3 (d) LiH According to Table 14.1 Rule 4, in compounds with metals the oxidation number of hydrogen is 1. The oxidation number of lithium is (a) NO 2 Net charge = + oxidation numbers oxidation numbers NO 2 N 2 O Net charge = 1 Total negative oxidation numbers = 2 ( 2) = 4 total positive N 1 = (4) N = 3 The oxidation numbers of N and O in NO 2 are 3+ and 2, respectively

17 (b) Cr 2 O 7 2 Net charge = + 2 oxidation numbers oxidation numbers Cr2O 7 2 Cr 7 O Net charge = 2 Total negative oxidation numbers = 7 ( 2) = 14 total positive 2 Cr 2 = (14) Cr = 6 The oxidation numbers of Cr and O in Cr 2 O 7 2 are 6+ and 2, respectively. (c) AgCl 2 From the periodic table we can predict that chlorine has an oxidation number of 1. Because there are two chloride ions in the species, and the overall charge of the species is 1, we can predict that silver has an oxidation number of 1+. To verify this, we can calculate the overall charge and see if it matches the net charge on the ion. Net charge = + oxidation numbers oxidation numbers AgCl 2 Ag 2 Cl Net charge = 1 Total negative oxidation numbers = 2 (1 ) = 2 Total positive oxidation numbers = +1 Overall charge = 1 2 = 1 The overall charge and the charge we calculated from the predicted oxidation numbers are the same. The oxidation numbers of Ag and Cl in AgCl 2 are 1+ and 1, respectively. (d) SO 3 2 Net charge = + 2 oxidation numbers oxidation numbers SO 3 1 S 3 O Net charge = 2 Total negative oxidation numbers = 3 ( 2) = 6. total positive S 2 = (6) S = 4 The oxidation numbers of S and O in SO 3 2 are 4+ and 2, respectively

18 (e) CO 3 2 Net charge = + 2 oxidation numbers oxidation numbers CO 3 1 C 3 O Net charge = 2 Total negative oxidation numbers = 3 ( 2) = 6. total positive C 2 = (6) C = 4 The oxidation numbers of C and O in CO 3 2 are 4+ and 2, respectively (a) H 2 PO 4 We determine the oxidation numbers of hydrogen (1+) and oxygen (2 ) from Table 14.1 Rules 4 and 6, and their positions on the periodic table. The oxidation number of phosphorus is 5+ as shown below: Net charge = + oxidation numbers oxidation numbers H2PO 4 2 H + 1 P 4 O Net charge = 1 Total negative oxidation numbers = 4 ( 2) = 8 total positive 2 + P 1 = (8) 1 = 2 + P 8 P = 5 (b) FeCl 6 3 We determine the oxidation number of chlorine (1 ) from Table 14.1 Rule 5c and its position on the periodic table. The oxidation number of iron in FeCl 6 3 is 3+, as shown: Net charge = + 3 oxidation numbers oxidation numbers FeCl 6 Fe 6 Cl Net charge = 3 Total negative oxidation numbers = 6 ( 1) =

19 total positive Fe 3 = (6) Fe = 3 (c) ClO 2 We determine the oxidation number of oxygen (2 ) from Table 14.1 Rule 6 and its position on the periodic table. The oxidation number of chlorine in ClO 2 is 3+, as shown: Net charge = + oxidation numbers oxidation numbers ClO 2 Cl 2 O Net charge = 1 Total negative oxidation numbers = 2 ( 2) = 4 total positive Cl 1 = (4) Cl = 3 (d) SiF 2 6 We determine the oxidation number of fluorine (1 ) from Table 14.1 Rule 5c and its position on the periodic table. The oxidation number of silicon in SiF 2 6 is 4+ as shown: Net charge = + 2 oxidation numbers oxidation numbers SiF 6 Si 6 F Net charge = 2 Total negative oxidation numbers = 6 ( 1) = 6 total positive Si 2 = (6) Si = 4 (e) AsO 4 3 We determine the oxidation number of oxygen (2 ) from Table 14.1 Rule 6 and its position on the periodic table. The oxidation number of arsenic in AsO 4 3 is 5+ as shown: Net charge = + 3 oxidation numbers oxidation numbers AsO 4 As 4 O Net charge = 3 Total negative oxidation numbers = 4 ( 2) =

20 total positive As 3 = (8) As = To be classified as a redox reaction, the oxidation number of at least one element must change. For each reaction, we calculate, or predict, the oxidation numbers of each element in each species and determine if a change occurs. If an element s oxidation number becomes more positive, that element loses electrons by reducing another substance and, the substance containing that element is the reducing agent. If an element s oxidation number becomes more negative, that element gains electrons by oxidizing another substance, and the substance containing that element is the oxidizing agent. (a) This is not a redox reaction because none of the oxidation numbers change BaCl 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2HCl(aq) (b) The oxidation number of N decreases because N atoms gain electrons from H atoms. N 2 is the oxidizing agent. H 2 is the reducing agent because the oxidation number of H increases as it supplies electrons to nitrogen H 2 (g) + N 2 (g) 2NH 3 (g) (c) This is not a redox reaction because none of the oxidation numbers change H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) (d) This is not a redox reaction because none of the oxidation numbers change AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) (e) The oxidation number of O decreases because it gains electrons by oxidizing C. O 2 is the oxidizing agent. C 2 H 6 is the reducing agent because the oxidation number of C increases when it supplies electrons to O C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) To be classified as a redox reaction, the oxidation number of at least one element must change. For each reaction, we calculate, or predict, the oxidation numbers of each element in each species and determine if a change occurs. If an element s oxidation number becomes more positive, that element loses electrons by reducing another substance and, the substance containing that element is the reducing agent. If an element s oxidation number becomes more negative, that element gains electrons by oxidizing another substance, and the substance containing that element is the oxidizing agent 14 20

21 (a) The oxidation number of H decreases because it gains electrons by oxidizing Na. H 2 O is the oxidizing agent. Na is the reducing agent because the oxidation number of Na increases when Na transfers electrons to H Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) (b) The oxidation number of F decreases because it gains electrons when it oxidizes H. F 2 is the oxidizing agent. H 2 is the reducing agent because its oxidation number increases when it provides electrons to fluorine H 2 (g) + F 2 (g) 2HF(g) (c) The oxidation number of H decreases because it gains electrons when it oxidizes C. H 2 O is the oxidizing agent. C is the reducing agent because its oxidation number increases when it supplies electrons to H C(s) + H 2 O(g) CO(g) + H 2 (g) (d) This is not a redox reaction because the oxidation numbers of the elements do not change Pb(NO 3 ) 2 (aq) + 2NaCl(aq) PbCl 2 (s) + 2NaNO 3 (aq) (e) The oxidation number of O in O 2 decreases because it gains electrons by oxidizing the C in C 2 H 5 OH. O 2 is the oxidizing agent. C 2 H 5 OH is the reducing agent because the oxidation number of C increases as it supplies electrons to O C 2 H 5 OH(g) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(g) (a) N 2 (g) + O 2 (g) 2O 2e/O = 4e 2NO(g) 2N (2e/N) = 4e (b) When N 2 is transformed into NO, the oxidation number of nitrogen increases by 2. Because there are two nitrogen atoms in N 2, a total of four electrons are transferred. Each N atom loses two electrons. (c) In any redox reaction, the numbers of electrons gained and lost must be equal. When O 2 reacts with N 2 to produce NO, the oxidation number of oxygen decreases by 2. Because there are two oxygen atoms in O 2, a total of four electrons are transferred from N to O atoms. Each O atom gains two electrons

22 14.32 (a) C(g) + O 2 (g) 2O 2e/O = 4e 2CO(g) 2C (2e/C) = 4e (b) When C is transformed into CO, the oxidation number of oxygen increases by 2. Because there are two oxygen atoms in the reactants, a total of four electrons are transferred. Each C loses two electrons. (c) In any redox reaction, the numbers of electrons gained and lost must be equal. When O 2 reacts with C and produces CO, the oxidation number of oxygen decreases by 2. Because there are two oxygen atoms in O 2, a total of four electrons are transferred from C to O atoms. Each O atom gains two electrons The oxidation numbers of both vanadium and chromium change in this reaction (see below) V 2+ ( a q) + Cr 2 O 7 2 ( a q) + 14H+ ( a q) 2Cr ( 3e/Cr ) = 6e 6V 3+ ( a q) + 2Cr3+ ( a q) + 7H 2 O (l) 6V ( +1e/V) = 6e (a) V 2+ is oxidized from 2+ to 3+. (b) Cr 2 O 7 2 is reduced (Cr changes from 6+ to 3+). (c) Because Cr (in Cr 2 O 7 2 ) gains electrons, Cr 2 O 7 2 is the oxidizing agent. (d) The reducing agent transfers electrons to reduce another element. Because vanadium (in V 2+ ) loses electrons, V 2+ is the reducing agent The oxidation numbers of both manganese and chlorine change in this reaction (see below) MnO 4 (aq) + 10Cl (aq) + 16H + (aq) 2Mn (+7e/Mn) = 14e 10Cl (1e/Cl) = 10e 2Mn 2+ (aq) + 5Cl 2 (aq) + 8H 2 O(l) (a) The oxidation number of chlorine changes from 1 to 0, so Cl is oxidized. (b) The oxidation number of manganese changes from 7+ to 2+, so it is reduced. (c) Because Mn (in MnO 4 ) gains electrons, MnO 4 is the oxidizing agent. (d) The reducing agent transfers electrons to reduce another element. Because Cl (in Cl ) loses electrons, Cl is the reducing agent

23 14.35 (a) In this reaction, I 2 is both the oxidizing agent and the reducing agent. The oxidation number of I atoms in molecular iodine (I 2 ) is zero. The oxidation number of iodine in I is 1, indicating a gain of one electron. In IO, iodine has an oxidation number of 1+, indicating a loss of one electron. In this reaction, 1 electron is transferred between I atoms I 2 (aq) + 2OH (aq) 1I (1e/I) = 1e I (aq) + IO (aq) + H 2 O(l) 1I (1e/I) = 1e (b) Cr = reducing agent, H + = oxidizing agent; 2 electrons are transferred Cr(s) + 2H + (aq) Cr (2e/Cr) = 2e Cr 2+ (aq) + H 2 (g) 2H (+1e/H) = 2e (c) Cr 2 O 7 2 is both the oxidizing and reducing agent, and 12 electrons are transferred. Note that this reaction is different than the one in 14.35(a) where the same element (iodine) is both oxidized and reduced. In the case of Cr 2 O 7 2, all of the chromium atoms are reduced (oxidation number changes from 6+ to 3+), while only 6 oxygen atoms are oxidized (oxidation number changes from 2 to 0). The remaining oxygen atoms contribute to the production of water Cr 2 O 2 7 (aq) + 16H + (aq) 4Cr 3+ (aq) + 3O 2 (g) + 8H 2 O(l) 4Cr (3e/Cr) = 12e 6O (+2e/O) = 12e (d) Fe 3+ = oxidizing agent, Al = reducing agent; 3 electrons are transferred Fe 3+ (aq) + Al(s) 3Fe (1e/Fe) = 3e 3Fe 2+ (aq) + Al 3+ (aq) 1Al (3e/Al) = 3e 14 23

24 14.36 (a) BrO 3 = oxidizing agent, I = reducing agent; 6 electrons are transferred I (aq) + BrO 3 (aq) + 6H + (aq) 1Br (6e/Cl) = 6e 3I 2 (aq) + Br (aq) + 3H 2 O(l) 6I (1e/I) = 6e (b) Br = reducing agent, BrO 3 = oxidizing agent, 5 electrons are transferred Br (aq) + BrO 3 (aq) + 6H + (aq) 1Br (5e/Br) = 5e 3Br 2 (aq) + 3H 2 O(l) 5Br (1e/Br) = 5e (c) XeF 4 = oxidizing agent, H 2 O = reducing agent; 4 electrons are transferred XeF 4 (s) + 2H 2 O(l) 1Xe (4e/Xe) = 4e 2O (2e/O) = 4e Xe(g) + 4HF(aq) + O 2 (g) (d) Pb = reducing agent, H + = oxidizing agent; 2 electrons are transferred Pb(s) + 2H + (aq) 1Pb (2e/Pb) = 2e Pb 2+ (aq) + H 2 (g) 2H (1e/H) = 2e 14 24

25 14.37 In an electrochemical cell, oxidation takes place at the anode and reduction takes place at the cathode. Iron is being oxidized, so the iron electrode in the Fe(NO 3 ) 2 solution is the anode half-cell. Nickel is being reduced, so the nickel electrode in the Ni(NO 3 ) 2 solution is the cathode half-cell. The half-reaction in the anode half-cell is: Fe(s) Fe 2+ (aq) + 2e The half-reaction in the cathode half-cell is: Ni 2+ (aq) + 2e Ni(s) Voltmeter Anode Cathode Fe Salt Bridge Ni Fe(NO 3 ) 2 (aq) Ni(NO 3 ) 2 (aq) In an electrochemical cell, oxidation takes place at the anode and reduction takes place at the cathode. Because magnesium is being oxidized, the Mg electrode in the Mg(NO 3 ) 2 solution is the anode half-cell. Tin is being reduced, so the Sn electrode in the Sn(NO 3 ) 2 solution is the cathode half-cell. The half-reaction in the anode half-cell is: Mg(s) Mg 2+ (aq) + 2e The half-reaction in the cathode half-cell is: Sn 2+ (aq) + 2e Sn(s) Voltmeter Anode Cathode Mg Salt Bridge Sn Mg(NO 3 ) 2 (aq) Sn(NO 3 ) 2 (aq) 14 25

26 14.39 As the cell runs, iron oxidizes causing the reduction of Ni 2+ ions. The iron electrode loses mass as Fe atoms on its surface oxidize and become Fe 2+ (aq). The nickel electrode gains mass as Ni 2+ ions from the solution are reduced to Ni(s) which deposits on the electrode. The nitrate ions are not shown. The concentration of Fe 2+ in the solution in the anode half-cell increases and the concentration of Ni 2+ in the solution in the cathode half-cell decreases. Fe 2+ Fe Ni 2+ Ni Iron Electrode Nickel Electrode As the cell runs, magnesium oxidizes causing the reduction of Sn 2+. The magnesium electrode loses mass as Mg atoms on its surface oxidize and become Mg 2+ (aq). The tin electrode gains mass as Sn 2+ ions from the solution are reduced to Sn(s) and deposit on the Sn electrode. The nitrate ions are not shown. The concentration of Mg 2+ in the solution in the anode half-cell increases, and the concentration of Sn 2+ in the solution in the cathode half-cell decreases. Mg Sn 2+ Sn Mg 2+ Iron Electrode Nickel Electrode To answer this question, we determine the oxidation numbers of the elements in each substance. The substances whose oxidation numbers change are shown below: Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq) 2PbSO 4 (s) + 2H 2 O(l) Pb(s) is oxidized, and the Pb in PbO 2 (s) is reduced. The oxidizing agent is PbO 2 (s), and the reducing agent is Pb(s) To answer this question, we determine the oxidation numbers of the elements in each substance. The substances whose oxidation numbers change are shown below: Zn(s) + 2MnO 2 (s) + 2NH 4 + (aq) Zn 2+ (aq) + Mn 2 O 3 (s) + 2NH 3 (aq) + 2H 2 O(l) Zn is oxidized, and the Mn in MnO 2 is reduced. The oxidizing agent is MnO 2, and the reducing agent is Zn

27 14.43 Note that in the cases of both half-reactions, there is a material balance (the atoms are balanced) but not a charge balance. To determine which half-reaction represents oxidation and which represents reduction, we begin by assigning oxidation numbers. In the Cd/Cd(OH) 2 half-reaction, the oxidation number of cadmium changes from 0 to 2+. This is the oxidation half-reaction because cadmium is oxidized Cd(s) + 2OH (aq) Cd(OH) 2 (s) By adding two electrons as reaction products, we balance the overall charge in the half-reaction. Both the product and reactant sides of the equation have a net 2 charge. When a half-reaction is properly balanced, both the atoms and charge balance. Cd(s) + 2OH (aq) Cd(OH) 2 (s) + 2e 2 2 In the NiO 2 /Ni(OH) 2 half-reaction, the oxidation number of nickel changes from 4+ to 2+. This is the reduction half-reaction because nickel is reduced NiO 2 (s) + 2H 2 O(l) Ni(OH) 2 (s) + 2OH (aq) By adding two electrons as reactants, we balance the overall charge in the half-reaction. Both the product and reactant sides of the equation have a net 2 charge. When a half-reaction is properly balanced, both the atoms and charge balance. 2e + NiO 2 (s) + 2H 2 O(l) Ni(OH) 2 (s) + 2OH (aq) Both half-reactions show a material balance, but not a charge balance. In the Zn/Zn(OH) 2 half-reaction, the oxidation number of Zn changes from 0 to 2+. This is the oxidation half-reaction because zinc is oxidized Zn(s) + 2OH (aq) Zn(OH) 2 (s) By adding two electrons as products, we balance the overall charge in the half-reaction. Both the product and reactant sides of the equation have a 2 charge. When a half-reaction is properly balanced, both the atoms and charge balance. Zn(s) + 2OH (aq) Zn(OH) 2 (s) + 2e 2 2 In the AgO/Ag 2 O half-reaction the oxidation number of silver changes from 2+ to 1+. This is the reduction half-reaction because silver is reduced AgO(s) + 2H 2 O(l) Ag 2 O(s) + 2OH (aq) We add two electrons as reactants to balance the charge, because two silver atoms undergo the change. Both the product and reactant sides of the equation have a 2 charge. When a half-reaction is properly balanced, both the atoms and charge balance

28 2e + 2AgO(s) + 2H 2 O(l) Ag 2 O(s) + 2OH (aq) To balance a simple half-reaction, we first balance the atoms and then add electrons to the side with more positive charge to balance the charge. (a) Fe 3+ (aq) Fe(s) The atoms are balanced. We balance the charge by adding three electrons to the reactant side of the equation. 3e + Fe 3+ (aq) Fe(s) 0 0 (b) Zn(s) Zn 2+ (aq) The atoms are balanced. We balance the charge by adding two electrons to the product side of the equation. Zn ( s ) Zn 2+ ( a q) + 2e 0 0 (c) Cl (aq) Cl 2 (g) The atoms are not balanced, so we add a coefficient of 2 in front of Cl. 2Cl (aq) Cl 2 (g) Then we balance the charge by adding two electrons to the product side of the equation. 2Cl ( a q) Cl 2 (g) + 2e 2 2 (d) Fe 2+ (aq) Fe 3+ (aq) The atoms are balanced. We balance the charge by adding one electron to the product side of the equation. Fe 2+ ( a q) Fe 3+ ( a q) + e (a) Ni(s) Ni 2+ (aq) The atoms are balanced. We balance the charge by adding two electrons to the product side of the equation. Ni(s) Ni 2+ (aq) + 2e 0 0 (b) Br 2 (l) Br (aq) The atoms are not balanced, so we add a coefficient of 2 in front of Br. Br 2 (l) 2Br (aq) Then we balance the charge by adding two electrons to the reactant side of the equation. Br 2Br 2 (l) + 2e (aq) 2 2 (c) Mg 2+ (aq) Mg(s) The atoms are balanced. We balance the charge by adding two electrons to the reactant side of the equation

29 Mg 2+ ( a q) + 2e Mg( s ) 0 0 (d) Cr 3+ (aq) Cr 2+ (aq) The atoms are balanced. We balance the charge by adding one electron to the reactant side of the equation. Cr 3+ ( a q) + e Cr 2+ ( a q) (a) Zn(s) + Fe(NO 3 ) 3 (aq) Zn(NO 3 ) 2 (aq) + Fe(s) Nitrate ions, NO 3, are spectator ions in this reaction so we can eliminate them in the first part of the balancing process, and add them back into the equation after we have balanced the half-reactions. Eliminating the nitrate ions gives us the ionic equation: Zn(s) + Fe 3+ (aq) Zn 2+ (aq) + Fe(s) We can write the oxidation half-reaction as: Zn(s) Zn 2+ (aq) skeletal ionic equation oxidation half-reaction The atoms are balanced. We can balance the charge by adding two electrons to the right side of the equation: Zn(s) Zn 2+ (aq) + 2e The reduction half-reaction is: Fe 3+ (aq) Fe(s) balanced oxidation half-reaction reduction half-reaction We can balance the charge by adding three electrons to the reactant side of the equation: 3e + Fe 3+ (aq) Fe(s) balanced reduction half-reaction Then we equalize the number of electrons lost and gained in the two half-reactions by multiplying each by an appropriate coefficient: 3 [Zn(s) Zn 2+ (aq) + 2e ] 2 [3e + Fe 3+ (aq) Fe(s)] 3Zn(s) 3Zn 2+ (aq) + 6e 6e + 2Fe 3+ (aq) 2Fe(s) Next we add the two half-reactions and cancel the six electrons that appear on both sides of the equation. 3Zn(s) 3Zn 2+ (aq) + 6e 6e + 2Fe 3+ (aq) 2Fe(s) 3Zn(s) + 2Fe 3+ (aq) 3Zn 2+ (aq) + 2Fe(s) Finally, we replace the nitrate ions to complete the balanced equation. 3Zn(s) + 2Fe(NO 3 ) 3 (aq) 3Zn(NO 3 ) 2 (aq) + 2Fe(s) (b) Mn(s) + HCl(aq) MnCl 2 (aq) + H 2 (g) balanced skeletal equation balanced Chloride ions, Cl, are spectator ions in this reaction so we can eliminate them in the first part of the balancing process, and add them back into the equation after we have balanced the half-reactions. Eliminating the chloride ions gives us the ionic equation: 14 29

30 Mn(s) + H + (aq) Mn 2+ (aq) + H 2 (g) The oxidation half-reaction is: Mn(s) Mn 2+ (aq) skeletal ionic equation oxidation half-reaction The atoms are balanced, and we can balance the charge by adding two electrons to the right side of the equation: Mn(s) Mn 2+ (aq) + 2e The reduction half-reaction is: H + (aq) H 2 (g) balanced oxidation half-reaction reduction half-reaction First, we balance the atoms by adding a coefficient of 2 in front of H +. Next we balance the charge by adding two electrons to the reactant side of the equation. 2e + 2H + (aq) H 2 (g) balanced reduction half-reaction The electrons exchanged between the two half-reactions are already balanced, so we add the two halfreactions and cancel the two electrons that appear on both sides of the equation. Mn(s) Mn 2+ (aq) + 2e 2e + 2H + (aq) H 2 (g) Mn(s) + 2H + (aq) Mn 2+ (aq) + H 2 (g) Finally, we replace the chloride ions to complete the balanced equation. Mn(s) + 2HCl(aq) MnCl 2 (aq) + H 2 (g) balanced skeletal equation balanced (a) Al(s) + Fe(NO 3 ) 3 (aq) Al(NO 3 ) 3 (aq) + Fe(NO 3 ) 2 (aq) Nitrate ions, NO 3, are spectator ions in this reaction, so we can eliminate them while we balance the two half-reactions and add them back when we complete the balanced chemical equation. Eliminating the nitrate ions gives us the following ionic equation: Al(s) + Fe 3+ (aq) Al 3+ (aq) + Fe 2+ (aq) skeletal ionic equation The oxidation half-reaction is: Al(s) Al 3+ (aq) oxidation half-reaction The atoms are balanced, so we can balance the charge by adding three electrons to the right side of the equation: Al(s) Al 3+ (aq) + 3e balanced oxidation half-reaction The reduction half-reaction is: Fe 3+ (aq) Fe 2+ (aq) reduction half-reaction The atoms are balanced. We can balance the charge by adding one electron to the reactant side of the equation: e + Fe 3+ (aq) Fe 2+ (aq) balanced reduction half-reaction To equalize the number of electrons lost and gained, we multiply each half-reaction by an appropriate coefficient: 1 [Al(s) Al 3+ (aq) + 3e ] 3 [e + Fe 3+ (aq) Fe 2+ (aq)] Al(s) Al 3+ (aq) + 3e 14 30

31 3e + 3Fe 3+ (aq) 3Fe 2+ (aq) Then we add the two half-reactions and cancel the three electrons that appear on both sides of the equation. Al(s) Al 3+ (aq) + 3e 3e + 3Fe 3+ (aq) 3Fe 2+ (aq) Al(s) + 3Fe 3+ (aq) Al 3+ (aq) + 3Fe 2+ ( aq) Finally, we replace the nitrate ions to complete the balanced equation. Al(s) + 3Fe(NO 3 ) 3 (aq) Al(NO 3 ) 3 (aq) + 3Fe(NO 3 ) 2 (aq) (b) Na(s) + HNO 3 (aq) NaNO 3 (aq) + H 2 (g) balanced skeletal equation balanced Nitrate ions, NO 3, are spectator ions in this reaction so we can eliminate them while we balance the two half-reactions and add them back when we complete the balanced chemical equation. Eliminating the nitrate ions gives us the following ionic equation: Na(s) + H + (aq) Na + (aq) + H 2 (g) The oxidation half-reaction is: Na(s) Na + (aq) skeletal ionic equation oxidation half-reaction The atoms are balanced; we can balance the charge by adding one electron to the right side of the equation: Na(s) Na + (aq) + e The reduction half-reaction is: H + (aq) H 2 (g) balanced oxidation half-reaction reduction half-reaction First, we balance the atoms by adding a coefficient of 2 in front of H +. Next we balance the charge by adding two electrons to the reactant side of the equation. 2e + 2H + (aq) H 2 (g) balanced reduction half-reaction To equalize the number of electrons exchanged by the two half-reactions, we multiply each halfreaction by the appropriate coefficient: 2 [Na(s) Na + (aq) + e ] 1 [2e + 2H + (aq) H 2 (g)] 2Na(s) 2Na + (aq) + 2e 2e + 2H + (aq) H 2 (g) Then we add the two half-reactions and cancel the two electrons that appear on both sides of the equation. 2Na(s) 2Na + (aq) + 2e 2e + 2H + (aq) H 2 (g) 2Na(s) + 2H + (aq) 2Na + (aq) + H 2 (g) Finally, we replace the nitrate ions to complete the balanced equation. 2Na(s) + 2HNO 3 (aq) 2NaNO 3 (aq) + H 2 (g) balanced skeletal equation balanced 14 31

32 14.49 From the problem we can write: Fe 2 (SO 4 ) 3 (aq) + KI(aq) FeSO 4 (aq) + K 2 SO 4 (aq) + I 2 (aq) Sulfate ions and potassium ions are spectator ions in this reaction, so we can eliminate them when we balance the half-reactions. The oxidation number of Fe in Fe 2 (SO 4 ) 3 is 3+. Fe 3+ (aq) + I (aq) Fe 2+ (aq) + I 2 (aq) The oxidation half-reaction is: I (aq) I 2 (aq) skeletal ionic equation oxidation half-reaction First, we balance the atoms by adding a coefficient of 2 in front of I. Next, we balance the charge by adding two electrons to the right side of the equation. 2I (aq) I 2 (aq) + 2e The reduction half-reaction is: Fe 3+ (aq) Fe 2+ (aq) balanced oxidation half-reaction reduction half-reaction The atoms are balanced, so we can balance the charge by adding one electron to the reactant side of the equation. e + Fe 3+ (aq) Fe 2+ (aq) balanced reduction half-reaction To balance the electrons between the two half-reactions we multiply each half-reaction by the appropriate coefficient: 1 [2I (aq) I 2 (aq) + 2e ] 2 [e + Fe 3+ (aq) Fe 2+ (aq)] 2I (aq) I 2 (aq) + 2e 2e + 2Fe 3+ (aq) 2Fe 2+ (aq) The electrons between the two half-reactions are already balanced, so we add the two half-reactions and cancel the two electrons that appear on both sides of the equation. 2I (aq) I 2 (aq) + 2e 2e + 2Fe 3+ (aq) 2Fe 2+ (aq) 2Fe 3+ (aq) + 2I (aq) 2Fe 2+ (aq) + I 2 (aq) balanced skeletal equation Finally, we replace the sulfate and potassium ions to complete the balanced equation. Because the formula for Fe 2 (SO 4 ) 3 includes two iron ions, we incorporate both Fe 3+ from the balanced skeletal equation into that chemical formula. Fe 2 (SO 4 ) 3 (aq) + 2KI(aq) 2FeSO 4 (aq) + K 2 SO 4 (aq) + I 2 (aq) From the problem we can write: SO 2 (g) + HNO 3 (aq) H 2 SO 4 (aq) + NO 2 (g) balanced To begin, we assign an oxidation number to each element in the reaction. The elements whose oxidation numbers change are shown below SO 2 (g) + HNO 3 (aq) H 2 SO 4 (aq) + NO 2 Sulfur is oxidized, so we write the oxidation half-reaction: SO 2 (g) H 2 SO 4 (aq) oxidation half-reaction 14 32

33 We add two electrons to the right side of the equation to account for the change in oxidation number. SO 2 (g) H 2 SO 4 (aq) + 2e add electrons Then we balance the charge by adding hydrogen ions (H + ). SO 2 (g) H 2 SO 4 (aq) + 2H + (aq) + 2e add H + Finally, we balance the atoms by adding water (H 2 O). 2H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) + 2H + (aq) + 2e The reduction half-reaction is: HNO 3 (aq) NO 2 (g) add H 2 O reduction half-reaction We add one electron to the left side of the equation to account for the change in oxidation number. e + HNO 3 (aq) NO 2 (g) add electrons We balance the charge by adding hydrogen ions (H + ). e + H + (aq) + HNO 3 (aq) NO 2 (g) add H + Then we balance the atoms by adding water (H 2 O). e + H + (aq) + HNO 3 (aq) NO 2 (g) + H 2 O(l) add H 2 O Next, we equalize the number of electrons exchanged in the two half-reactions by multiplying each halfreaction by the appropriate coefficient: 1 [2H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) + 2H + (aq) + 2e ] 2 [e + H + (aq) + HNO 3 (aq) NO 2 (g) + H 2 O(l)] 2H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) + 2H + (aq) + 2e 2e + 2H + (aq) + 2HNO 3 (aq) 2NO 2 (g) + 2H 2 O(l) Then we add the two half-reactions and cancel the two electrons that appear on both sides of the equation. 2H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) + 2H + (aq) + 2e 2e + 2H + (aq) + 2HNO 3 (aq) 2NO 2 (g) + 2H 2 O(l) 2H 2 O(l) + 2H + (aq) + SO 2 (g) + 2HNO 3 (aq) H 2 SO 4 (aq) + 2NO 2 (g) + 2H + (aq) + 2H 2 O(l) Finally, we complete the balancing process by eliminating substances that appear on both the reactant and product sides of the equation. SO 2 (g) + 2HNO 3 (aq) H 2 SO 4 (aq) + 2NO 2 (g) Balanced reaction For simple half-reactions (see part (a) below) we determine the oxidation numbers of the atoms and add the number of electrons needed to balance the charge. To balance more complicated half-reactions (in acidic conditions) we follow the steps outlined below. Because these reactions occur in acidic solution, we add H + to balance the charge. (a) Ba(s) Ba 2+ (aq) We balance the charge by adding two electrons to the product side: Ba(s) Ba 2+ (aq) + 2e

34 (b) HNO 2 (aq) NO(g) HNO 2 (aq) NO(g) determine oxidation numbers e + HNO 2 (aq) NO(g) e + H + (aq) + HNO 2 (aq) NO(g) e + H + (aq) + HNO 2 (aq) NO(g) + H 2 O (c) H 2 O 2 (aq) H 2 O(l) add H + to balance charge add H 2 O to balance atoms 1 2 H 2 O 2 (aq) H 2 O(l) determine oxidation numbers H 2 O 2 (aq) 2H 2 O(l) 2e + H 2 O 2 (aq) 2H 2 O(l) 2e + 2H + (aq) + H 2 O 2 (aq) 2H 2 O(l) (d) Cr 3+ (aq) Cr 2 O 2 7 (aq) balance atoms being oxidized or reduced add H + to balance charge, equation balanced Cr 3+ 2 Cr 2 O 7 2Cr 3+ (aq) Cr 2 O 7 2 (aq) 2Cr 3+ (aq) Cr 2 O 7 2 (aq) + 6e 2Cr 3+ (aq) Cr 2 O 7 2 (aq) + 14H + + 6e 2Cr 3+ (aq) + 7H 2 O(l) Cr 2 O 7 2 (aq) + 14H + + 6e determine oxidation numbers balance atoms being oxidized or reduced add H + to balance charge add H 2 O to balance atoms For simple half-reactions (see part (a) below) we determine the oxidation numbers of the atoms and add the number of electrons needed to balance the charge. To balance more complicated half-reactions (in acidic conditions) we follow the steps outlined below. Because these reactions occur in acidic conditions, we add H + to balance the charge. (a) Br 2 (aq) Br (aq) 0 1 Br 2( a q) Br ( a q) Br 2 (aq) 2Br (aq) 2e + Br 2 (aq) 2Br (aq) (b) H 2 O(l) O 2 (g) 0 2 determine oxidation numbers balance atoms being oxidized or reduced H 2 O(l) O 2 (g) determine oxidation numbers 2H 2 O(l) O 2 (g) balance atoms being oxidized or reduced 14 34

35 2H 2 O(l) O 2 (g) + 4e 2H 2 O(l) O 2 (g) + 4e + 4H + (aq) (c) IO 3 (aq) I 2 (aq) add H + to balance charge, equation balanced IO 3 ( aq) I 2( a q) 2IO 3 (aq) I 2 (aq) 10e + 2IO 3 (aq) 2I 2 (aq) 10e + 12H + (aq) + 2IO 3 (aq) I 2 (aq) 10e + 12H + (aq) + 2IO 3 (aq) I 2 (aq) + 6H 2 O(l) (d) MnO 2 (s) Mn 2+ (aq) determine oxidation numbers balance iodine add H + to balance charge add H 2 O to balance atoms MnO 2( s ) Mn 2+ ( a q) 2e + MnO 2 (s) Mn 2+ (aq) 2e + 4H + (aq) + MnO 2 (s) Mn 2+ (aq) 2e + 4H + (aq) + MnO 2 (s) Mn 2+ (aq) + 2H 2 O(l) determine oxidation numbers add H + to balance charge add H 2 O to balance atoms To balance more complicated half-reactions that occur in basic solutions, we follow the steps outlined below. Because these reactions occur in basic solutions, we add OH to balance the charge. (a) La(s) La(OH) 3 (s) 0 3+ La(s) Li(OH) 3 (s) La(s) La(OH) 3 (s) + 3e 3OH (aq) + La(s) La(OH) 3 (s) + 3e (b) NO 3 (aq) NO 2 (aq) determine oxidation numbers add OH to balance charge NO 3 (aq) NO 2 (aq) determine oxidation numbers 2e + NO 3 (aq) NO 2 (aq) 2e + NO 3 (aq) NO 2 (aq) + 2OH (aq) add OH to balance charge 2e + H 2 O(l) + NO 3 (aq) NO 2 (aq) + 2OH (aq) add H 2 O to balance atoms (c) H 2 O 2 (aq) O 2 (g) 1 0 H 2 O 2 (aq) O 2 (g) determine oxidation numbers 14 35

36 H 2 O 2 (aq) O 2 (g) + 2e H 2 O 2 (aq) + 2OH (aq) O 2 (g) + 2e H 2 O 2 (aq) + 2OH (aq) O 2 (g) + 2e + 2H 2 O(l) (d) Cl 2 O 7 (aq) 2ClO 2 (aq) add OH to balance charge add H 2 O to balance atoms Cl 2 O 7 (aq) ClO 2 (aq) Cl 2 O 7 (aq) 2ClO 2 (aq) 8e + Cl 2 O 7 (aq) 2ClO 2 (aq) 8e + Cl 2 O 7 (aq) 2ClO 2 (aq) + 6OH (aq) determine oxidation numbers balance atoms being oxidized or reduced add OH to balance charge 8e + 3H 2 O(l) + Cl 2 O 7 (aq) 2ClO 2 (aq) + 6OH (aq) add H 2 O to balance atoms To balance more complicated half-reactions that occur in basic solutions, we follow the steps outlined below. Because these reactions occur in basic solution, we add OH to balance the charge. (a) CrO 2 4 (aq) Cr(OH) 3 (s) CrO 4 2 ( a q) Cr ( OH)3( s ) 3e + CrO 4 2 (aq) Cr(OH) 3 (s) 3e + CrO 4 2 (aq) Cr(OH) 3 (s)+ 5OH (aq) 3e + 4H 2 O(l) + CrO 4 2 (aq) Cr(OH) 3 (s) + 5OH (aq) (b) ClO 2 (aq) ClO (aq) determine oxidation numbers add OH to balance charge add H 2 O to balance atoms ClO 2( aq) ClO ( a q) 3e + ClO 2 (aq) ClO (aq) 3e + ClO 2 (aq) ClO (aq) + 2OH (aq) 3e + H 2 O(l) + ClO 2 (aq) ClO (aq) + 2OH (aq) (c) MnO 4 (aq) MnO 2 (s) determine oxidation numbers add OH to balance charge add H 2 O to balance atoms MnO 4 (aq) MnO 2 (s) 3e + MnO 4 (aq) MnO 2 (s) 3e + MnO 4 (aq) MnO 2 (s) + 4OH (aq) 3e + 2H 2 O(l) + MnO 4 (aq) MnO 2 (s) + 4OH (aq) (d) Br (aq) BrO 2 (aq) determine oxidation numbers add OH to balance charge add H 2 O to balance atoms 14 36