Animated Solution for Chemistry - Redox Reactions: For the reaction ${\text{I}}^{-}+{\text{ClO}}_3^{-}+{\text{H}}_2{\text{SO}}_4\to {\text{Cl}}^{-}+{\text{HSO}}_4^{-}+{\text{I}}_2$ The correct statement(s) in the balanced equation is / are :

When we first look at the unbalanced chemical equation:

It might seem like a standard redox reaction, but there is a subtle trick hidden within the sulfuric acid. To truly understand what is happening, we must break the reaction down into its fundamental electron transfers. Let's start by assigning oxidation states to the key elements.

For iodine, it starts as an iodide ion (${\text{I}}^{-}$) with an oxidation state of $-1$ and ends up as elemental iodine (${\text{I}}_2$) with an oxidation state of $0$. Since its oxidation state increases, iodine is oxidized.

For chlorine, it begins in the chlorate ion (${\text{ClO}}_3^{-}$) with an oxidation state of $+5$ and is converted to a chloride ion (${\text{Cl}}^{-}$) with an oxidation state of $-1$. A decrease in oxidation state means chlorine is reduced.

Sulfur, on the other hand, starts in sulfuric acid (${\text{H}}_2{\text{SO}}_4$) at $+6$ and remains at $+6$ in the bisulfate ion (${\text{HSO}}_4^{-}$). It is merely a spectator in the electron exchange. Right away, we can confidently say that Option (B) is correct and Option (C) is incorrect.

Half-Reactions

To balance the equation, we split it into two half-reactions.

The oxidation half-reaction is straightforward:

The reduction half-reaction requires a bit more care. We start with the core transformation:

To balance the three oxygen atoms on the reactant side, we must add three water molecules to the product side. This immediately proves that water is a product, making Option (D) correct.

Now, we have six hydrogen atoms on the right, so we add six hydrogen ions (${\text{H}}^{+}$) to the left. Finally, to balance the charge (which is $+5$ on the left and $-1$ on the right), we add six electrons to the reactant side:

Equalizing and Combining

Notice that the oxidation half-reaction produces $2$ electrons, while the reduction half-reaction consumes $6$ electrons. To ensure the total number of electrons transferred is equal, we multiply the entire oxidation half-reaction by $3$:

Adding the two balanced half-reactions together, the electrons cancel out perfectly:

Here is where the trap lies. The source of the ${\text{H}}^{+}$ ions is sulfuric acid (${\text{H}}_2{\text{SO}}_4$). However, look at the products: we form bisulfate ions (${\text{HSO}}_4^{-}$), not sulfate ions (${\text{SO}}_4^{2-}$). This means that in this specific reaction, each sulfuric acid molecule acts as a monoprotic acid, releasing only one ${\text{H}}^{+}$ ion.

Since we need $6{\text{H}}^{+}$ ions, we must use $6$ molecules of ${\text{H}}_2{\text{SO}}_4$, which will subsequently leave behind $6$ bisulfate ions:

Substituting this back into our ionic equation gives the final, fully balanced molecular equation:

The stoichiometric coefficient of ${\text{HSO}}_4^{-}$ is indeed $6$, which means Option (A) is also correct.

By carefully tracking the electrons and understanding the specific role of the acid in the reaction, we have successfully navigated all the options!