Question

Balance the following equations by ion electron method.
a. ${Cr}_2{O}_7^{2-}+C_2{H}_{4}O+H^{\oplus }⟶2{Cr}^{3+}+C_2{H}_4{O}_2+H_{2}O$
b. ${Cu}_{2}O+H^{\oplus }+N{O}_3^{\ominus }⟶{Cu}^{2+}+NO+H_{2}O$

• A. a. ${Cr}_2{O}_7^{2-}+8H^{\oplus }+3C_2{H}_{4}O⟶2{Cr}^{3+}+4H_{2}O+3C_2{H}_4{O}_2$
  b. $3{Cu}_{2}O+14H^{\oplus }+2N{O}_3^{\ominus }⟶6{Cu}^{2+}+7H_{2}O+2NO$
• B. a. ${Cr}_2{O}_7^{2-}+4H^{\oplus }+3C_2{H}_{4}O⟶2{Cr}^{3+}+2H_{2}O+3C_2{H}_4{O}_2$
  b. $3{Cu}_{2}O+14H^{\oplus }+4N{O}_3^{\ominus }⟶6{Cu}^{2+}+7H_{2}O+4NO$
• C. a. $2{Cr}_2{O}_7^{2-}+8H^{\oplus }+3C_2{H}_{4}O⟶4{Cr}^{3+}+4H_{2}O+3C_2{H}_4{O}_2$
  b. $6{Cu}_{2}O+14H^{\oplus }+2N{O}_3^{\ominus }⟶6{Cu}^{2+}+7H_{2}O+2NO$
• D. None of these

Answer 1

The correct option is A a. ${Cr}_2{O}_7^{2-}+8H^{\oplus }+3C_2{H}_{4}O⟶2{Cr}^{3+}+4H_{2}O+3C_2{H}_4{O}_2$

b. $3{Cu}_{2}O+14H^{\oplus }+2N{O}_3^{\ominus }⟶6{Cu}^{2+}+7H_{2}O+2NO$
(a) The unbalanced redox equation is as follows:
${Cr}_2{O}_7^{2-}+C_2{H}_{4}O+H^{\oplus }⟶2{Cr}^{3+}+C_2{H}_4{O}_2+H_{2}O$
The oxidation numbers for all the atoms are assigned.
$\underset{6}{{Cr}_2}{O}_7^{2-}+\underset{-1}{C_2}{H}_{4}O+H^{\oplus }⟶\underset{3}{2{Cr}^{3+}}+\underset{0}{C_2}{H}_4{O}_2+H_{2}O$
Oxidation half equation: $C_2{H}_{4}O\to C_2{H}_4{O}_2$
Reduction half equation: ${Cr}_2{O}_7^{2-}\to 2{Cr}^{3+}$
All atoms other than H and O are balanced.
Balance O atoms by adding water molecules on the side containing less number of O atoms.
Balance hydrogen atoms by ading $H^{+}$ ions to the side containing less number of H atoms.
Oxidation half equation: $C_2{H}_{4}O+H_{2}O\to C_2{H}_4{O}_2+2H^{+}$
Reduction half equation: ${Cr}_2{O}_7^{2-}+14H^{+}\to 2{Cr}^{3+}+7H_{2}O$
Balance charges by adding appropriate number of electrons.
Oxidation half equation: $C_2{H}_{4}O+H_{2}O+2e^{-}\to C_2{H}_4{O}_2+2H^{+}$
Reduction half equation: ${Cr}_2{O}_7^{2-}+14H^{+}\to 2{Cr}^{3+}+7H_{2}O+6e^{-}$
Multiply the half equations by suitable factors to equalize the number of electrons in the two half equations.
Oxidation half equation: $3C_2{H}_{4}O+3H_{2}O+6e^{-}\to 3C_2{H}_4{O}_2+6H^{+}$
Reduction half equation: ${Cr}_2{O}_7^{2-}+14H^{+}\to 2{Cr}^{3+}+7H_{2}O+6e^{-}$
Add the two half equations and cancel the number of electrons on both sides.
${Cr}_2{O}_7^{2-}+3C_2{H}_{4}O+8H^{\oplus }⟶2{Cr}^{3+}+3C_2{H}_4{O}_2+4H_{2}O$
This is the balanced chemical equation.
(b) The unbalanced redox equation is as follows:
${Cu}_{2}O+H^{\oplus }+N{O}_3^{\ominus }⟶{Cu}^{2+}+NO+H_{2}O$
The oxidation numbers for all the atoms are assigned
$\underset{+1}{{Cu}_2}O+H^{\oplus }+\underset{+5}{N}{O}_3^{\ominus }⟶\underset{+2}{{Cu}^{2+}}+\underset{2}{NO}+H_{2}O$
Oxidation half equation: ${Cu}_{2}O\to {Cu}^{2+}$
Reduction half equation: $N{O}_3^{\ominus }\to NO$
All atoms other than H and O are balanced.
Oxidation half equation: ${Cu}_{2}O\to 2{Cu}^{2+}$
Reduction half equation: $N{O}_3^{\ominus }\to NO$
Balance O atoms by adding water molecules on the side containing less number of O atoms.
Balance hydrogen atoms by ading $H^{+}$ ions to the side containing less number of H atoms.
Oxidation half equation: ${Cu}_{2}O+2H^{+}\to 2{Cu}^{2+}+H_{2}O$
Reduction half equation: $N{O}_3^{\ominus }+4H^{+}\to NO+2H_{2}O$
Balance charges by adding appropriate number of electrons.
Oxidation half equation: ${Cu}_{2}O+2H^{+}\to 2{Cu}^{2+}+H_{2}O+2e^{-}$
Reduction half equation: $N{O}_3^{\ominus }+4H^{+}+3e^{-}\to NO+2H_{2}O$
Multiply the half equations by suitable factors to equalize the number of electrons in the two half equations.
Oxidation half equation: $3{Cu}_{2}O+6H^{+}\to 6{Cu}^{2+}+3H_{2}O+6e^{-}$
Reduction half equation: $2N{O}_3^{\ominus }+8H^{+}+6e^{-}\to 2NO+4H_{2}O$
Add the two half equations and cancel the number of electrons on both sides.
$3{Cu}_{2}O+14H^{\oplus }+2N{O}_3^{\ominus }⟶6{Cu}^{2+}+7H_{2}O+2NO$
This is the balanced chemical equation.