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Transcript
Daniel L. Reger Scott R. Goode David W. Ball www.cengage.com/chemistry/reger Lecture 03B (Chapter 18, sections 18.1, 18.2) Balancing Redox Reactions Oxidation and Reduction • Oxidation is the loss of electrons by a chemical process. • When sodium forms a compound, Na+ is formed. Sodium is oxidized. • Reduction is the gain of electrons by a chemical process. • When Cl- ions are formed from elemental chlorine, chlorine is reduced. Oxidation-Reduction (“Redox”) • An oxidation-reduction reaction, or redox reaction, is one in which electrons are transferred from one species to another. • In every redox reaction, at least one species is oxidized and at least one species is reduced. • 2Na(s) + Cl2(g) → 2NaCl(s) is a redox reaction because Na is oxidized and Cl is reduced. Oxidizing and Reducing Agents • An oxidizing agent is the reactant that accepts electrons, causing an oxidation to occur. • The oxidizing agent is reduced. • A reducing agent is the reactant that supplies electrons, causing a reduction to occur. • The reducing agent is oxidized. • In the reaction of sodium with chlorine, Na is the reducing agent and Cl2 is the oxidizing agent. Half-reactions • In a half-reaction, either the oxidation or reduction part of a redox reaction is given, showing the electrons explicitly. • Half-reactions emphasize the transfer of electrons in a redox reaction. • For 2Na(s) + Cl2(g) → 2NaCl(s) : • Na → Na+ + 1eoxidation half-reaction • Cl2 + 2e- → 2Clreduction half-reaction Oxidation States • The oxidation state is the charge on the monatomic ion, or the charge on an atom when the shared electrons are assigned to the more electronegative atom. • Electron pairs shared by atoms of the same element are divided equally. • In CaCl2, an ionic compound: • calcium has an oxidation state of +2. • chlorine has an oxidation state of -1. Rules for Determining Oxidation Numbers Rule 1: The O.N. of an atom in its (uncombined) elemental state is zero. Exs. Al(0), O2(0), H2(0), Br2(0), Na(0) Rule 2: +The O.N. 2+ of a monatomic ion2-is equal-to the charge on the ion. 3+ Exs. Na (+1), Mg (+2), Al (+3), O (-2), Br (-1) Rule 3: Atoms in polyatomic ions or molecular compounds usually have O.N.s identical to the charges they would have as ions. Exceptions: -Hydrogen (except as H2) is usually +1, but sometimes -1 (ex. H-Ca-H) -Oxygen (except as O2) is usually -2, but can be -1 when –O-O- bonds exist (peroxides) -Halogens usually -1, unless combine with more EN element (dichlorine monoxide, ClO-Cl, Cl is +1; F is always -1) Rule 4: The algebraic sum of all O.N.s of all atoms in a neutral compound or polyatomic ion is equal to the net charge. - For neutral compounds, the sum of the O.N.s is 0 - For a charged polyatomic ion, the sum of the O.N.s is equal to the charge. Assigning Oxidation Numbers • Assign oxidation numbers for each atom in K2CrO4. • K = +1 by rule 2. • O = -2 by rule 5. • Cr = +6 by rule 6. • 2(+1) + 6 + 4(-2) = 0, the overall charge on the substance. Assigning Oxidation Numbers K2CO3 CH2O NO3 - 2(O.N of K) + (O.N. of C) + 3(O.N. of O) = 0 2(+1) + ? + 3(-2) = -4 2(+1) + (+4) + 3(-2) = 0 2(O.N. of H) + (O.N. of O) = 0 2(+1) + (-2) = 0 + (-2) = 0 (O.N of C) + (0) + 2(+1) (O.N of N) + 3(O.N. of O) = -1 + 3(-2) = -6 + 3(-2) = -1 (+5) Test Your Skill • Assign oxidation numbers to each atom in the following substances. (a) PF3 (b) CO (c) NH4Cl Test Your Skill • Assign oxidation numbers to each atom in the following substances. (a) PF3 (b) CO (c) NH4Cl • Answers:(a) P = +3, F = -1 (b) C = +2, O = -2 (c) N = -3, H = +1, Cl = -1 Redox Rxns Determined by Change in O.N. S(s) + (O.N of uncombined element = 0) O2(g) (O.N of uncombined element = 0) SO2(g) (O.N. of S) + 2(O.N. of O) = 0 (+4) + 2(-2) = 0 Sulfur is oxidized, Oxygen is reduced 4Al(s) (O.N of uncombined element = 0) + 3O2(g) (O.N of uncombined element = 0) 2Al2O3(s) 2(O.N. of Al) + 3(O.N. of O) = 0 + 3(-2) = -6 2(+3) Al is oxidized (O.N. increased by 3), Al is reducing agent Oxygen is reduced (O.N. decreases by 2), O is oxidizing agent + 3(-2) = 0 Redox e- Transfer: Examples SO2(g) + H2O(l) H2SO3(aq) S +4 O -2 H +4 -2 -2 +1 +1 4Fe(s) + 3O2 (g) 2Fe2O3(s) Balancing Net Ionic Equations Using Redox Half-reactions Cr2O72-(aq) + Cl-(aq) Cr3+(aq) + Cl2(aq) With more complex net ionic equations, it is not always easy to see how the equation should be balanced. We have another set of steps to address this!!!! Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 1: Write the unbalanced net ionic equation Step 2: Determine which atoms are oxidized and reduced, and write the two unbalanced half-reactions Step 3: Balance both half-reactions for all atoms except H, O Step 4: Balance each half-reactions for O - Add H2O to side with less O - Add H+ to side with less H Step 5: Balance each half-reactions for charge - Add electrons to side with greater positive charge - Multiply half-reaction as needed so the electron count is the same on both sides Step 6: Add the two half-reactions - Cancel electrons and other species (spectator ions) that appear on both sides Step 7: Confirm that equation balanced for atoms and charge Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 1: Write the unbalanced net ionic equation Cr2O72-(aq) + Cl-(aq) Cr3+(aq) + Cl2(aq) Step 2: Determine which atoms are oxidized and reduced, and write the two unbalanced half-reactions Cl is oxidized (loses e-) from -1 to 0 Reducing agent Cr is reduced (gains e-) from +6 to +3 Oxidizing agent Oxidation half-rxn 2 Cl-(aq) Cl2(aq) Reduction half-rxn Cr2O72-(aq) 2 Cr3+(aq) Step 3: Balance both half-reactions for all atoms except H, O Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 4: Balance each half-reactions for O, H - Add H2O to side with less O - Add H+ to side with less H Oxidation half-rxn 2Cl-(aq) Cl2(aq) Reduction half-rxn Cr2O72-(aq) 2Cr3+(aq) +7H20 (l) Cr2O72-(aq) + 14H+ (aq) 2Cr3+(aq) + 7H20 (l) Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 5: Balance each half-reactions for charge - Add electrons to side with greater positive charge - Multiply half-reaction as needed so the electron count is the same on both sides Oxidation half-rxn 2Cl-(aq) Cl2(aq) + 2 eReduction half-rxn Cr2O72-(aq) + 14H+ (aq) 2Cr3+(aq) + 7H20 (l) Cr2O72-(aq) + 14H+ (aq) + 6 e- 2Cr3+(aq) + 7H20 (l) Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 5: Balance each half-reactions for charge - Add electrons to side with greater positive charge - Multiply half-reaction as needed so the electron count is the same on both sides Oxidation half-rxn 3 X [2Cl-(aq) Cl2(aq) + 2e-] = Reduction half-rxn 6Cl-(aq) 3Cl2(aq) + 6e- Cr2O72-(aq) + 14H+ (aq) + 6e- 2Cr3+(aq) + 7H20 (l) Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 6: Add the two half-reactions - Cancel electrons and other species (spectator ions) that appear on both sides 6Cl-(aq) 3Cl2(aq) + 6e(Ox) Cr2O72-(aq) + 14H+ (aq) + 6e- 2Cr3+(aq) + 7H20 (l) (Red) Cr2O72-(aq) + 14H+ (aq) + 6Cl-(aq) 3Cl2(aq) + 2Cr3+(aq) + 7H20 (l) Step 7: Confirm that equation balanced for atoms and charge Balancing Redox Reactions under basic conditions • In basic solutions, follow all of the same steps. When finished, add an equivalent amount of OH- to each side of the reaction to eliminate any H+ by combining them to make H2O. Test Your Skill • Balance the following equation in acid solution: • Cr2O72- + C2H5OH → Cr3+ + CO2 Test Your Skill • Balance the following equation in acidic solution: • Cr2O72- + C2H5OH → Cr3+ + CO2 • Answer: 2Cr2O72- + C2H5OH + 16H+ → 4Cr3+ + 2CO2 + 11H2O Test Your Skill • Balance the following equation in basic solution: • Zn + ClO- → Zn(OH)42- + Cl- Test Your Skill • Balance the following equation in basic solution: • Zn + ClO- → Zn(OH)42- + Cl- • Answer: Zn + ClO- + H2O + 2OH→ Zn(OH)42- + Cl-