# Balance the following reaction showing both correct half-reactions: {eq}Mn^{+7} + ClO_3^- \to ClO_4^- + Mn^{+2} {/eq}

## Question:

Balance the following reaction showing both correct half-reactions:

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

## Redox Reaction:

In a redox reaction two or more different atoms undergo changes in oxidation state through the processes of oxidation and reduction. One type of atom loses electrons through oxidation and increases its oxidation state, while another type of atom gains these electrons through reduction and decreases its oxidation state. These processes are represented in the form of half-reaction equations, which are balanced according to the type of reaction solution present. In all cases they must sum together to an overall redox reaction equation that is balanced in mass and charge with no net electron species present.

{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}