Question

The complete balanced equation in acid medium is:

$H_{2}S{O}_3+C{r}_2{O}_7^{2-}\to H_{2}S{O}_4+C{r}^{3+}+H_{2}O$

• A. $3H_{2}S{O}_3+C{r}_2{O}_2^{7-}+8H^{+}\to 3H_{2}S{O}_4+2C{r}^{3+}+4H_{2}O$
• B. $4H_{2}S{O}_3+2C{r}_2{O}_2^{7-}+8H^{+}\to 5H_{2}S{O}_4+4C{r}^{3+}+4H_{2}O$
• C. $7H_{2}S{O}_3+3C{r}_2{O}_2^{7-}+3H^{+}\to 3H_{2}S{O}_4+5C{r}^{3+}+2H_{2}O$
• D. none of the above

Answer 1

Answer: (A)

$3H_{2}S{O}_3+C{r}_2{O}_2^{7-}+8H^{+}\to 3H_{2}S{O}_4+2C{r}^{3+}+4H_{2}O$

The unbalanced chemical equation is $H_{2}S{O}_3+C{r}_2{O}_7^{2-}\to H_{2}S{O}_4+C{r}^{3+}+H_{2}O$.
Balance all atoms except H and O.
$H_{2}S{O}_3+C{r}_2{O}_7^{2-}\to H_{2}S{O}_4+2C{r}^{3+}+H_{2}O$
The oxidation number of $S$ changes from +4 to +6.
The net increase in the oxidation number is 2.
The oxidation number of $Cr$ changes from +6 to +3.
Hence, the net decrease in the oxidation number for 1 $Cr$ atom is 3.
The total decrease in the oxidation number for 2 $Cr$ atoms is 6.
To balance the net increase in the oxidation number with net decrease in the oxidation number, multiply the species containing $S$ with 3.
$3H_{2}S{O}_3+C{r}_2{O}_7^{2-}\to 3H_{2}S{O}_4+2C{r}^{3+}+H_{2}O$
LHS contains 16 O atoms and RHS contains 13 O atoms. Add 3 water molecules on RHS.
$3H_{2}S{O}_3+C{r}_2{O}_7^{2-}\to 3H_{2}S{O}_4+2C{r}^{3+}+4H_{2}O$
To balance H atoms, add 8 protons on LHS.
$3H_{2}S{O}_3+C{r}_2{O}_7^{2-}+8H^{+}\to 3H_{2}S{O}_4+2C{r}^{3+}+4H_{2}O$
This is the balanced chemical equation.