Question

118 HNO: Mn + H2O + NO MnO2 + H + Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent 4. PbO2 + Mn?' + SO42- + H P bSO4 + MnO4 + H2O Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent 5._HNO: + Cr2O72- + H → Cr + NO + H2O Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent 6. Mn?+ CIO, MnO2 + CIO Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent 129

Answer 1

1)    MnO₂ + H⁺ + HNO₂ ------> Mn⁺² + H₂O + NO₃^-

Oxidation half reaction : Addition of oxygen and removal of electrons :

HNO₂ ---> NO₃^- + 2e^-

N =+3    N = +5

Less number of oxygen atoms on lefts side so oxygen is added in the form of water

  HNO₂ + H₂O ---> NO₃^- + 3 H⁺ + 2e^-    --------- 1

The substance that undergo oxidation is reducing agent - HNO₂   

Reduction half reaction : Removal of oxygen and addition of electrons :

MnO₂ ----> Mn⁺²

   +4 +2

   Oxygen atoms are none on the right side so added on the left side in the form of water

MnO₂ + 2e^- + 4H⁺ ----> Mn⁺² + 2H₂O ------ 2

The substance that undergo reduction is oxidising agent - MnO₂

Combine equations 1 and 2

  HNO₂ + H₂O ---> NO₃^- + 3 H⁺ + 2e^-

  MnO₂ + 2e^- + 4H⁺ ----> Mn⁺² + 2H₂O

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  HNO₂ +  MnO₂ + H⁺ -------> Mn⁺² + H₂O + NO₃^-

2)    HNO₂ + Cr₂O₇^-2 + H⁺ -------> Cr⁺² + NO₃^- + H₂O

Oxidation half reaction :

Addition of oxygen and removal of electrons :

HNO₂ ---> NO₃^- + 2e^-

N =+3    N = +5

Less number of oxygen atoms on lefts side so oxygen is added in the form of water

  HNO₂ + H₂O ---> NO₃^- + 3 H⁺ + 2e^-    --------- 1

The substance that undergo oxidation is reducing agent - HNO₂

Reduction half reaction : Removal of oxygen and addition of electrons :

  Cr₂O₇^-2 ----> Cr⁺²

   Cr = +6 +2

There are no oxygen atoms on right hand side so added in the form of water

Cr₂O₇^-2 ----> 2Cr⁺² + 7H₂O

  Cr₂O₇^-2 + 14H⁺ + 8e^- ----> 2Cr⁺² + 7H₂O ------- 2

The substance that undergo reduction is oxidising agent - Cr₂O₇^-2

Combine 1 and 2 :

HNO₂ + H₂O ---> NO₃^- + 3 H⁺ + 2e^- x 4    ( to balance electrons)

   4HNO₂ + 4H₂O ---> 4NO₃^- + 12 H⁺ + 8e^-

Cr₂O₇^-2 + 14H⁺ + 8e^- ----> 2Cr⁺² + 7H₂O

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   4HNO₂ + Cr₂O₇^-2 + 2H⁺ ----> 4NO₃^- + 2Cr⁺² + 3H₂O