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)

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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)

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Answer 1 · 2014-07-17T21:17:23

$8 {HNO}_{3} + 3 C_{2} H_{6} O + K_{2} {Cr}_{2} O_{7} \to 2 {KNO}_{3} + 3 C_{2} H_{4} O + 7 H_{2} O + 2 Cr {({NO}_{3})}_{3}$ $8"HNO"_3 + 3"C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → 2"KNO"_3 + 3"C"_2"H"_4"O"+ 7"H"_2"O" + 2"Cr"("NO"_3)_3$

Explanation:

You can find the steps for balancing redox equations here.

Your unbalanced equation is

${HNO}_{3} + C_{2} H_{6} O + K_{2} {Cr}_{2} O_{7} \to {KNO}_{3} + C_{2} H_{4} O + H_{2} O + Cr {({NO}_{3})}_{3}$ $"HNO"_3 + "C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → "KNO"_3 + "C"_2"H"_4"O"+ "H"_2"O" + "Cr"("NO"_3)_3$

Step 1. Calculate the oxidation numbers of every atom.

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

Step 2. Calculate the changes in oxidation number:

$C: -2 \to -1; Change = +1$ $"C: -2 → -1; Change = +1"$
$Cr: +6 \to +3; Change = -3$ $"Cr: +6 → +3; Change = -3"$

Step 3. Balance changes in oxidation number.

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

You have 2 atoms of $Cr$ $"Cr"$ , so you need 6 atoms of $C$ $"C"$ .

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

Step 4. Insert coefficients to get these numbers.

${HNO}_{3} + 3 C_{2} H_{6} O + 1 K_{2} {Cr}_{2} O_{7} \to {KNO}_{3} + 3 C_{2} H_{4} O + H_{2} O + 2 Cr {({NO}_{3})}_{3}$ $"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → "KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3$

Step 5a. Balance $K$ $"K"$ .

${HNO}_{3} + 3 C_{2} H_{6} O + 1 K_{2} {Cr}_{2} O_{7} \to 2 {KNO}_{3} + 3 C_{2} H_{4} O + H_{2} O + 2 Cr {({NO}_{3})}_{3}$ $"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3$

Step 5b. Balance $N$ $"N"$ .

$8 {HNO}_{3} + 3 C_{2} H_{6} O + 1 K_{2} {Cr}_{2} O_{7} \to 2 {KNO}_{3} + 3 C_{2} H_{4} O + H_{2} O + 2 Cr {({NO}_{3})}_{3}$ $color(green)(8)"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3$

Step 6. Balance $O$ $"O"$ .

$8 {HNO}_{3} + 3 C_{2} H_{6} O + 1 K_{2} {Cr}_{2} O_{7} \to 2 {KNO}_{3} + 3 C_{2} H_{4} O + 7 H_{2} O + 2 Cr {({NO}_{3})}_{3}$ $color(green)(8)"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ color(orange)(7)"H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3$

Step 7. Balance $H$ $"H"$ .

Done.

Step 8. Check that all atoms balance.

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

The balanced equation is

$8 {HNO}_{3} + 3 C_{2} H_{6} O + K_{2} {Cr}_{2} O_{7} \to 2 {KNO}_{3} + 3 C_{2} H_{4} O + 7 H_{2} O + 2 Cr {({NO}_{3})}_{3}$ $8"HNO"_3 + 3"C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → 2"KNO"_3 + 3"C"_2"H"_4"O"+ 7"H"_2"O" + 2"Cr"("NO"_3)_3$

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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)

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