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10/20/2011 Acid/Base Titrations • Read pages 114-117 about titrations. • We’ll come back to titrations in much more detail in Ch 8 . • What do you need to know now? – titration is the volumetric analysis of one solution using another – equivalence point (stoichiometric point)…you’ll encounter these terms in Lab #2 • Examples of titration problems you should be able to solve now: – 25.0 mL of HCl is titrated with 50.0 mL 0.20 M OH-. What is the [HCl]? (generic acid is HA, generic base is B-) – How many mL of 0.35 M HBr are required to neutralize 45 mL of 0.15 M Ca(OH)2? Reaction Classes • Precipitation: synthesis of an ionic solid – a solid precipitate forms when aqueous solutions of certain ions are mixed • Acid-Base: proton transfer reactions – acid donates a proton to a base, forming a molecule (water or another weak acid) and an aqueous salt – Acid: proton-donor; Base: proton-acceptor • Oxidation-Reduction: electron transfer reactions – electron transfer from one species to another, causing a change in the oxidation state of the two species – OIL RIG: Oxidation Is Loss (of e-), Reduction Is Gain (of e-) – includes combustion, the reaction of a substance with oxygen 2 1 10/20/2011 Oxidation-Reduction Reactions Many important chemical reactions involve oxidation and reduction. In fact, most reactions used for energy production are redox reactions: In humans the oxidation of sugars, fats, and proteins provides the energy necessary for life. Glucose + O2 Energy Combustion reactions, which provide most of the energy to power our civilization, also involve oxidation and reduction. CxHy + O2 CO2 + H2O + Energy 3 Redox Reactions • “Redox” Chemistry: Reduction and Oxidation • Oxidation: Loss of electrons • Reduction: Gain of electrons (a reduction in oxidation number) OIL RIG Oxidation Is Loss of electrons Reduction Is Gain of electrons 2 10/20/2011 Redox Reactions • In a redox reaction, one species loses electrons and another species accepts those electrons. • Electrons are neither created nor destroyed during the reaction…charges are conserved. Example: Na (s) and Cl2 (g) reacting to form NaCl (s), which contains Na+ and Cl-. Oxidation Numbers (or States) • How can you tell which species is gaining electrons and which is losing them? • We need a way to look at the effective charge on each species before and after the reaction. • This is accomplished by assigning oxidation numbers (states) to atoms on both the reactant and product sides of the reaction, then looking for the species that gains electrons (reduction) and loses electrons (oxidation). 3 10/20/2011 Oxidation Numbers (or States) • An “accounting” of the electrons in a chemical species. Remember: we are not making ions within covalent compounds!! • Oxidation number (state) represents the number of electrons required to produce the “effective charge” on a species. • Oxidation numbers (states) are chosen so that: – charges are conserved – in ionic compounds, the sum of the oxidation numbers on the atoms coincides with the charge on the ion Bid idea: the oxidation numbers of the atoms/ions in a species must sum to the total charge of the species. See Table 4.3 in Zumdahl 4 10/20/2011 Period 1 IA H +1 -1 IIA IIIA Li Be B +1 +2 +3 Na Mg Al Si +1 +2 +3 +4,-4 K Ca Ga +1 +2 Rb Sr +1 +2 Cs +1 2 3 4 5 6 Common Oxidation Numbers IVA VA C N +4,+2 all from -1,-4 +5 to-3 Ge +4,+2 +3, +2 -4 VIA VIIA O F -1,-2 -1 VIIIA He Ne P +5,+3 -3 S -1 Cl +6,+4 +7,+5 +2,-2 +3,+1 Ar As +5,+3 -3 Se -1 Br +6,+4 +7,+5 -2 +3,+1 Kr Sb Te -1 I +5,+3 +6,+4 +7,+5 -3 -2 +3,+1 Xe In +3,+2 +1 Sn +4,+2, -4 Ba Tl Pb Bi +2 +3,+1 +4,+2 +3 Po -1 At +6,+4 +7,+5 +2,-2 +3,+1 +2 +6,+4 +2 Rn +2 Transition Metals Common Oxidation Numbers VIIIB IVB VB VIB VIIB Ti V Cr +2 Mn Fe Co +3 +4,+3 +5,+4 +6,+3 +7,+6 +3,+2 +3,+2 +2 +3+2 +2 +4,+3 IIIB Sc Y IIB Zn +2 +2,+1 +2 Nb Mo Tc Ru Rh Pd Ag +5,+4 +6,+5 +7,+5 +8,+5 +3 +4,+3 +4,+3 +4,+2 +1 +2 +4,+3 +4 +4,+3 La Zr Ni IB Cu Cd +2 Hf Ta W Re +2 Os Ir Pt Au Hg +5,+4 +6,+5 +7,+5 +8,+6 +4,+3 +3 +4,+3 +4,+2 +3,+1 +2,+1 +3 +4 +4 +4,+3 +1 5 10/20/2011 Oxidation Numbers Determine the oxidation number (O.N.) of each element in the following compounds. Ex 1: iron(III) chloride Ex 2: nitrogen dioxide Ex 3: sulfuric acid Ex 4: barium iodide Ex 5: ammonium nitrate Strategy: We apply the rules in Table 4.3, always making sure that the O.N. values add up to zero in a compound or, for a polyatomic ion, to the ion’s charge. Oxidation Numbers Ex 1: FeCl3 This compound is composed of monoatomic ions. The O.N. of Cl- is -1, for a total of -3. Therefore, the O.N. of Fe is +3. Ex 2: NO2 The O.N. of oxygen is -2 for a total of -4. Since the O.N. in a compound must add up to zero, the O.N. of N is +4. Ex 3: 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. Therefore S has the O.N. +6. Ex 4: BaI2 Ba = +2 I = -1 Ex 5: NH4NO3 H = +1 O = -2 N = -3 N = +5 6 10/20/2011 Mnemonic devices “LEO says GER” “OIL RIG” Redox Examples Fe2O3(s) + +3 -2 iron is REDUCED Mg(s) + 2Al(s) 2Fe(s) + 0 0 aluminum is OXIDIZED 2HCl(aq) Al2O3(s) +3 -2 nothing happens to oxygen H2(g) + MgCl2(aq) +1 -1 +2 -1 0 0 hydrogen is REDUCED magnesium is OXIDIZED nothing happens to chlorine Cu(s) + 2AgNO3(aq) 0 silver is REDUCED +1 2Ag(s) + Cu(NO3)2(aq) -1 copper is OXIDIZED 0 +2 -1 nothing happens to nitrate 7 10/20/2011 Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the rxns a) Zn(s) + 2 HCl(aq) ZnCl2 (aq) + H2 (g) b) S8 (s) + 12 O2 (g) 8 SO3 (g) c) NiO(s) + CO(g) Ni(s) + CO2 (g) First we assign an oxidation number (O.N.) to each atom (or ion). The reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction). The reactant is the oxidizing agent if it contains an atom that is reduced (O.N. decreased). a) Assigning oxidation numbers: -1 +1 0 Zn(s) + 2 HCl(aq) -1 0 +2 ZnCl2 (aq) + H2 (g) HCl is the oxidizing agent, and Zn is the reducing agent! Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: 0 +6 0 S8 (s) + 12 O2 (g) -2 8 SO3 (g) S [0] S[+6] S is Oxidized O[0] O[-2] O is Reduced S8 is the reducing agent and O2 is the oxidizing agent c) Assigning oxidation numbers: -2 -2 +2 +2 NiO(s) + CO(g) 0 +4 Ni[+2] Ni[0] Ni is Reduced C[+2] C[+4] C is Oxidized -2 Ni(s) + CO2 (g) CO is the reducing agent and NiO is the oxidizing agent 8 10/20/2011 Demo: Activity Series Mg(s) + 2 HCl(aq) → Mg2+(aq) + 2 Cl–(aq) + H2(g). Mg is a very active metal (reduces H in both water and acid). Zn(s) + 2 HCl(aq) → Zn2+(aq) + 2 Cl–(aq) + H2(g). Zn is an active metal (reduces H in acid but not in water). Cu(s) + HCl(aq) → no reaction Cu is an inactive metal (does not reduce H in acid nor water). Activity Series Lab 3: You use Mg instead of Zn 9 10/20/2011 A “Classic” Redox Example Solid copper is immersed in aqueous silver nitrate. Blue color is formation of Cu2+ Ag is formed from solution of Ag+ net charge = +2 Balanced Net Ionic eqn.: Ox. #: net charge = +2 Cu(s) + 2Ag+(aq) 0 +1 Cu2+(aq) + 2Ag(s) +2 oxidation 0 reduction At the particulate level… Before rxn occurs... During rxn 10 10/20/2011 Writing Half-Reactions We can separate an electron-transfer reaction between Cu(s) and Ag+(aq) into two “half-reactions,” where the electron loss or gain is explicitly stated. The half-reaction idea provides us with a reliable method of balancing redox reactions in aqueous solution. Electrons just 2+ change Cu(s) Cu (aq) + 2e places… Ag+(aq) + eAg(s) x 2 charge is + conserved!!! Cu(s) + 2Ag+(aq) net charge = +2 Cu2+(aq) + 2Ag(s) net charge =+2 Balancing by Half-Reactions • Recall, a properly balanced redox reaction means both mass and charge are conserved. • Oftentimes the solvent (in this course, we will only consider water) will be explicitly involved in the redox reaction. • For example, consider the following reaction: CuS(s) + NO3-(aq) Cu2+(aq) + SO42-(aq) + NO(g) 11 10/20/2011 Balancing REDOX Equations: The Half-Reaction Method Step 1) Write the half-reactions for the chemical equation. Step 2) For each reaction, balance the atoms other than O and H. Step 3) Add H2O to balance O, then H+ to balance H. Step 4) Balance the charge by adding electrons. The net charge of the reactants should equal the net charge of the products. Step 5) Add the two half-reactions together, making sure e- lost equal e- gained, and canceling any species that appear on both sides of the reaction. The reaction is now balanced in an acidic solution. Step 6) If you need to balance in a basic solution, first balance in acidic solution (!), then add OH- to both sides to neutralize any H+ present. Cancel any species that appear on both sides of the reaction. Balancing by Half-Reactions (cont.) • Step 1: Identify and write down the unbalanced half-reactions. CuS(s) + NO3-(aq) -2 Cu2+(aq) + SO42-(aq) + NO(g) +5 +6 oxidation +2 reduction CuS(s) Cu2+(aq) + SO42-(aq) oxidation NO3-(aq) NO(g) reduction 12 10/20/2011 Balancing by Half-Reactions (cont.) • Step 2: Balance atoms (other than H and O). • Step 3: Use H2O to balance O, and H+ to balance H (assume acidic media). • Step 4: Use e- to balance charge (won’t always be 0). 4H2O + CuS(s) Cu2+(aq) + SO42-(aq) + 8H+ + 8e- 3e- + 4H+ + NO3-(aq) NO(g) + 2H2O Balancing by Half-Reactions (cont.) • Step 5: Add the reactions together - multiply each half-reaction by an integer such that the number of electrons cancels. 4H2O + CuS(s) Cu2+(aq) + SO42-(aq) + 8 H+ +8e- x 3 3e- + 4 H++ NO3-(aq) 8 3CuS + 8H++ 8NO3- NO(g) + 2H2O x 8 3Cu2+ 4 + 3SO42- + 8NO + 4H2O 13 10/20/2011 Balancing by Half-Reactions (cont.) • Step 6: For reactions that occur in basic solution, first balance in acid. At the end, add OH- to both sides for every H+ present, neutralizing acid present and leaving OH- on the other side. 3CuS + 8H++ 8 NO3+ 8OH- 3Cu2+ + 3SO42- + 8NO + 4H2O + 8OH- 3CuS + 8H2O+ 8 NO34 3Cu2+ + 3SO42- + 8NO + 4H2O + 8OH- Example Balance the following redox reaction occurring in basic media: BH4- + ClO3- H2BO3- + Cl- BH4- H2BO3- ClO3- Cl- -5 +5 +3 -1 oxidation reduction 14 10/20/2011 Example (cont) 3H2O + BH4- H2BO3- + 8H+ + 8e- x 3 6e- + 6H+ + ClO3- Cl- + 3H2O x 4 3 3 BH4- +4 ClO3- 3H2BO3- +4 Cl- + 3H2O Done! Demo: Thermite Reaction __ Fe2O3(s) + __ Al(s) __ Fe(s) + __ Al2O3(s) ΔHo = –849 kJ/mol ΔSo = –37.48 J/mol·K ΔGo = –838 kJ/mol Uses: welding, purifying an ore (U in the Manhattan Project) In this reaction, unfavorable entropy (ΔS) is offset by a very large negative ΔH (heat is released). The heat is sufficient to raise the temperature of the products past the melting point of Fe (1530 oC). Can go horribly wrong…the heat has to go somewhere and a sand pit is the best. “Mythbusters”: thermite and ice…ice chunks flew 150 feet, or half a football field; in another episode, they split a car in half 15 10/20/2011 Balancing REDOX Equations: The Oxidation Number 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 transfers. Step 4) Choose coefficients for these species to make the electrons lost equal the electrons gained; in other words… total increase in O.N. = total decrease in O.N. Step 5) Complete the balancing by inspection. If you need to add O and H, use the same process as in the half-reaction method: add H2O for O and H+ for H REDOX Balancing using Ox. No. Method Balancing the Thermite Reaction -2 -2 Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) +3 0 → +3 2 x (- 3e- per Al) 0 2 x (+3e- per Fe) 32 16 10/20/2011 REDOX Balancing by Oxidation Number Method +2 -1e- lost per Fe +3 +1 -2 Fe+2(aq) + MnO4-(aq) + H+(aq) +7 +1 -2 Fe+3(aq) + Mn+2(aq) + H2O(l) +5 e- gained per Mn +2 Balance the number of each redox element and then # electrons. Multiply Fe+2 & Fe+3 by five to balance the electrons gained by Mn: 5 Fe+2(aq) + MnO4-(aq) + H+(aq) 5 Fe+3(aq) + Mn+2(aq) + H2O(l) Balance O: Need 4 H2O on right to balance 4 O from the MnO4-. Balance H: Need 8 H+ on the left to balance 8 H in the 4 H2O. 5 Fe+2(aq) + MnO4-(aq) +8 H+(aq) 5 Fe+3(aq) + Mn+2(aq) +4 H2O(l) REDOX Balancing by Half-Reaction Method Fe+2(aq) + MnO4-(aq) Fe+3(aq) + Mn+2(aq) (acid solution) Identify Oxidation and Reduction Half Reactions Fe+2(aq) Fe+3(aq) + e- MnO4-(aq) Mn+2(aq) what shall we do with oxygen? iron is oxidized manganese is reduced Add H2O to the products to balance O, add H+ to balance the H in the water, add electrons to balance the charge. 5e- + MnO4-(aq) + 8H+(aq) Mn+2(aq) + 4H2O(l) Sum the two half-reactions (multiply Fe rxn by 5 to cancel electrons). { Fe+2(aq) MnO4-(aq) + 8H+(aq) +5eMnO4-(aq)+ 8H+(aq)+5e- +5Fe+2(aq) Fe+3(aq) +e- } x5 Mn+2(aq) + 4H2O(l) 5Fe+3(aq)+5e- + Mn+2(aq)+ 4H2O(l) 34 17