# Question 3d7dc

Oct 15, 2015

$2 {\text{HBr"_text((aq]) + 2"H"_2"SO"_text(4(aq]) -> "Br"_text(2(l]) + "SO"_text(2(g]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

#### Explanation:

Start by assigning oxidation numbers to all the atoms that take part in the reaction

stackrel(color(blue)(+1))("H") stackrel(color(blue)(-1))("Br")_text((aq]) + stackrel(color(blue)(+1))("H")_2stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_text(4(aq]) -> stackrel(color(blue)(+4))("S")stackrel(color(blue)(-2))("O")_text(2(g]) + stackrel(color(blue)(0))("Br"_text(2(l])

Notice that sulfur's oxidation number changes from $\textcolor{b l u e}{+ 6}$ on the reactants' side, to $\textcolor{b l u e}{+ 4}$ on the products' side, which means that it is being reduced.

Bromine's oxidation number changes from $\textcolor{b l u e}{- 1}$ on the reactants' side, to $\textcolor{b l u e}{0}$ on the products' side, which means that it is being oxidized.

The two half-equations will look like this

• oxidation half-equation

2stackrel(color(blue)(-1))("Br"^(-)) -> stackrel(color(blue)(0))"Br"_2 + 2e^(-)

Two bromine atoms will lose a total of two electrons.

• reduction half-equation

$\stackrel{\textcolor{b l u e}{+ 6}}{\text{S")O_4^(2-) + 2e^(-) -> stackrel(color(blue)(+4))("S}} {O}_{2}$

Notice that the oxygen atoms are not balanced. Since you're in acidic solution, you can use water molecules to balance oxygen atoms and protons, ${\text{H}}^{+}$, to balance hydrogen atoms.

stackrel(color(blue)(+6))("S")O_4^(2-) + 2e^(-) -> stackrel(color(blue)(+4))("S")O_2 + 2"H"_2"O"#

Now balance the hydrogen atoms

$4 \text{H"^(+) + stackrel(color(blue)(+6))("S")O_4^(2-) + 2e^(-) -> stackrel(color(blue)(+4))("S")O_2 + 2"H"_2"O}$

In any redox reaction, the number of electrons lost during oxidation must be equal to the number of electrons gained during reduction.

In your case, the numbe of electrons gained is equal to the number of electrons lost, so you don't have to change the half-equations.

So, you have

$\left\{\begin{matrix}\textcolor{w h i t e}{\times \times \times \times \times} 2 \text{Br"^(-) -> "Br"_2 + 2e^(-) \\ 4"H"^(+) + "SO"_4^(2-) + 2e^(-) -> "SO"_2 + 2"H"_2"O}\end{matrix}\right.$

Add these two half-equations to get

$2 \text{Br"^(-) + 4"H"^(+) + "SO"_4^(2-) + color(red)(cancel(color(black)(2e^(-)))) -> "Br"_2 + color(red)(cancel(color(black)(2e^(-)))) + "SO"_2 + 2"H"_2"O}$

Finally, the balanced chemical equation will be - keep an eye out for those $4$ protons, they will be distributed to $\text{HBr}$ and ${\text{H"_2"SO}}_{4}$!

$2 {\text{HBr"_text((aq]) + 2"H"_2"SO"_text(4(aq]) -> "Br"_text(2(l]) + "SO"_text(2(g]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$