Group 17: Question 5

Worked solution

Step 1: Find the oxidation number of sulfur in $\text{H}_2\text{SO}_4$

Hydrogen is $+1$ (two of them, total $+2$) and oxygen is $-2$ (four of them, total $-8$). Since the compound is neutral overall:

$(+2) + x + (-8) = 0 \implies x = +6$

So sulfur has oxidation number $\boxed{+6}$ in concentrated sulfuric acid.

Step 2: Find the oxidation number of sulfur in $\text{H}_2\text{S}$

Hydrogen is again $+1$ (two of them, total $+2$), and the compound is neutral overall:

$(+2) + y = 0 \implies y = -2$

So sulfur has oxidation number $\boxed{-2}$ in hydrogen sulfide.

Step 3: Interpret the change

Sulfur’s oxidation number falls from $+6$ to $-2$, a very large decrease of $8$, meaning each sulfur atom gains 8 electrons overall. This is a reduction of sulfur. Since the iodide ions are what caused this reduction (by supplying the electrons, and being oxidised to iodine, $\text{I}_2$, themselves in the process), the iodide ion has acted as a reducing agent.

This large reduction of sulfur, all the way down to $\text{H}_2\text{S}$, reflects the fact that iodide is the strongest reducing agent among the halide ions, because its large ionic radius means its outer electrons are held only loosely and are easily given away. (By contrast, bromide can only reduce sulfur as far as $\text{SO}_2$ ($+4$), and chloride cannot reduce $\text{H}_2\text{SO}_4$ at all.)

Why the other options are wrong

• A: incorrectly starts from $+4$ (the oxidation number of sulfur in $\text{SO}_2$) rather than $+6$ (in $\text{H}_2\text{SO}_4$); it also wrongly calls the iodide ion “oxidised” when the question asks about sulfur’s change and iodide’s role as reducing agent.
• C: correctly identifies the oxidation number change ($+6$ to $-2$) but wrongly labels iodide as an “oxidising agent”. Iodide is reducing sulfur, so iodide itself is the reducing agent, not the oxidising agent.
• D: only takes sulfur as far as $\text{SO}_2$ ($+4$), which is the level of reduction caused by the weaker reducing agent bromide, not the much stronger reducing agent iodide, which reduces sulfur all the way to $\text{H}_2\text{S}$ ($-2$).

Final answer

• Sulfur changes from $+6$ to $-2$; the iodide ion has acted as a reducing agent, option B.