Complete and balance the following redox reaction in basic solution. Be sure to include the proper phases for all species within the reaction.

Question

Cr2O7^2-(aq)+I2(s)==>Cr^3+(aq)+IO3^-(aq)

Answer 1

I'm too lazy to include the phases, so I'll leave that up to you. The phases will be the same as those shown in the original reaction with H2O being (l) and OH^- being (aq)

Cr₂O₇^2- ==> Cr³⁺ ... reduction half reaction

I₂(s) ==> IO₃^- ... oxidation half reaction

Cr₂O₇^2- ==> 2Cr³⁺ + 7H₂O ... balanced for Cr and O

Cr₂O₇^2- + 14H₂O ==> 2Cr³⁺ + 7H₂O + 14OH^- ... balanced for Cr, O and H in basic solution (OH-).

Cr₂O₇^2- + 14H₂O + 6e- ==> 2Cr³⁺ + 7H₂O + 14OH^- ... balanced reduction reaction

I₂(s) + 6H2O ==> 2IO₃^- ... balanced for I and O

I₂(s) + 6H2O + 12OH- ==> 2IO₃^- + 12H₂O ... balanced for I, O and H in basic solution (OH-)

I₂(s) + 12OH- ==> 2IO₃^- + 6H2O + 10e- ... balanced oxidation reaction

Multiply reduction reaction by 5 and oxidation reaction by 3 to equalize electrons:

5Cr₂O₇^2- + 70H₂O + 30e- ==> 10Cr³⁺ + 35H₂O + 70 OH^-

3I₂(s) + 36OH- ==> 6IO₃^- + 18H₂O + 30e-

Combine and/or cancel like terms:

5Cr₂O₇^2- + 17H₂O + 3I₂(s) ==> 10Cr³⁺ + 34OH^- + 6IO₃^- ... BALANCED REDOX EQUATION