Cr2o2−7(aq)+i−(aq)→cr3+(aq)+io−3(aq ) (acidic solution) express your answer as a chemical equation. identify all of the phases in your answer.

Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either. Start with your 2 half reactions: I- --> IO3- Cr2O72- --> 2 Cr3+ Balance O by adding H2O: I- + 3 H2O --> IO3- Cr2O72- --> 2 Cr3+ + 7H2O Balance H by adding H+: I- + 3 H2O --> IO3- + 6 H+ Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O Balance charge by adding e-: I- + 3 H2O --> IO3- + 6 H+ + 6 e- Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give: Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O

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