# How would you balance the following equation: "S" + "HNO"_3 -> "H"_2"SO"_4 + "NO"_2 + "H"_2"O" ?

Sep 26, 2016

By the standard method for redox reactions we get:

$\text{S"+6"HNO"_3 \rarr "H"_2"SO"_4+6"NO"_2+2"H"_2"O}$

#### Explanation:

Use the standard method for redox reactions.

Oxidation:

Sulfur goes from 0 oxidation state in the element to +6 in sulfuric acid, so it gives off six (moles of) electrons per (mole of) atoms:

"S"^0 \rarr "S"^{"VI"}+6e^-

Reduction:

Nitrogen goes from +5 oxidation state in nitric acid to +4 in nitrogen dioxide, so it takes up one (mole of) electron(s) per (mole of) atoms:

"N"^"V"+e^- \rarr "N"^{"IV"}

Balancing:

For a redox eeaction to be balanced, the electrons given up must be matched with the electrons taken up. Here, we need six moles of notrogen atoms to take up tge electrons given off by one mole of sulfur atoms:

"S"^0+6"N"^"V" \rarr "S"^{"VI"}+6"N"^{"IV"}

Then we put those coefficients back into the original compounds. ${\text{S}}^{0}$ is the elementsl sulfur, $\text{N"^"V}$ is the nitrogen in nitric scid, etc.:

$\text{S"+6"HNO"_3 \rarr "H"_2"SO"_4+6"NO"_2+"H"_2"O}$

And don't forget the easy part:

The reaction is still not balanced because the elements that were not oxidized or reduced, hydrogen and oxygen, haven't been balanced. But.by balancing electrons in the oxidation and reduction components, we now have to balance only one other element; the last element is forced to fall into place. So.we selwct hydrogen and to keep the sulfur and nitrogen balanced, we adjust the coefficient on the water. Then:

$\text{S"+6"HNO"_3 \rarr "H"_2"SO"_4+6"NO"_2+2"H"_2"O}$

Sep 27, 2016

Here's what I got.

#### Explanation:

You're dealing with a redox reaction in which nitric acid oxidizes elemental sulfur to sulfuric acid, ${\text{H"_2"SO}}_{4}$, while being reduced to nitrogen dioxide, ${\text{NO}}_{2}$, in the process.

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

${\stackrel{\textcolor{b l u e}{0}}{\text{S")_ ((s)) + stackrel(color(blue)(+1))("H") stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_ (3(aq)) -> stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_ (4(aq)) + stackrel(color(blue)(+4))("N") stackrel(color(blue)(-2))("O")_ (2(g)) + stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(-2))("O}}}_{\left(l\right)}$

Notice that the oxidation state of nitrogen goes from $\textcolor{b l u e}{+ 5}$ on the products' side to $\textcolor{b l u e}{+ 4}$ on the reactants' side, which means that nitrogen is being reduced.

On the other hand, the oxidation state of sulfur goes from $\textcolor{b l u e}{0}$ on the reactants' side to $\textcolor{b l u e}{+ 6}$ on the products' side, which means that sulfur is being oxidized.

The oxidation half-reaction looks like this

stackrel(color(blue)(0))("S")_ ((s)) -> "H"stackrel(color(blue)(+6))("S") "O"_ (4(aq))^(-) + 6"e"^(-)

Balance the oxygen atoms by using water molecules.

$4 {\text{H"_ 2"O"_ ((l)) + stackrel(color(blue)(0))("S")_ ((s)) -> "H"stackrel(color(blue)(+6))("S") "O"_ (4(aq))^(-) + 6"e}}^{-}$

To balance the hydrogen atoms, add protons, ${\text{H}}^{+}$, to the side that needs hydrogen atoms.

$4 {\text{H"_ 2"O"_ ((l)) + stackrel(color(blue)(0))("S")_ ((s)) -> "H"stackrel(color(blue)(+6))("S") "O"_ (4(aq))^(-) + 6"e"^(-) + 7"H}}_{\left(a q\right)}^{+}$

The reduction half-reaction looks like this

stackrel(color(blue)(+5))("N")"O"_ (3(aq))^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N") "O"_ (2(g))

Once again, balance the oxygen atoms by adding water molecules.

stackrel(color(blue)(+5))("N")"O"_ (3(aq))^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N") "O"_ (2(g)) + "H"_ 2"O"_ ((l))

Balance the hydrogen atoms by adding protons.

$2 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+5))("N")"O"_ (3(aq))^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N") "O"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

Now, in any redox reaction ,the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

To balance out the number of electrons transferred, multiply the reduction half-reaction by $6$. Add the two half-reactions to get

$\left\{\begin{matrix}\textcolor{w h i t e}{a a a a a a a .} 4 \text{H"_ 2"O"_ ((l)) + stackrel(color(blue)(0))("S")_ ((s)) -> "H"stackrel(color(blue)(+6))("S") "O"_ (4(aq))^(-) + 6"e"^(-) + 7"H"_ ((aq))^(+) \\ 2"H"_ ((aq))^(+) + stackrel(color(blue)(+5))("N")"O"_ (3(aq))^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N") "O"_ (2(g)) + "H"_ 2"O"_ ((l))" } | \times 6\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

$4 {\text{H"_ 2"O"_ ((l)) + "S"_ ((s)) + 12"H"_ ((aq))^(+) + 6"NO"_ (3(aq))^(-) + color(red)(cancel(color(black)(6"e"^(-)))) -> "HSO"_ (4(aq))^(-) + 6"NO"_ (2(g)) + color(red)(cancel(color(black)(6"e"^(-)))) + 7"H"_ ((aq))^(+) + 6"H"_ 2"O}}_{\left(l\right)}$

This will be equivalent to

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{S"_ ((s)) + 6"NO"_ (3(aq))^(-) + 5"H"_ ((aq))^(+) -> "HSO"_ (4(aq))^(-) + 6"NO"_ (2(g)) uarr + 2"H"_ 2"O}}_{\left(l\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$