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

Question

IO3- + Re > ReO4- + IO-

Answer 1

Half reactions:

IO₃^-(aq) ==> IO^-(aq) .. unbalanced reduction reaction

IO₃^-(aq) ==> IO^-(aq) + 2H₂O(l) .. balanced for I and O

IO₃^-(aq) + 4H₂O(l) ==> IO^-(aq) + 2H₂O(l) + 4OH^-(aq) .. balanced for I, O and H (using base, OH^-)

IO₃^-(aq) + 4H₂O(l) + 4e- ==> IO^-(aq) + 2H₂O(l) + 4OH^-(aq) .. balanced for mass and charge

Re(s) ==> ReO₄^-(aq) .. unbalanced oxidation reaction

Re(s) + 4H₂O(l) ==> ReO₄^-(aq) .. balanced for Re and O

Re(s) + 4H₂O(l) + 8OH^-(aq) ==> ReO₄^-(aq) + 8H₂O(l) .. balanced for Re, O and H (using base, OH^-)

Re(s) + 4H₂O(l) + 8OH^-(aq) ==> ReO₄^-(aq) + 8H₂O(l) + 7e- .. balanced for mass and charge

Multiply reduction rxn by 7 and oxidation rxn by 4 in order to balanced electrons:

7 IO₃^-(aq) + 28 H₂O(l) + 28e- ==> 7 IO^-(aq) + 14 H₂O(l) + 28 OH^-(aq)

4 Re(s) + 16H₂O(l) + 32 OH^-(aq) ==> 4 ReO₄^-(aq) + 32 H₂O(l) + 28e-

Add the two together and combine/cancel like terms to end up with the final balance equation:

7 IO₃^-(aq) + 4 Re(s) + 4 OH^-(aq) ==> 7 IO^-(aq) + 2 H₂O(l) + 4 ReO₄^-(aq) .. balanced equation