Balance each of the following redox recctions occurring in acidic solution.PbO2(s)+NO(g)→Pb2+(aq)+NO3−(aq S2O32−(aq)+Cl2(g)→SO42−(aq)+Cl−(aq?

3 Answers

• Balancing redox reaction….

  Chemical reactions must be both mass balanced and charge balanced. Use the half-reaction method to balance redox reactions. Balance oxygen with H2O and balance hydrogen with H+. Adjust the coefficients so that the number of electrons “lost” equals the number of electrons “gained.”

  PbO2(s) + NO(g) –> Pb2+ + NO3^- ……… not balanced

  3(PbO2 + 4H+ + 2e- –> Pb2+ + 2H2O) ……… reduction half-reaction

  2(NO + 2H2O –> NO3^- + 4H+ + 3e-) ……….. oxidation half-reaction

  ————— —————- ————— —————-

  3PbO2 + 2NO + 12H+ + 4H2O –> 3Pb2+ + 2NO3^- + 6H2O + 8H+

  simplify

  3PbO2 + 2NO + 4H+ –> 3Pb2+ + 2NO3^- + 2H2O ……….balanced net ionic equation

  S2O3^2-(aq) + Cl2(g) –> SO4^2-(aq) + Cl-(aq) ………….. not balanced

  S2O3^2- + 5H2O –> 2SO4^2- + 10H+ + 8e- ……………… oxidation half-reaction

  4(Cl2 + 2e- –> 2Cl-) ………………….. …………………. ……… reduction half-reaction

  ———————– ———————– ——————— ———-

  S2O3^2- + 4Cl2 + 5H2O –> 2SO4^2- + 8Cl- + 10H+ ….. balanced net ionic equation

• PbO2(s) + NO(g) → Pb2+ (aq) + NO3−(aq)

  Pb(4+) + 2e- —> Pb(2+) <<< X3

  N(2+) —> N(5+) + 3e- <<<< X2

  3Pb(4+) + 6e- —> 3Pb(2+)

  2N(2+) —> 2N(5+) + 6e-

  3PbO2(s) + 2NO(g) + 4H+ → 3Pb2+ (aq) + 2NO3−(aq) + 2H2O <<<< this needs an anion, depending on the ACID involved.

  ================================================

  S2O32−(aq)+Cl2(g)→SO42−(aq)+Cl−(aq)

  2S(2+)(aq) → 2S(6+)(2−)(aq) + 8e-

  Cl2(g) + 2e -→ 2Cl−(aq) <<<< X4

  2S(2+)(aq) → 2S(6+)(2−)(aq) + 8e-

  4 Cl2(g) + 8e- -→ 8Cl−(aq)

  S2O3(2−)(aq) + 4Cl2(g) → 2SO4(2−)(aq) + 8Cl(−)(aq) + 10 H+

• 6 PbO2(s) + 7 NO(g) + 2 H2O(l) → 3 Pb2{+}(aq) + 7 NO3{-}(aq) + 4 H{+}(aq)

  S2O3{2-}(aq) + 4 Cl2(g) + 5 H2O(l) → 2 SO4{2-}(aq) + 8 Cl{-}(aq) + 10 H{+}(aq)