General Chemistry I Redox EquationsBalancing Redox EquationsStrategy for redox equations in acidic solutionsProblem: When dilute nitric acid is poured on a piece of copper metal, copper(II) ionsand the gas nitric oxide, NO, are formed. Write the balanced equation for the reaction.Step 0: Write the “skeleton equation”Cu(s) + H+(aq) + NO3−(aq) −→ Cu2+(aq) + NO(g)and determine the oxidation state (the oxidation state is per atom!)0 +1 +5 −2 +2 +2 −2Cu(s) + H+(aq) + NO3−(aq) −→ Cu2+(aq) + NO(g)So we haveCu : 0 → +2 oxN : +5 → +2 redStep 1: Write the “skeleton” half-reactionsox: Cu −→ Cu2+red: NO3−−→ NOStep 2: Balance each half-reaction “atomically”• all atoms other than H and O (You can use any of the species that appear in theskeleton equation—Step 0—for this purpose.)• balance O atoms by adding H2O• balance H atoms by adding H+1 of 5General Chemistry I Redox Equationsox: Cu −→ Cu2+red: NO3−−→ NO + 2 H2Ored: NO3−+ 4 H+−→ NO + 2 H2OStep 3: Balance the electric charges by adding electronsox: Cu −→ Cu2++ 2 e−red: NO3−+ 4 H++ 3 e−−→ NO + 2 H2OThe electrons have to appear on the right hand side for the oxidation half-reaction, andon the left hand side for the reduction half-reaction.Step 4: Prepare the two half-equations for summation by making the number of electronsthe same in both, i.e, find the least common multiple3 × ox: 3 Cu −→ 3 Cu2++ 6 e−2 × red: 2 NO3−+ 8 H++ 6 e−−→ 2 NO + 4 H2OStep 5: Combine the two half-reactions3 Cu + 2 NO3−+ 8 H++ 6 e−−→ 3 Cu2++ 6 e−+ 2 NO + 4 H2OStep 6: Simplify3 Cu + 2 NO3−+ 8 H+−→ 3 Cu2++ 2 NO + 4 H2OStep 7: Indicate the state of each species3 Cu(s) + 2 NO3−(aq) + 8 H+(aq) −→ 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l)This is the fully balanced net ionic equation.Check by writing a table:2 of 5General Chemistry I Redox Equationsspecies left rightCu 3 3N 2 2O 6 6H 8 8charge +6 +6Strategy for redox equations in basic solutionsIn basic solutions, no H+are available to balance H!Strategy: Pretend the solution is acidic and carry out a “neutralization reaction” at theend.Problem: The reaction of permanganate ions with bromide ions in basic solution yieldssolid manganese(IV) oxide and bromate ions. Write the fully balanced net equation.Step 0: Write the “skeleton equation” and determine the oxidation state (the oxidationstate is per atom!)+7 −2 −1 −2 +1 +4 −2 +5 −2MnO4−(aq) + Br−(aq) + OH−−→ MnO2(s) + BrO3−(aq)So we haveBr : −1 → +5 oxMn : +7 → +4 redStep 1: Write the “skeleton” half-reactionsox: Br−−→ BrO3−red: MnO4−−→ MnO2Step 2: Balance each half-reaction “atomically”• all atoms other than H and O (You can use any of the species that appear in theskeleton equation—Step 0—for this purpose.)• balance O atoms by adding H2O3 of 5General Chemistry I Redox Equations• balance H atoms by adding H+ox: Br−+ 3 H2O −→ BrO3−ox: Br−+ 3 H2O −→ BrO3−+ 6 H+red: MnO4−−→ MnO2+ 2 H2Ored: MnO4−+ 4 H+−→ MnO2+ 2 H2OStep 3: Balance the electric charges by adding electronsox: Br−+ 3 H2O −→ BrO3−+ 6 H++ 6 e−red: MnO4−+ 4 H++ 3 e−−→ MnO2+ 2 H2OThe electrons have to appear on the right hand side for the oxidation half-reaction, andon the left hand side for the reduction half-reaction.Step 4: Prepare the two half-equations for summation by making the number of electronsthe same in both, i.e, find the least common multipleox: Br−+ 3 H2O −→ BrO3−+ 6 H++ 6 e−2 × red: 2 MnO4−+ 8 H++ 6 e−−→ 2 MnO2+ 4 H2OStep 5: Combine the two half-reactionsBr−+ 3 H2O + 2 MnO4−+ 8 H++ 6 e−−→ BrO3−+ 6 H++ 6 e−+ 2 MnO2+ 4 H2OStep 6: SimplifyBr−+ 2 MnO4−+ 2 H+−→ BrO3−+ +2 MnO2+ H2OStep 6a: Change to basic solution by adding as many OH−to both sides as thereare H+4 of 5General Chemistry I Redox EquationsBr−+ 2 MnO4−+ 2 H++ 2 OH−−→ BrO3−+ +2 MnO2+ H2O + 2 OH−“neutralization”: Combine the H+and the OH−to form H2OBr−+ 2 MnO4−+ 2 H2O −→ BrO3−+ +2 MnO2+ H2O + 2 OH−simplifyBr−+ 2 MnO4−+ H2O −→ BrO3−+ +2 MnO2+ 2 OH−Step 7: Indicate the state of each speciesBr−(aq) + 2 MnO4−(aq) + H2O(l) −→ BrO3−(aq) + +2 MnO2(s) + 2 OH−(aq)This is the fully balanced net ionic equation.Check by writing a table:species left rightBr 1 1Mn 2 2O 9 9H 2 2charge −3 −35 of
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