Question

${\mathrm{S}}_8+{\mathrm{bOH}}^{-}\to {\mathrm{cS}}^{2-}+{\mathrm{sS}}_2{\mathrm{O}}_3^{2-}+{\mathrm{H}}_2\mathrm{O}$. Find the value of c.

Answer 1

Step 1: Writing the half-reaction

1. The reduction reaction of sulfur is ${\mathrm{S}}_8\to {\mathrm{S}}^{2-}$
2. The oxidation reaction of Sulfur is ${\mathrm{S}}_8\to {\mathrm{S}}_2{\mathrm{O}}_3^{2-}$

Step 2: Balancing the change in oxidation state and other atoms.

1. For reduction reaction balancing the sulfur atom we get, ${\mathrm{S}}_8\to 8{\mathrm{S}}^{2-}$
2. Adding electron equal to the charge on the left-hand side and right-hand side. ${\mathrm{S}}_8+16{\mathrm{e}}^{-}\to 8{\mathrm{S}}^{2-}$
3. For oxidation reaction, balancing the sulfur atom and charge on both sides by adding electrons we get, ${\mathrm{S}}_8\to 4{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+16{\mathrm{e}}^{-}$
4. Adding hydroxide ion as the reaction takes place in a basic medium ${\mathrm{S}}_8+{\mathrm{OH}}^{-}\to 4{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+16{\mathrm{e}}^{-}$
5. Balancing all the other atoms by adding water molecule we get, ${\mathrm{S}}_8+24{\mathrm{OH}}^{-}\to 4{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+12{\mathrm{H}}_2\mathrm{O}+16{\mathrm{e}}^{-}$

Step 3: Adding both the half-reactions

1. Adding both the reactions, electrons get eliminated, and we get $2{\mathrm{S}}_8+24{\mathrm{OH}}^{-}\to 4{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+8{\mathrm{S}}^{2-}+12{\mathrm{H}}_2\mathrm{O}$
2. Dividing the reaction by $2$, we get ${\mathrm{S}}_8+12{\mathrm{OH}}^{-}\to 2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+4{\mathrm{S}}^{2-}+6{\mathrm{H}}_2\mathrm{O}$
3. Hence, the coefficient of ${\mathrm{S}}^{2-}$is $4$

Therefore, the value of c is $4$