Question

Balance the following equation.

$H_{2}S+C{r}_2{O}_7^{2-}+H^{+}\to C{r}_2{O}_3+S_8+H_{2}O$

• A. $24H_{2}S+8C{r}_2{O}_7^{2-}+16H^{+}\to 8C{r}_2{O}_3+3S_8+32H_{2}O$
• B. $24H_{2}S+3C{r}_2{O}_7^{2-}+6H^{+}\to 7C{r}_2{O}_3+3S_8+32H_{2}O$
• C. $24H_{2}S+8C{r}_2{O}_7^{2-}+5H^{+}\to 6C{r}_2{O}_3+4S_8+32H_{2}O$
• D. None of these

Answer 1

The correct option is A $24H_{2}S+8C{r}_2{O}_7^{2-}+16H^{+}\to 8C{r}_2{O}_3+3S_8+32H_{2}O$

The unbalanced redox equation is as follows:

$H_{2}S+C{r}_2{O}_7^{2-}+H^{+}\to C{r}_2{O}_3+S_8+H_{2}O$

Balance all atoms other than H and O.

$8H_{2}S+C{r}_2{O}_7^{2-}+H^{+}\to C{r}_2{O}_3+S_8+H_{2}O$

The oxidation number of chromium changes from +6 to +3. The change in the oxidation number of one chromium atom is 3. For 2 Cr atoms, the change in the oxidation number is 6.
The oxidation number of sulphur changes from -2 to 0. The change in the oxidation number of one S atom is 2. For 8 S atoms, the change in the oxidation number is 16.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $C{r}_2{O}_7^{2-}$ and $C{r}_2{O}_3$ with 8 and by multiplying $H_{2}S$ and $S_8$ with 3.

$24H_{2}S+8C{r}_2{O}_7^{2-}+H^{+}\to 8C{r}_2{O}_3+3S_8+H_{2}O$

O atoms are balanced by adding 32 $H_{2}O$ molecules on RHS.

$24H_{2}S+8C{r}_2{O}_7^{2-}+H^{+}\to 8C{r}_2{O}_3+3S_8+32H_{2}O$

Hydrogen atoms are balanced by adding 15 $H^{+}$ on LHS.

$24H_{2}S+8C{r}_2{O}_7^{2-}+16H^{+}\to 8C{r}_2{O}_3+3S_8+32H_{2}O$

This is the balanced chemical equation.