# How do you balance Cu + H_2SO_4 -> CuSO_4 + SO_2 + H_2O and identify the element oxidized and the oxidizing agent?

Mar 15, 2016

One way is to use the method of oxidation numbers.
$\text{Cu}$ is oxidized; $\text{S}$ is reduced.

#### Explanation:

$\text{Cu" + "H"_2"SO"_4 → "CuSO"_4 + "SO"_2 + "H"_2"O}$

Step 1. Identify the atoms that change oxidation number

$\stackrel{\textcolor{b l u e}{0}}{\text{Cu") + stackrelcolor(blue)(+1)("H")_2stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 → stackrelcolor(blue)(+2)("Cu")stackrelcolor(blue)(+6)("S")stackrelcolor(blue)(-2)("O")_4 + stackrelcolor(blue)(+4)("S")stackrelcolor(blue)("-2")("O")_2 + stackrelcolor(blue)(+1)("H")_2stackrelcolor(blue)("-2")("O}}$

The changes in oxidation number are:

"Cu: 0 → +2; Change =+2" (bb "oxidation")
"S: +6 → +4 in SO"_2; "Change = -2" (bb"reduction")

$\text{Cu}$ is oxidized; $\text{S}$ is reduced.

Step 2. Equalize the changes in oxidation number

We need 1 atom of $\text{Cu}$ for every 1 atom of $\text{S}$ in $\text{SO"_2}$. This gives us total changes of +2 and -2.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{1} \text{Cu" + "H"_2"SO"_4 → color(red)(1)"CuSO"_4 + color(red)(1)"SO"_2 + "H"_2"O}$

Step 4. Balance $\text{S}$

We have fixed 2 $\text{S}$ atoms on the right, so we need 2 $\text{S}$ atoms on the left. Put a 2 before ${\text{H"_2"SO}}_{4}$.

$\textcolor{red}{1} \text{Cu" + color(blue)(2)"H"_2"SO"_4 → color(red)(1)"CuSO"_4 + color(red)(1)"SO"_2 + "H"_2"O}$

Step 5. Balance $\text{O}$.

We have fixed 8 $\text{O}$ atoms on the left and 6 $\text{O}$ atoms on the right.

We need 2 more $\text{O}$ atoms on the right. Put a 2 before $\text{H"_2"O}$.

$\textcolor{red}{1} \text{Cu" + color(blue)(2)"H"_2"SO"_4 → color(red)(1)"CuSO"_4 + color(red)(1)"SO"_2 + color(green)(2)"H"_2"O}$

Every formula now has a coefficient. The equation should be balanced.

Step 6. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l)bb "On the right}}$
$\textcolor{w h i t e}{m m} \text{1 Cu"color(white)(mmmm) "1 Cu}$
$\textcolor{w h i t e}{m m} \text{4 H"color(white)(mmmmll) "4 H}$
$\textcolor{w h i t e}{m m} \text{2 S"color(white)(mmmmm) "2 S}$
$\textcolor{w h i t e}{m m} \text{8 O"color(white)(mmmmll) "8 O}$

The balanced equation is

$\textcolor{red}{\text{Cu"color(white)(l) + "2H"_2"SO"_4 → "CuSO"_4 + "SO"_2 +2"H"_2"O}}$