# How do you balance this redox reaction using the oxidation number method? HNO3(aq) + C2H6O(l) + K2Cr2O7(aq) → KNO3(aq) + C2H4O(l) + H2O(l) + Cr(NO3)3(aq)

Jul 17, 2014

8"HNO"_3 + 3"C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → 2"KNO"_3 + 3"C"_2"H"_4"O"+ 7"H"_2"O" + 2"Cr"("NO"_3)_3

#### Explanation:

You can find the steps for balancing redox equations here.

"HNO"_3 + "C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → "KNO"_3 + "C"_2"H"_4"O"+ "H"_2"O" + "Cr"("NO"_3)_3

Step 1. Calculate the oxidation numbers of every atom.

Left hand side: $\text{H= +1; N= +5; O = -2; C = -2; K = +1; Cr = +6}$
Right hand side: $\text{K = +1; N = +5; O = -2; C = -1; H = +1; Cr = +3}$

Step 2. Calculate the changes in oxidation number:

$\text{C: -2 → -1; Change = +1}$
$\text{Cr: +6 → +3; Change = -3}$

Step 3. Balance changes in oxidation number.

You need 3 atoms of $\text{C}$ for every 1 atom of $\text{Cr}$.

You have 2 atoms of $\text{Cr}$, so you need 6 atoms of $\text{C}$.

This gives you total changes of +6 and -6.

Step 4. Insert coefficients to get these numbers.

"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → "KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3

Step 5a. Balance $\text{K}$.

"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3

Step 5b. Balance $\text{N}$.

color(green)(8)"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3

Step 6. Balance $\text{O}$.

color(green)(8)"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ color(orange)(7)"H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3

Step 7. Balance $\text{H}$.

Done.

Step 8. Check that all atoms balance.

$\text{Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side}$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m m} 26 \textcolor{w h i t e}{m m m m m m m} 26$
$\textcolor{w h i t e}{m} \text{N} \textcolor{w h i t e}{m m m m m l l} 8 \textcolor{w h i t e}{m m m m m m m l} 8$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m m} 34 \textcolor{w h i t e}{m m m m m m m} 34$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{m m m m m l l} 6 \textcolor{w h i t e}{m m m m m m m l} 6$
$\textcolor{w h i t e}{m} \text{Cr} \textcolor{w h i t e}{m m m m m l} 2 \textcolor{w h i t e}{m m m m m m m l} 2$

The balanced equation is

8"HNO"_3 + 3"C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → 2"KNO"_3 + 3"C"_2"H"_4"O"+ 7"H"_2"O" + 2"Cr"("NO"_3)_3