Balancing reaction equations,oxidation state, andreduction-oxidation reactionsSummary of Types of Reactionsand Equations***Net Ionic EquationsIf a solution of lead nitrate is added to a solution ofsodium chloride, lead chloride precipitates:Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)(This is a conventional equation.)To better describe the reaction, the formulas of thedissolved substances are replaced with their solutioninventories:Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Cl-(aq) →PbCl2(s) + 2 Na+(aq) + 2 NO3-(aq)(This is an ionic equation.)We can then eliminate the “spectators”.Net Ionic EquationsThe result is a net ionic equation, which tells exactlywhat chemical change took place, and nothing else:Pb2+(aq) + 2 Cl-(aq) → PbCl2(s)Steps in writing a net ionic equation:• Write the conventional equation, including designations of state[(g), (l), (s), (aq)]. Balance the equation.• Write the ionic equation by replacing each dissolved substance(aq) with its solution inventory species. Never change states inthis step. Be sure the equation is balanced for both atoms andcharge.• Write the net ionic equation by removing the spectators.Reduce coefficients to lowest terms. Be sure the equation isbalanced for both atoms and charge.Electron-transfer ReactionsExample: H2 combining with O2 to form water:2 H2 + O2 → 2H2OAn electron is transferred from H to O: the H2 isoxidized and the O2 is reduced.We use the oxidation number (oxidation state) to keeptrack of electron shifts in chemical reactions. It isdefined as “the charge which an atom appears tohave when the net electric charge on a chemicalspecies is apportioned according to certain rules”.Important because: the binding of atoms results fromthe transfer or sharing of electrons.Common Oxidation States of ElementsDetermining Oxidation Number ofElements & Molecules1. In uncombined or free elements (not ionized), eachatom has an oxidation number of 0. E.g., all of theatoms in these molecules: H2, Na, S8, O2, P4.2. In simple ions (i.e., charged species which containonly one atom), the oxidation number is equal to thecharge on the ion. E.g., Na and K only form +1 ions;thus, their oxidation numbers are +1 in allcompounds.3. The oxidation number of oxygen is generally –2.Exceptions: case #1, and oxygen atoms directlybonded to other oxygen atoms (peroxides) and tofluorine.Determining Oxidation Number ofElements & Molecules4. The oxidation number of hydrogen is generally +1.Exceptions: case #1, and hydrogen atoms combinedwith elements such as Na or Ca (hydrides).5. Oxidation numbers must be consistent withconservation of charge. E.g., The sum of chargemust be 0 for neutral molecules. For H2O:H: oxid. number is +1O : oxid. number is –2Net charge = 2(+1) + 1(-2) = 0Determining Oxidation Number ofElements & Molecules6. Fractional oxidation numbers are possible. E.g., inNa2S4O6 (sodium tetrathionate), S has an oxidationnumber of +10/4:O: 6(-2) = -12Na: 2(+1) = 2Residual = -10, which must be balance by S:S: 4(+10/4) = +107. The oxidation number is designated by:• Arabic number below the atom, or• Roman numeral or arabic number after theatom (in parentheses)Determining Oxidation Number ofElements & MoleculesOxidation states for important N, S and C compounds:Reduction-Oxidation ReactionsExample: 2 H2 + O2 → 2H2O• The hydrogen changes oxidation number from 0 to +1(is oxidized)• The oxygen changes oxidation number from 0 to –2 (is reduced)0-210Oxidation is a chemical process in which an atomshows an increase in oxidation numberReduction is a chemical process in which an atomshows an decrease in oxidation numberOxidantReductantReduction-Oxidation ReactionsOxidant(Oxidizing agent)Reduced productReduction(gain of e-)Reductant(Reducing agent)Oxidized producte-Oxidation(loss of e-)2 H++1H2 (g)0Zn (s)0Zn2++2e-Oxidation of zinc: Zn (s) + 2 H+ → Zn2+ + H2 (g)Reduction-Oxidation ReactionsNote: Strengths not necessarily related to# of electrons transferredReduction-Oxidation ReactionsProcedure for balancing redox reactions in aqueous solutions:Reduction-Oxidation ReactionsReduction-Oxidation ReactionsA: Gibbs freeenergycalculationQ: How dowe knowthat