Question

Balance the following equation by ion electron method:
$C{r}_2{O}_7^{2-}+C_2{H}_{4}O+H^{+}\to 2C{r}^{3+}+C_2{H}_4{O}_2+H_{2}O$

• A. $C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+12H^{+}\to 2C{r}^{3+}+3C_2{H}_4{O}_2+6H_{2}O$
• B. $C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+8H^{+}\to 2C{r}^{3+}+3C_2{H}_4{O}_2+4H_{2}O$
• C. $C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+4H^{+}\to 2C{r}^{3+}+3C_2{H}_4{O}_2+2H_{2}O$
• D. $2C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+8H^{+}\to 4C{r}^{3+}+3C_2{H}_4{O}_2+4H_{2}O$

Answer 1

The correct option is B $C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+8H^{+}\to 2C{r}^{3+}+3C_2{H}_4{O}_2+4H_{2}O$

Since, the reaction is unbalanced,
The oxidation number for all the atoms can be:
$C{r}_2{O}_7^{2-}+C_2{H}_{4}O+H^{+}\to 2C{r}^{3+}+C_2{H}_4{O}_2+H_{2}O$
Oxidation half equation: $C_2{H}_4{O}_2\to C_2{H}_4{O}_2$
Reduction half equation: $C{r}_2{O}_7^{2-}\to 2C{r}^{3+}$
In both the half reactions, hydrogen and oxygen atoms are not balanced

Balance O atoms by adding water molecules on the side containing less number of O atoms.
Balance hydrogen atoms by adding $H^{+}$ ions to the side containing less number of H- atoms.
Oxidation half equation: $C_2{H}_4{O}_2+H_{2}O\to C_2{H}_4{O}_2+2H^{+}$
Reduction half equation: $C{r}_2{O}_7^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$
Balancing charges by adding 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: $C{r}_2{O}_7^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O+6e^{-}$
Multiply the half equation 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: $C{r}_2{O}_7^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O+6e^{-}$
Now, adding the two half reactions,
$C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+8H^{+}\to 2C{r}^{3+}+3C_2{H}_4{O}_2+4H_{2}O$