Complete and balance the following redox reaction in basic solution

Question

IO₃^-_(aq) + Re_{(s) -->} ReO₄^-_(aq) + IO^-_(aq)

I got 7IO₃^-_(aq) + 4Re_{(s) -->} 4ReO₄^-_(aq) + 7IO^-_(aq) +4H⁺_(aq) but it says I got it wrong

Answer 1

NOTE: The following is based on the equation as you have posted it, but it may be that you meant the have I^- as a product and not IO^-. If that's the case, the answer will obviously be different, and you should re-post or make a comment to the current question/answer for follow up. Also, note in your answer that the oxygens, hydrogens and charge are not balanced, so it couldn't possibly be correct. You should always check that mass and charge are the same on both sides of the equation.

IO₃^- ===> IO^-

IO₃^- ===> IO- + 2H2O ... to balance oxygens

IO₃^- + 4H+ ===> IO- + 2H2O ... to balance hydrogens

IO₃^- + 4H+ + 4OH- ===> IO- + 2H2O + 4OH- ... because it is in basic solution

IO₃^- + 4H+ + 4OH- + 4e- ===> IO- + 2H2O + 4OH- ... to balance charge

IO₃^- + 2H2O + 4e- ===> IO- + 4OH- ... REDUCTION HALF REACTION

Re ===> ReO4

Re + 4H2O ===> ReO4 ... to balance oxygens

Re + 4H2O ===> ReO4 + 8H+ ... to balance hydrogens

Re + 4H2O + 8 OH- ===> ReO4 + 8H+ + 8OH- ... because it is in basic solution

Re + 4H2O + 8 OH- ===> ReO4 + 8H+ + 8OH- + 8e- ... to balance charge

Re + 8OH- ===> ReO4 + 4H2O + 8e- ... OXIDATION HALF REACTION

Multiply reduction half reaction by 2 to equalize the electrons and add the half reactions to get:

2IO3- + 4H2O + 8e- + Re + 8OH- ===> 2IO- + 8OH- + ReO4 + 4H2O + 8e-

Finally, cancel common items on each side and condense to get ...

2IO₃^- + Re ==> 2IO^- + ReO₄ which is balanced for both mass and charge