How do I solve this question? Bromate(V) ions act as oxidizing agents in acidic conditions to form bromide ions. Can you deduce the half-equation for this reduction reaction?

Write down the unbalanced equation using correct formulae and adding electrons to the left and water to the right (since it is an oxidation - addition of electrons and addition of oxygen to hydrogen)…

BrO3^(-) + H^(+) + e^(-) → Br^(-) + H2O

Balance Oxygens…

BrO3(-) + H^(+) + e^(-) → Br^(-) + 3H2O

Balance Hydrogens…

BrO3^(-) + 6H^(+) + e^(-) → Br^(-) + 3H2O

Balance Br - already done so nothing more for this stage…

Balance for electron charges…

BrO3^(-) + 6H^(+) + 6e^(-) → Br^(-) + 3H2O

Always check everything balances in your final half reaction…

NOTE: When learning these processes, it is always a good idea (though laborious) to show the working out like this. In that way if you make a mistake, the teacher should be able to spot it easily and correct your working more effectively. If you scribble a lot of scrawl to fit on one line with crossings out all over the place, the teacher can only tell you if it is correct or incorrect.

Well, you got bromate, i.e. [math]BrO_{3}^{-}[/math], [math]Br(+V)[/math], and this undergoes 6-electron reduction to give [math]Br^{-}[/math], i.e. [math]Br(-I)[/math]….and these 6 electrons feature in the formal reduction equation…

[math]underbrace{BrO_{3}^{-} + 6e^{-} ightarrow Br^{-} + 3H_{2}O(l)}_{ ext{Almost balanced...}}[/math]

This is not quite balanced with respect to mass, and so we add [math]6H^{+}[/math] to the left-hand side as we face the page….

[math]underbrace{BrO_{3}^{-}+6H^{+} + 6e^{-} ightarrow Br^{-} + 3H_{2}O(l)}_{ ext{Charge and mass balanced as required...}}[/math]