Balance each redox reaction occuring in acidic aqueous solution Use the half reaction method

Question

I^−(aq)+NO2^−(aq)→I2(s)+NO(g)

BrO3^−(aq)+N2H4(g)→Br^−(aq)+N2(g)

NO3^−(aq)+Sn^2+(aq)→Sn^4+(aq)+NO(g)

Answer 1

I've done several in previous questions, so since this is in acid, I'll do one, and hopefully you can follow the procedure and do the others.

The two half reactions will be:

I^- ==> I₂ reduction half reaction

NO₂^- ==> NO oxidation half reaction

2I^- ==> I₂ .. to balance I atoms

2I^- ==> I₂ + 2e- .. to balance the charge & this is the balanced 1/2 reaction for reduction

NO₂^- ==> NO .. N atoms already balanced

NO₂^- ==> NO + H₂O .. to balance the O atoms

NO₂^- + 2H⁺ ==> NO + H₂O .. to balance the H atoms using acid (H⁺)

NO₂^- + 2H⁺ + e- ==> NO + H₂O .. to balance the charge & this is the balanced 1/2 reaction for oxidation

Since reduction reaction has 2 electrons, and oxidation has only 1 electron, we must multiply oxidation by 2

2NO₂^- + 4H⁺ + 2e- ==> 2NO + 2H₂O

2I^- ==> I₂ + 2e-

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2NO₂^- + 2I^- + 4H⁺ ==> 2NO + I₂ + 2H₂O ... balanced overall equation