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Redox Basics

Redox Basics. #68. Assign OX# a. UO 2 2+ Net charge is 2+ and OX#(O) = -2. Therefore 2+ = OX#(U) +2(-2) = 2+. Or OX#(U) = +6 c. NaBiO 3 Net charge = 0. OX#(O) = -2, OX#(Na) = +1. Therefore 0 = +1 + OX#(Bi) + 3(-2). Or OX#(Bi) = +5

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Redox Basics

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  1. Redox Basics #68. Assign OX# a. UO22+ Net charge is 2+ and OX#(O) = -2. Therefore 2+ = OX#(U) +2(-2) = 2+. Or OX#(U) = +6 c. NaBiO3 Net charge = 0. OX#(O) = -2, OX#(Na) = +1. Therefore 0 = +1 + OX#(Bi) + 3(-2). Or OX#(Bi) = +5 d. As4 Net charge = 0, As4 is an element. Therefore OX#(As) = 0 g. Na2S2O3 OX#(Na) & OX#(O) known. OX#(S) =

  2. Redox Basics #72.Redox?, oxidized?, reduced? OA?, RA? CH4(g) + H2O(g)  CO(g) + 3H2(g) • -4, +1 +1, -2 +2, -2 0 • OX# of carbon increases from -4 to +2; C loses e-s and is oxidized; methane = RA • OX# of hydrogen decreases from +1 to 0; H gains e-s and is reduced; water = OA

  3. Redox Basics #72.Redox?, oxidized?, reduced? OA?, RA? c. Zn(s) +2HCl(aq)  ZnCl2(aq) + H2(g) • 0 +1, -1 +2, -1 0 • Zn is oxidized; Zn loses e-s; Zn = RA • H is reduced; H gains e-s; HCl = OA • 2H+(aq)+2CrO42-(aq)Cr2O72(aq)+H2O(l) • +1 +6, -2 +6, -2 +1, -2

  4. Balance Redox Rxn in Acid #74a. Cu(s) +NO3-(aq)  Cu2+(aq)+ NO(g) 0 +5, -2 +2 +2, -2 • Cu is oxidized; N is reduced • Cu(s)  Cu2+(aq) oxidation ½ rxn • No O or H to balance in oxid ½ rxn • NO3-(aq)  NO(g) reduction ½ rxn • NO3-(aq)  NO(g) +2H2O(l) Balance O • 4H+(aq) + NO3-(aq)  NO(g) +2H2O(l) Balance H

  5. Balance Redox Rxn in Acid(2) #74a. Cu(s) +NO3-(aq)  Cu2+(aq)+ NO(g) • Cu(s)  Cu2+(aq) + 2e- Balance e • 3e- + 4H+(aq) + NO3-(aq)  NO(g) +2H2O(l) Balance e • Note that 2 e are lost and 3 e are gained. • Multiple oxidation ½ rxn by 3 and reduction ½ rxn by 2 to balance e-s

  6. Balance Redox Rxn in Acid(3) #74a. Cu(s) +NO3-(aq)  Cu2+(aq)+ NO(g) • 3Cu(s)  3Cu2+(aq) + 6e- • 6e- + 8H+ + 2NO3- 2NO(g) + 4H2O(l) • Add two half-rxns • 3Cu + 6e- + 8H+ + 2NO3- 3Cu2+ + 6e- + 2NO(g) +4H2O(l) Cancel as needed • 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO(g) +4H2O(l)

  7. Balance Redox Rxn in Acid(4) #74a. • Note that all electrons cancel • 3Cu + 8H+ + 2NO3-  3Cu2+ + 2NO(g) + 4H2O(l) • Check

  8. Balance Redox Rxn in Base MnO4- + NO2- MnO2 + NO3- +7, -2 +3, -2 +4, -2 +5, -2 • N is oxidized; Mn is reduced • Work on oxidation ½ rxn • NO2- NO3- balance O • NO2- + H2O  NO3- balance H • NO2- + H2O  NO3- + 2H+ balance e • NO2- + H2O  NO3- + 2H+ + 2e-

  9. Balance Redox Rxn in Base (2) • Work on reduction ½ reaction • MnO4- MnO2 Balance O • MnO4- MnO2 + 2H2O Balance H • MnO4- + 4H+ MnO2 + 2H2O Balance e- • MnO4- + 4H+ + 3e-  MnO2 + 2H2O • Notice that in oxid ½ rxn, 2 e- are lost, but in red ½ rxn, 3 e- are gained. • #e- lost must equal #e- gained

  10. Balance Redox Rxn in Base (3) • Multiply oxid ½ rxn by 3 • 3NO2- + 3H2O  3NO3- + 6H+ + 6e- • Multiply red ½ rxn by 2 • 2MnO4- + 8H+ + 6e-  2MnO2 + 4H2O • Add ½ rxns together • 3NO2- + 3H2O + 2MnO4- + 8H+ + 6e-  3NO3- + 6H+ + 6e- + 2MnO2 + 4H2O

  11. Balance Redox Rxn in Base (3) • 3NO2- + 3H2O + 2MnO4- + 8H+ + 6e-  3NO3- + 6H+ + 6e- + 2MnO2 + 4H2O • Cancel • 3NO2- + 2MnO4- + 2H+ 3NO3- + 2MnO2 + H2O Now add steps to balance in base. • Add 2 OH- to each side to make soln basic. Note that 2H+ + 2OH- 2H2O

  12. Balance Redox Rxn in Base (4) • 3NO2- + 2MnO4- + 2H+ + 2OH- 3NO3- + 2MnO2 + H2O + 2OH- • 3NO2- + 2MnO4- + 2H2O  3NO3- + 2MnO2 + H2O + 2OH- • Cancel waters • 3NO2- + 2MnO4- + H2O  3NO3- + 2MnO2 + 2OH-

  13. Balance Redox Rxn in Base (5) • 3NO2- + 2MnO4- + H2O  3NO3- + 2MnO2 + 2OH- • A redox rxn in base must have OH- left over NOT H+ (which means acid) • Check

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