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) =

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

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

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

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

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)

Balance Redox Rxn in Acid(4) Note that all electrons cancel 3Cu + 8H+ + 2NO3-  3Cu2+ + 2NO(g) + 4H2O(l) Check R P 3 Cu 8 H 2 N 6 O +6 Charge

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-

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

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

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

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-

Balance Redox Rxn in Base (5) P 3 N 15 O 2 Mn H -5 Ch 3NO2- + 2MnO4- + H2O  3NO3- + 2MnO2 + 2OH- A redox rxn in base must have OH- left over NOT H+ (which means acid) Check