Consider the following unbalanced redox reaction:

Question

ClO₃^-_(aq) + I^-_(aq) --> I_{2 (s)} + Cl^-_(aq)

A) Balance the redox reaction using water (H+ and OH-) as needed. Show work for each step.

B) Identify the oxidizing reagent and the reducing reagent for the redox reaction.

Answer 1

The easiest way to approach these problems is to separate the half reactions and work on them separately.

A)

ClO3- ==> Cl- and balance O by adding H2O to get

ClO3- ==> Cl- + 3H2O and now balance H by adding H+

ClO3- + 6H+ ==> Cl- + 3H2O and now add equal amounts of OH- to both sides (it says to use OH-)

ClO3- + 6H+ + 6OH- ==> Cl- + 3H2O + 6OH- and finally balancing charge by adding electrons

ClO3- + 6H+ + 6OH- + 6e- ==> Cl- + 3H2O + 6OH- balanced reduction half reaction

I- ===> I2 and balance the I to get

2I- ==> I2 and now balance charge by adding electrons

2I- ==> I2 + 2e- balance oxidation half reaction ... MULTIPLY BY 3 TO EQUALIZE ELECTRONS & ADD

ClO₃- + 6H⁺ + 6OH^- + 6e- ==> Cl- + 3H₂O + 6OH^-

6I^- ==> 3 I₂ + 6e-

---------------------------------------------------------------------

ClO₃- + 6I^- + 3H₂O ===> Cl^- + 3I₂ + 6OH^- ... BALANCED OVERALL REACTION

B)

ClO₃^- is the oxidizing reagent since the Cl atom is being reduced (ox.no.+5 to ox.no. -1)

I^- is the reducing reagent since it is being oxidized (ox.no. -1 to ox.no. zero)