Question

Balance the following reaction.

$C{r}_2{O}_7^{2-}+C_2{H}_{4}O+H^{+}\to C_2{H}_4{O}_2+C{r}^{3+}$

• A. $C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+8H^{+}\to 3C_2{H}_4{O}_2+2C{r}^{3+}+4H_{2}O$
• B. $C{r}_2{O}_7^{2-}+C_2{H}_{4}O+8H^{+}\to 3C_2{H}_4{O}_2+5C{r}^{3+}+H_{2}O$
• C. $C{r}_2{O}_7^{2-}+4C_2{H}_{4}O+4H^{+}\to C_2{H}_4{O}_2+2C{r}^{3+}+4H_{2}O$
• D. None of these

Answer 1

Answer: (A)

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

The unbalanced redox equation is as follows:

$C{r}_2{O}_7^{2-}+C_2{H}_{4}O+H^{+}\to C_2{H}_4{O}_2+C{r}^{3+}$

Balance all atoms other than H and O.
$C{r}_2{O}_7^{2-}+C_2{H}_{4}O+H^{+}\to C_2{H}_4{O}_2+2C{r}^{3+}$

The oxidation number of Cr changes from 6 to 3. So the change in oxidation number of one Cr atom is 3.
For 2 Cr atoms, the oxidation number changes by 6.

The oxidation number of C changes from -1 to 0. So the change in oxidation number of 1 C atom is 1.
For 2 C atoms, the oxidation number changes by 2.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying
$C_2{H}_{4}O$ and $C_2{H}_4{O}_2$ with 3.
$C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+H^{+}\to 3C_2{H}_4{O}_2+2C{r}^{3+}$

O atoms are balanced by adding 4 water molecules on RHS.
$C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+H^{+}\to 3C_2{H}_4{O}_2+2C{r}^{3+}+4H_{2}O$

Hydrogen atoms are balanced by adding 7 $H^{+}$ atoms on RHS.
$C{r}_2{O}_7^{2-}+3C_2{H}_{4}O+8H^{+}\to 3C_2{H}_4{O}_2+2C{r}^{3+}+4H_{2}O$

This is the balanced chemical equation.