Question

Balance the followinh equation for the oxidation-reduction reaction occurring in an acidic solution, shoe all work, steps.

IO3- + N2H4 -> I- + N2

Answer 1

Answer -

IO₃^- + N₂H₄ ----> I^- + N₂

Step 1) Write the two half reaction and assign the oxidation state to each element and

    IO₃^- + N₂H₄ ----> I^- + N₂

I = +5                          I = -1

N = -2                          N = 0

O = -2                         

N₂H₄ gets oxidized and IO₃^- gets reduced.

So,

                      N₂H₄ ----> N₂ ………oxidation half reaction

                     IO₃^- ----> I^- ……….reduction half reaction

Step 2) Balance the element other than O and H

                      N₂H₄ ----> N₂   

                      IO₃^- ----> I^-

Step 3) Balance the O by adding 1 H₂O for 1 O

                      N₂H₄ ----> N₂   

                      IO₃^- ----> I^- + 3H₂O

Step 4) Balance the H by adding H⁺

                      N₂H₄ ----> N₂ + 4H⁺

                      IO₃^- + 6H⁺ ----> I^- + 3H₂O

Step 5) Balance the charge by adding electron

                      N₂H₄ ----> N₂ + 4H⁺ +4e^-

                      IO₃^- + 6H⁺ +6e^- ----> I^- + 3H₂O

Step 6) Balance the electron in both half reaction                      

                      6 N₂H₄ ----> 6 N₂ + 24H⁺ +24e^-

                     4 IO₃^- + 24H⁺ +24e^- ----> 4I^- + 12H₂O

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   4 IO₃^- + 6 N₂H₄ -----> 4I^- + 6 N₂ + 12H₂O

So balanced equation for the oxidation-reduction reaction as follow –

   4 IO₃^- + 6 N₂H₄ -----> 4I^- + 6 N₂ + 12H₂O