Redox Reaction Theory Redox Reaction Theory I Facts About Redox R x ns A Def 1 Oxidation loss of e ‘s 2 Reduction gain of e ‘s B Reduction Always Accompanies Oxidation C How To Detect Redox Rxns Chang[.]
Redox Reaction Theory I Facts About Redox R x ns A Def Oxidation- loss of e-‘s Reduction- gain of e-‘s B Reduction Always Accompanies Oxidation C How To Detect Redox Rxns- Change in Oxidation Number Reduction- Oxidation No becomes More negative Less posititve Oxidation- Oxidation No becomes Less negative More positive Example HCl + MnO2 > MnCl2 + Cl2 + H2O Example Sn + HNO3 > SnO2 + NO2 + H2O Example NaCl + H2SO4 > Na2SO4 + HCl II Half Reactions A Net Ionic Equations that are redox in nature may be divided into two rxns (half rxns)- one for reduction and one for oxidation Example Overall HCl + MnO2 > MnCl2 + Cl2 + 2H2O Ionic H+ + Cl- + MnO2 > Mn+2+ Cl- + Cl2 + H2O Net Ionic H+ + Cl- + MnO2 > Mn+2 + Cl2 +2 H2O Oxid Half Rxn : Cl- —> Cl2 + eReduction Half Rxn: H+ + MnO2 + e- —> Mn+2 + H2O _ _ H+ + Cl- + MnO2 + e- —> Mn+2 + Cl2 + H2O + e- H+ + Cl- + MnO2 —> Mn+2 + Cl2 + H2O Example Overall Sn + HNO3 –> SnO2 + NO2 + H2O Ionic Sn + H+ + NO3- —> SnO2 + NO2 + H2O Net Ionic Sn + H+ + NO3- —> SnO2 + NO2 + H2O Oxidation Half Rxn Sn + H2O —> SnO2 + H+ + eReduction Half Rxn NO3- + H+ + e- —> NO2 + H2O Sn + NO3- + H+ + e- + H2O SnO2 + H+ + e- + NO2 + H2O Sn + NO3- + H+ SnO2 + NO2 + H2O Example Ce+4 + Fe+2 —> Ce+3 + Fe+3 Fe+2 —> Fe+3 + eCe+4 + e- —> Ce+3 Ce+4 + Fe+2 + e- —> Fe+3 +1 e- + Ce+3 Ce+4 + Fe+2 —> Fe+3 + Ce+3 B Use of half Reactions To Balance Redox Equations In Acid Solution a MnO2 + V+3 Mn+2 + VO+2 (in acid) b IO3- + Sn I2 + Sn +4 (in acid) In base a Procedure Balance each half as though in acid Add enough OH-‘s to side containing H+ to consume them Add same number of OH-‘s to other side Cancel out waters b Example Bi+3 + SnO2-2 —> Bi + SnO3-2 ( in base) BALANCING OXIDATION # REDUCTION EQUATIONS BY THE ION-ELECTRON METHOD Equation: Write the equation in ionic form Pick our the ions that are changed, (mark them) Write the half-step equations in skeleton form Balance the above equations with respect to atoms Add up the charges on each side of the equations and subtract the total on the right from the total on the left Enter the resulting signed number as the coefficient of the “e” term on the left side of the equation 6 When both half-steps have been balanced with respect to atoms and charges, then determine the lowest common factor for the “e” term coefficients and multiply each half-step by the appropriate number Add the above half-steps Check the sources of each reactant ion and determine the ionic partner (spectator ion) which must be added to the redox portion of the equation For every addition to the left make a corresponding addition to the right Finally combine the ion pairs to give a balanced molecular equation III Electrode Potentials A Define and electrode- A conducting metal in contact with a solution its ions Pb (s) / Pb+2(aq) Half Rxn Pb+2 + 2e- > Pb or Pb > Pb+2 + 2e- B Define an electrode- An inert metal, Pt, in contact with an oxidation- reduction couple in solution Pt (s)/ Fe+3(aq), Fe+2(aq) Half Fe+3 + e- -> Fe+2 Rxn or Fe+2 -> Fe+3 + e- C Substances (Pb, Pb+2, Fe+2, Fe+3) differ in their ability to combine with electrons and become reduced or to lose electrons and become oxidized Electrode’s Potential is a measure of its oxidation- reduction ability Can not measure a single electrode potential Must determine the potential of various half cells relative to one another a Couple with the greatest tendency to be reduced will be reduced b Couple with the greatest tendency to be oxidized will be oxidized c Measure the difference of the two half cell potentialsProblem D To Solve the Problem- take an electrode and assign it a half cell potential of 0.000 V Connect every other cell to it & measure Edifference Hydrogen Electrode a Pt sheet coated with amorphous Pt metal which will absorb H2 gas b Bubble H2 onto electrode holding the Pressure = atm c Set [H+] in cell= 1.0 M pH 0.0 d EE = 0.000 V for H+ + 2e > H2 EE = 0.000 V for H2 > H+ + e- Write all half rxns as though they were reduction i.e Reduction Potentials (assumes H2 > H+ + e-) a Cu+2 + e- > Cu To compare Potentials Cu+2 + e- > Cu+ of reduction half rxns Fe+3 + e- > Fe+2 b If metal ion has a greater tendency to be reduced than H+ c get E = + goes spontaneously as reduction If metal ion has less tendency to be reduced than H+ get E = (-) does not go spontaneously as reduction - will go spontaneously as oxidation if connected to the H+ electrode Standard Electrode Potential EE a E for a half rxn varies with [ ]’s b Necessary To define a Standard State for ions a= M c~1M = γι Ci for metal As an element for gas fugacity= atm p~1atm c Connect cell up to Standard H2 electrode KNO3 - Agar atm 66 H2 gas 9 9 9 966 Pt Fe+2 = M H+ = M Standard hydrogen Electrode H2 > H+ + e- Standard Couple Fe+3 + e- > Fe+2 Emeas = EEFe+3/ Fe+2 - EEH+/H Emeas = EEFe+3/Fe+2 d See text for EE’s Pt Fe+3 = M