Balance the following reaction showing both correct half-reactions:

{eq}Mn^{+7} + ClO_3^- \to ClO_4^- + Mn^{+2} {/eq}

{eq}Mn^{+7} + ClO_3^- \to ClO_4^- + Mn^{+2} {/eq}

This is a redox reaction equation. To balance this equation we need to identify changes in oxidation states occurring between elements. To do this we need to remember these rules:

• A neutral element on its own in its standard state has an oxidation number of zero.
• The sum of the oxidation numbers for a neutral molecule must be zero.
• The sum of the oxidation numbers for an ion is equal to the net charge on the ion.
• In a compound: hydrogen prefers +1, oxygen prefers -2, fluorine prefers -1.
• In a compound with no oxygen present the other halogens will also prefer -1.

We assume the reaction is occurring in acidic solution, so we need to balance charge, hydrogens and oxygens with {eq}H^+ {/eq} and {eq}H_2O {/eq}.

• Chlorine is oxidized from +5 in chlorate anion to +7 in perchlorate anion.
• Oxidation half-reaction: {eq}ClO_3^- (aq) \rightarrow ClO_4^- (aq) + 2e^- \\ ClO_3^- (aq) \rightarrow ClO_4^- (aq) + 2e^- + 2H^+ (aq) \\ H_2O (l) + ClO_3^- (aq) \rightarrow ClO_4^- (aq) + 2e^- + 2H^+ (aq) \\ 5H_2O (l) + 5ClO_3^- (aq) \rightarrow 5ClO_4^- (aq) + 10e^- + 10H^+ (aq) {/eq}

• Manganese is reduced from +7 in the cation to +2 in the cation.
• Reduction half-reaction: {eq}Mn^{7+} (aq) + 5e^- \rightarrow Mn^{2+} (aq) \\ 2Mn^{7+} (aq) + 10e^- \rightarrow 2Mn^{2+} (aq) {/eq}

• Overall reaction: {eq}\boxed{5H_2O (l) + 2Mn^{7+} (aq) + 5ClO_3^- (aq) \rightarrow 5ClO_4^- (aq) + 10H^+ (aq) + 2Mn^{2+} (aq) }{/eq}