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Aqueous Reactions and Solution Stoichiometry (continuation) 1. Oxidation-Reduction Reactions 2. The Activity Series 3. Balancing REDOX Equations Terminology for Redox Reactions • OXIDATION—loss of electron(s) by a species; increase in oxidation number. • REDUCTION—gain of electron(s); decrease in oxidation number. • OXIDIZING AGENT—electron acceptor; species are reduced. • REDUCING AGENT—electron donor; species are oxidized. Oxidation-Reduction Reactions - reaction where electrons are exchanged. 2 Na(s) + 2 H2O(l) Æ 2 NaOH(aq) + H2(g) Æ + 2 oxidation – loss of electrons 2Na0(s) Na+(aq) e- These processes occur simultaneously – you cannot have an oxidation without a 2 H+(g) + 2 e- Æ H20(g) reduction and vicereduction – gain of electrons versa!!! GeneralRules for Assigning an Oxidation Number (O.N.) O.N 1) For an atom in its elemental neutral form (Na, O2 Cl2, etc.): O.N. = 0 2) For a monoatomic ion: O.N. = ion charge 3) The sum of O.N. values for the atoms in a compound equals zero. zero 4) The sum of the O.N. values for the atoms in a polyatomic ion equals the ion’s charge. charge Rules for specific atoms or periodic table groups: 1) For Group IA(1): O.N. = +1 in all compounds 2) For Group IIA(2): O.N. = +2 in all compounds 3) For hydrogen: O.N. = +1 in combination with nonmetals O.N. = -1 in combination with metals and boron 4) For fluorine: O.N. = -1 in all compounds 5) For oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compounds (except with F ) 6) For Group VIIA(17) O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group. +1 -1 1 Period Highest and Lowest Oxidation Numbers of Chemically Reactive Main-Group Elements H 1A 2A +1 +2 2 Li Be 3A 4A 5A 6A 7A +3 +4-4 +5-3 +6-2 +7-1 B C N O F Al Si P S Cl Ga Ge As Se Br metalloids 5 Rb Sr In Sn Sb Te I 6 Cs Ba Tl Pb Bi 3 Na Mg non-metals 4 K Ca metals 7 Fr Ra Po At Transition Metals Possible Oxidation States VIIIB IIIB IVB VB VIB VIIB IB IIB Sc Ti V Cr +2Mn Fe Co Ni Cu Zn +3 +4,+3 +5,+4 +6,+3 +7,+6+3,+2 +3,+2 +2 +2,+1 +2 +4,+3 +2 +3+2 +2 Y Zr Nb Mo Tc Ru Rh Pd Ag +3 +4,+3 +5,+4 +6,+5 +7,+5 +8,+5 +4,+ +4,+2 +1 +2 +4,+3 +4 +4,+3 3 Cd +2 La Hf Ta W Re +2 Os Ir Pt Au Hg +3 +4,+3 +5,+4 +6,+5 +7,+5 +8,+6 +4,+3 +4,+2 +3,+1 +2,+1 +3 +4 +4 +4,+3 +1 Determining the Oxidation Number of an Element in a Compound Problem: Determine the oxidation number (O.N.) O.N of each element in sulfuric acid H2SO4: The O.N. of H is +1, so the SO42- group must sum to -2. The O.N. of each O is -2 for a total of -8. So the Sulfur atom is +6. Recognizing Oxidizing and Reducing Agents NiO(s) + CO(g) → C[+2] C[+4] C is oxidized -2 -2 +2 +2 NiO(s) + CO(g) Ni(s) + CO2 (g) Ni[+2] Ni[0] Ni is Reduced 0 Ni(s) +4 -2 + CO2 (g) CO is the reducing agent and NiO is the oxidizing agent. Types of Redox Reactions Redox Combination Reactions (Intermolecular)when reducing and oxidizing agents are in different compounds 0 0 +3 -1 P4(s) + 6 Cl2(g) → 4 PCl3(l) 0 0 +5 -1 P4(s) + 10 Cl2(g) → 4 PCl5(l) 0 0 +3 -2 Fe(s) + O2 (g) → Fe2O3 (s) Rusting Redox Decomposition Reactions (Intramolecular reactions) -when both reducing and oxidizing agents are in the same compound +2 -2 0 0 2 HgO (s) → 2 Hg(l) + O2(g) +1 -2 0 0 2 Hg2O(s) → 4 Hg(l) + O2(g) +1 -1 +1 -2 0 2 H2O2 (l) → 2 H2O(l) + O2(g) Oxygenation and Hydrogenation reactions 0 0 +1 -2 4 Li (s) + O2(g) → 2 Li2O(s) +2 -2 0 +2 -2 +4 -2 2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g) 0 0 +1 -1 2 Na(l) + H2(g) → 2 NaH(s) +3 -2 0 0 +1 -2 Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g) Disproportionation reactions (self-reduction/self oxidation) +1 -1 +1 -2 0 2 H2O2 (l) → 2 H2O(l) + O2(g) +1 +2 0 2 Cu+(aq) → Cu2+(aq) + Cu(s) (net ionic equation) Displacement Reactions +1 +5 -2 0 +2 +5 -2 2 AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2 Ag(s) +1 0 +2 2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) 0 +1 -1 0 -1 0 (overall equation) 0 (net ionic equation) +1 -1 Cl2(g) + 2 KI(aq) → I2(s) + 2 KCl(aq) 0 0 (overall equation) -1 Cl2(g) + 2 I-(aq) → I2(s) + 2 Cl-(aq) (net ionic equation) electrolyte electrode - + + - + + + Redox potential depends on the atomic structure and hydratation tendency of the element The Activity Series is based on the ability of element to lose electrons in aqueous solutions - Some metals are more easily oxidized than others. Activity series - a list of metals arranged in decreasing ease of oxidation - The higher the metal on the activity series, the more active that metal. - Any metal can be oxidized by the ions of elements below it Hydrogen is in the activity series even though it is a nonmetal EN<1.4 1.4< EN <1.9 1.9< EN <2.54 most can be dissolved by oxidizing acids (HNO3) Tips on Balancing Equations • Never add O2, O atoms, or O2- to balance oxygen. • Never add H2 or H atoms to balance hydrogen. • Be sure to write the correct charges on all the ions. • Check your work at the end to make sure mass and charge are balanced. Balancing REDOX Equations: The Oxidation Number (O.N.)Method Step 1) Assign oxidation numbers to all elements in the equation. Step 2) From the changes in oxidation numbers, identify the oxidized and reduced species. Step 3) Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. Draw tie-lines between these atoms to show electron changes. Step 4) Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients. Step 5) Complete the balancing by inspection, adding states of matter. REDOX Balancing Using O. N. +2 -1e- +3 Fe+2(aq+ MnO4-(aq)+ H3O+(aq) +7 +5 e- Fe+3(aq)+ Mn+2(aq)+H2O(aq +2 Multiply Fe+2 & Fe+3 by five to correct for the electrons gained by the manganese. 5 Fe+2(aq) + MnO4-(aq) + H3O+(aq) 5 Fe+3(aq) + Mn+2(aq) + H2O(aq) Make four water molecules from the 4 oxygen atoms from the MnO4- and protons from the acid, and this will require 8 protons, or 8 hydronium ions. This will give a total of 12 water molecules formed. 5 Fe+2(aq) + MnO4-(aq) + 8 H3O+(aq) 5 Fe+3(aq) + Mn+2(aq) + 12 H2O(aq) 5 FeCl2 + KMnO4 + 8HCl=5 FeCl3 + MnCl2 +KCl + 4 H2O Balancing Redox Equations by half-reactions method •Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product •Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation. •Step 3 Balance oxygen by adding H2O to the side deficient in O in each halfequation •Step 4 Balance hydrogen. –For half-reaction in acidic solution, add H+ on to the side deficient in hydrogen. –For a half-reaction in basic solution, add H2O to the side that is deficient in hydrogen and an equal amount of OH- to the other side •Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction. •Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication •Step 7 Check for balance Example: Balancing Redox Reactions Dithionate ion reacting with chlorous acid in aqueous acidic conditions S2O62-(aq) +HClO2(aq) --> SO42-(aq) + Cl2(g) Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product 5+ 6+ S2O62- --> SO42OXIDATION 3+ 0 HClO2 --> Cl2 REDUCTION Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation. S2O62- --> 2 SO42- 2 HClO2 --> Cl2 Step 3 Balance oxygen by adding H2O to the side deficient in O in each halfequation 2 H2O + (Need 2 Os) S2O62- --> 2 SO42(6Os) 2 HClO2 --> Cl2 (4Os) (8Os) + 4 H2O (Need 4 Os) Step 4 Balance hydrogen. For half-reaction in acidic solution, add H+ on to the side deficient in hydrogen. For a half-reaction in basic solution, add H2O to the side that is deficient in hydrogen and an equal amount of OH- to the other side 2H2O + S2O62- → 2SO42- + 4H+ (4 Hs) Need 4 Hs 6H+ + Need 6 Hs 2HClO2 → Cl2 + 4H2O (2 Hs) (8 Hs) Step 5 Balance charge by inserting e(electrons) as a reactant or product in each half-reaction. 2H2O + S2O62- --> 2SO42- + 4H+ + 2 eOxidation reaction, electrons are lost on the reactant side (gained on the product side) 6H+ + 2HClO2 + 6 e- --> Cl2 + 4H2O Reduction reaction, electrons are gained on the reactant side Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Oxidation Reaction has lost 2 electrons Reduction reaction has gained 6 electrons ∴ Multiply the oxidation reaction by 3. This will give 6 electrons on both sides of the reaction 6H2O + 3S2O62- → 6SO42- + 12H+ + 6e- Step 6 Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication 6H2O + 3S2O62- --> 6SO42- + 12H+ + 6e+ 6H+ + 2HClO2 + 6e- --> Cl2 + 4H2O 2H2O + 3S2O62- + 2HClO2 --> 6SO42- + 6H+ + Cl2 Step 7 Check for balance 2H2O + 3S2O62- + 2HClO2 --> 6SO42- + 6H+ + Cl2 Left side Right side H 6 H 6 O 24 O 24 S 6 S 6 Cl 2 Cl 2