Question

balance the following redox reaction in basic solution:

ClO2 → ClO3- + Cl-

Answer 1

You have to split the reaction into 2 half-reactions: one where oxidation happens, and one where reduction happens. This is actually easier than it looks in this case, because there's only one reactant, it has to do both, but form different products in each case\[ClO_2 \rightarrow ClO_3^{-1}\] and \[ClO_2 \rightarrow Cl^{-1}\] Each reaction has to be balanced independently first, then we can add them back together to get the whole reaction.

1) Balance the atoms other than H and O first. This is already done, since there's only one Cl on each side already.
2) Balance oxygens by adding water (H2O) to the side that needs it
\[H_2O + ClO_2 \rightarrow ClO_3^{-1}\] and \[ClO_2 \rightarrow Cl^{-1} + 2H_2O\]
3) balance hydrogens by adding H+ to the side that needs it \[H_2O + ClO_2 \rightarrow ClO_3^{-1} + 2H^{+1}\] and \[4H^{+1} + ClO_2 \rightarrow Cl^{-1} + 2H_2O\]
4) if this were acidic, there would be lots of H+ ions floating around in the solution, but since this is basic, every H+ ion used has to be paired with an OH-, forming water. Because the O's and H's have already been balanced, adding OH to only one side would unbalance the half-reactions, so we add OH's to both sides
\[2OH^{-1} + H_2O + ClO_2 \rightarrow ClO_3^{-1} + 2H^{+1} + 2OH^{-1}\] and \[4OH^{-1} + 4H^{+1} + ClO_2 \rightarrow Cl^{-1} + 2H_2O + 4OH^{-1}\]
5) Now each half reaction is balanced for mass, but not for charge. We have to add electrons to each laf-reaction so that the total charge on each side is the same \[2OH^{-1} + H_2O + ClO_2 \rightarrow ClO_3^{-1} + 2H^{+1} + 2OH^{-1} + 1e^{-1}\] and \[5e^{-1} + 4OH^{-1} + 4H^{+1} + ClO_2 \rightarrow Cl^{-1} + 2H_2O + 4OH^{-1}\]
6) each half-reaction is balanced itself, but if we try to add them back together again now, the electrons won't cancel. So we multiply one half-reaction so that the TOTAL number of electrons in each half-reaction is the same \[5(2OH^{-1} + H_2O + ClO_2 \rightarrow ClO_3^{-1} + 2H^{+1} + 2OH^{-1} + 1e^{-1})\] and \[5e^{-1} + 4OH^{-1} + 4H^{+1} + ClO_2 \rightarrow Cl^{-1} + 2H_2O + 4OH^{-1}\]
7) distribute the multiplication, add the half-reactions back together, cancel like terms on opposite sides, combine terms on the same side, and you have a a balanced redox reaction in a basic solution

Answer 2

6OH^-(aq) +6ClO_2(g) --> 3H_2O(l)+5ClO_3^-(aq) +Cl^-(aq)