# Question #ebe14

Apr 24, 2015

With the assumption that you're meaning:

$B r {O}^{3 -} \left(a q\right) + S {b}^{3 +} \left(a q\right) \to B {r}^{-} \left(a q\right) + S {b}^{5 +} \left(a q\right)$

I imagine the balanced version would be:

$B r {O}^{3 -} \left(a q\right) + S {b}^{3 +} \left(a q\right) + 2 {H}^{+} \left(a q\right) \to B {r}^{-} \left(a q\right) + S {b}^{5 +} \left(a q\right) + {H}_{2} O \left(l\right) + 2 {e}^{-}$

The first step is to recognise that one of the products do not contain Oxygen, but it is present in the reactants.

Therefore, we add a number of moles of water (${H}_{2} O$) that gives a number of moles of Oxygen, corresponding to the number of moles of Oxygen in the reactants.

In this example, $B r {O}^{3 -}$ contains one mole of oxygen, so we add one mole of water to the opposite side (in this case the products) to balance the oxygen, giving:

$B r {O}^{3 -} \left(a q\right) + S {b}^{3 +} \left(a q\right) \to B {r}^{-} \left(a q\right) + S {b}^{5 +} \left(a q\right) + {H}_{2} O \left(l\right)$

Now the problem becomes that there is no balance in hydrogen. This is corrected by adding Hydrogen ions (${H}^{+}$) to the opposite side (in this case the reactants) of the water (${H}_{2} O$).

We add a sufficient number of them so that it balances the number of moles of hydrogen in the water molecule.

Remember: ${H}_{2} O$ contains 2 moles of hydrogen, so for every one water molecule, you must add 2 ${H}^{+}$ ions.

This gives us the following:

$B r {O}^{3 -} \left(a q\right) + S {b}^{3 +} \left(a q\right) + 2 {H}^{+} \left(a q\right) \to B {r}^{-} \left(a q\right) + S {b}^{5 +} \left(a q\right) + {H}_{2} O \left(l\right)$

Notice theat one mole of ${H}_{2} O$ is present on the products side and 2 ${H}^{+}$ ions are present on the reactants side.

The final problem we need to deal with is the charge on each side. Right now it is unbalanced, but this is a simple fix!

We simply total up the charge on each side, and the side that has a more positive charge, gets a sufficient number of electrons added to it, in order to bring it down to the same charge as the opposite side.

When totalling the charge, we do not include ${H}_{2} O$. So for this example:
$B r {O}^{3 -} \left(a q\right) + S {b}^{3 +} \left(a q\right) + 2 {H}^{+} \left(a q\right) \to B {r}^{-} \left(a q\right) + S {b}^{5 +} \left(a q\right) + {H}_{2} O \left(l\right)$

we total the $3 -$, $3 +$ and the $2 +$ charges on the reactants side. Note that as there are 2 $H +$ ions, so we count a 2$+$ charge. On the products side we total the $-$ and $5 +$ charges, giving us a total of $2 +$ on reactant's side and $4 +$ on the product's.

Finally, as it has the more positive charge, we add the difference in charge ($2 -$) in the form of electrons:

$B r {O}^{3 -} \left(a q\right) + S {b}^{3 +} \left(a q\right) + 2 {H}^{+} \left(a q\right) \to B {r}^{-} \left(a q\right) + S {b}^{5 +} \left(a q\right) + {H}_{2} O \left(l\right) + 2 {e}^{-}$

Apr 24, 2015

Start by writing the oxidation numbers of all the atoms that take part in the reaction

$\stackrel{\textcolor{b l u e}{+ 5}}{B r}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{3 \left(a q\right)}^{-}} + \stackrel{\textcolor{b l u e}{+ 3}}{S {b}_{\left(a q\right)}^{3 +}} \to \stackrel{\textcolor{b l u e}{- 1}}{B {r}_{\left(a q\right)}^{-}} + \stackrel{\textcolor{b l u e}{+ 5}}{S {b}_{\left(a q\right)}^{5 +}}$

Now identify the oxidation and reduction half-reactions

$\stackrel{\textcolor{b l u e}{+ 5}}{B r} {O}_{3}^{-} + 6 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 1}}{B r}$ $\to$ the reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 3}}{S {b}^{3 +}} \to \stackrel{\textcolor{b l u e}{+ 5}}{S {b}^{5 +}} + 2 {e}^{-}$ $\to$ the oxidation half-reaction

Now take them one at a time and try to balance the oxygens and hydrogens (if needed).

SInce you're in acidic solution, you can balance the oxygen by adding one molecule of water for every oxygen atom you lack, and a proton, or ${H}^{+}$, for every hydrogen atom that you lack.

$\stackrel{\textcolor{b l u e}{+ 5}}{B r} {O}_{3}^{-} + 6 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 1}}{B r}$

You have three oxygens on the reactants' side, but none on the products' side $\to$ add three water molecules to balance the oxygen out

$\stackrel{\textcolor{b l u e}{+ 5}}{B r} {O}_{3}^{-} + 6 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 1}}{B r} + 3 {H}_{2} O$

Now you have six hydrogen atoms on the products' side, so add 6 protons on the reactants' side to balance them out

$6 {H}^{+} + \stackrel{\textcolor{b l u e}{+ 5}}{B r} {O}_{3}^{-} + 6 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 1}}{B r} + 3 {H}_{2} O$

The oxidation half-reaction has no oxygen and hydrogen atoms, so you're good to go.

Now it's time to balance the electrons. Oxidation and reduction reactions take place at the same time, so the number of electrons lost by one species must be equal to the number of electrons gained by another.

To balance the two half-reactions, multiply the oxidation half-reaction by 3 to get a total number of ${6}^{-}$ lost by the antimony.

$\stackrel{\textcolor{b l u e}{+ 3}}{S {b}^{3 +}} \to \stackrel{\textcolor{b l u e}{+ 5}}{S {b}^{5 +}} + 2 {e}^{-}$ $| \cdot 3$

You'll get

$\stackrel{\textcolor{b l u e}{+ 3}}{3 S {b}^{3 +}} \to \stackrel{\textcolor{b l u e}{+ 5}}{3 S {b}^{5 +}} + 6 {e}^{-}$

Now add the two half-reactions together

$6 {H}^{+} + B r {O}_{3}^{-} + 3 S {b}^{3 +} + \cancel{6 {e}^{-}} \to 3 S {b}^{5 +} + \cancel{6 {e}^{-}} + B {r}^{-} + 3 {H}_{2} O$

The balanced chemical equation will be

$6 {H}^{+} + B r {O}_{3}^{-} + 3 S {b}^{3 +} \to 3 S {b}^{5 +} + B {r}^{-} + 3 {H}_{2} O$

Apr 24, 2015

$B r {O}_{3}^{-} + 3 S {b}^{3 +} + 6 {H}^{+} \rightarrow B {r}^{-} + 3 S {b}^{5 +} + 3 {H}_{2} O$

We can use oxidation numbers to help balance this equation.

Consider first the bromate(V) ion $B r {O}_{3}^{-}$. It is converted to $B {r}^{-}$

$B r {O}_{3}^{-} \rightarrow B {r}^{-}$

$+ 5 \rightarrow - 1$

Bromine has changed from +5 to -1. This will require the addition of 6 electrons:

$B r {O}_{3}^{-} + 6 e \rightarrow 6 B {r}^{-}$

To take account of 3 oxygens we will need 6H+ ions to form water:

$B r {O}_{3}^{-} + 6 {H}^{+} + 6 e \rightarrow B {r}^{-} + 3 {H}_{2} O$ $\textcolor{red}{\left(1\right)}$

This is our first half - equation.

Now for the second:

$S {b}^{3 +} \rightarrow S {b}^{5 +}$

$+ 3 \rightarrow + 5$

Antimony has gone from +3 to +5. This means it needs to give out 2 electrons:

$S {b}^{3 +} \rightarrow S {b}^{5 +} + 2 e$ $\textcolor{red}{\left(2\right)}$

To get the full balanced equation we can add $\textcolor{red}{\left(1\right)}$ to $\textcolor{red}{\left(2\right)}$

The problem here is that the electrons don't balance.

If we x equation $\textcolor{red}{\left(2\right)}$ by 3 you can see we get 6 electrons given out and 6 taken in.

So we x $\textcolor{red}{\left(2\right)}$ by 3 $\Rightarrow$

$3 S {b}^{3 +} \rightarrow 3 S {b}^{5 +} + 6 e$

Now we can add both sides of each equation together $\Rightarrow$

$B r {O}_{3}^{-} + 6 {H}^{+} + \cancel{6 e} + 3 S {b}^{3 +} \rightarrow B {r}^{-} + 3 {H}_{2} O + 3 S {b}^{5 +} + \cancel{6 e}$

You can see here that the 6 electrons have now cancelled from both sides to give us:

$B r {O}_{3}^{-} + 3 S {b}^{3 +} + 6 {H}^{+} \rightarrow B {r}^{-} + 3 S {b}^{5 +} + 3 {H}_{2} O$

This is a good general method that can be applied to balancing redox equations.