Question

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)

Answer 1

`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.

Your unbalanced equation is

`"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: `"H= +1; N= +5; O = -2; C = -2; K = +1; Cr = +6"`
Right hand side: `"K = +1; N = +5; O = -2; C = -1; H = +1; Cr = +3"`

Step 2. Calculate the changes in oxidation number:

`"C: -2 → -1; Change = +1"`
`"Cr: +6 → +3; Change = -3"`

Step 3. Balance changes in oxidation number.

You need 3 atoms of `"C"` for every 1 atom of `"Cr"`.

You have 2 atoms of `"Cr"`, so you need 6 atoms of `"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 `"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 `"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 `"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 `"H"`.

Done.

Step 8. Check that all atoms balance.

`"Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side"`
`color(white)(m)"H"color(white)(mmmmm)26color(white)(mmmmmmm)26`
`color(white)(m)"N"color(white)(mmmmmll)8color(white)(mmmmmmml)8`
`color(white)(m)"O"color(white)(mmmmm)34color(white)(mmmmmmm)34`
`color(white)(m)"C"color(white)(mmmmmll)6color(white)(mmmmmmml)6`
`color(white)(m)"Cr"color(white)(mmmmml)2color(white)(mmmmmmml)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`