Balance redox reaction include H+; ClO2 (aq) + Cr2O7 (aq )⇄ClO4 (aq) + Cr^(3+) (aq). (half reaction method)

Answer 1

ClO₂^- + Cr₂O₇^2- <==> ClO₄ + Cr³⁺ (in acid) NOTE: it's important to include the charge on each ion

Half reactions:

ClO₂^- ===> ClO₄^- ... oxidation half reaction b/c Cl goes from 3+ to 7+

Cr₂O₇^2- ===> Cr³⁺ ... reduction half reaction b/c Cr goes from 6+ to 3+

Step 1: balance elements other than oxygen:

ClO₂^- ===> ClO₄^-

Cr₂O₇^2- ===> 2Cr³⁺

Step 2: balance the oxygens by adding H₂O:

ClO₂^- + 2H₂O ===> ClO₄^-

Cr₂O₇^2- ===> 2Cr³⁺ + 7H₂O

Step 3: balance hydrogens by adding H⁺:

ClO₂^- + 2H₂O ===> ClO₄^- + 4H⁺

Cr₂O₇^2- + 14H⁺ ===> 2Cr³⁺ + 7H₂O

Step 4: balance charge by adding electrons:

ClO₂^- + 2H₂O ===> ClO₄^- + 4H⁺ + 4e-

Cr₂O₇^2- + 14H⁺ + 6e- ===> 2Cr³⁺ + 7H₂O

Step 5: equalize electrons in both half reactions (oxidation x 3 and reduction x 2 to get 12 electrons):

3ClO₂^- + 6H₂O ===> 3ClO₄^- + 12H⁺ + 12e-

2Cr₂O₇^2- + 28H⁺ + 12e- ===> 4Cr³⁺ + 14H₂O

Step 6: add the half reactions and add and/or cancel like terms to obtain final balanced equation:

3ClO₂^- + 2Cr₂O₇^2- + 16H⁺ ==> 3ClO₄^- + 4Cr³⁺ + 8H₂O ... FINAL BALANCED EQUATION