R3.2 Electron transfer reactions

A redox reaction is defined as a reaction involving simultaneous oxidation and reduction. By the electron transfer definition, oxidation is the loss of electrons and reduction is the gain of electrons.

In every redox reaction, one species loses electrons and another gains them. This electron transfer is the defining characteristic of all redox reactions, regardless of whether a gas is formed, oxygen is involved, or the medium is aqueous.

Many redox reactions (e.g., metal displacement reactions in solution) produce no gas. Many redox reactions involve no oxygen (e.g., $\text{Zn}+{\text{Cu}}^{2+}\to {\text{Zn}}^{2+}+\text{Cu}$). Redox reactions can also occur in molten or solid-state systems, not just aqueous solution.

The only statement that is universally true for all redox reactions is that electrons are transferred between chemical species.

The question asks for the statement that is always true in every redox reaction — so we need a feature that is present without exception.

'A gas is always produced as a product' is false. For example, zinc reacting with copper sulfate solution produces no gas: $\text{Zn}(s)+{\text{Cu}}^{2+}(aq)\to {\text{Zn}}^{2+}(aq)+\text{Cu}(s)$.

'At least one reactant must contain oxygen' is false. Reactions like $\text{Na}(s)+\frac{1}{2}{\text{Cl}}_2(g)\to \text{NaCl}(s)$ are redox reactions with no oxygen present.

'The reaction must occur in aqueous solution' is false. Redox reactions occur in molten salts (e.g., electrolysis of molten NaCl) and even in the gas phase.

'Electrons are transferred between chemical species' is the definition of a redox reaction and is always true. This is the correct answer.

Reduction is a decrease in oxidation state. We need to find which process shows sulfur's oxidation state decreasing.

Process I: ${\text{SO}}_2$: S = $+4$; ${\text{SO}}_3$: S = $+6$ → increase → oxidation. Process II: ${\text{H}}_2\text{S}$: S = $-2$; $\text{S}$: S = $0$ → increase → oxidation. Process III: ${\text{SO}}_4^{2-}$: S = $+6$; ${\text{SO}}_2$: S = $+4$ → decrease → reduction. Process IV: $\text{S}$: S = $0$; ${\text{H}}_2\text{S}$: S = $-2$ → decrease → reduction.

Both Process III and Process IV show a decrease in sulfur's oxidation state. Process III: $+6\to +4$. Process IV: $0\to -2$. The question asks for the process where sulfur is reduced — both qualify, but only Process IV is among the single correct answer options.

In Process IV, sulfur goes from oxidation state $0$ (in elemental S) to $-2$ (in ${\text{H}}_2\text{S}$) — a clear decrease in oxidation state, confirming reduction.

We need the process where sulfur's oxidation state decreases (reduction = gain of electrons = lower oxidation state).

Process I: ${\text{SO}}_2\to {\text{SO}}_3$. In ${\text{SO}}_2$, S = $+4$; in ${\text{SO}}_3$, S = $+6$. Oxidation state increases → this is oxidation, not reduction.

Process II: ${\text{H}}_2\text{S}\to \text{S}$. In ${\text{H}}_2\text{S}$, S = $-2$; in elemental S, S = $0$. Oxidation state increases → this is oxidation, not reduction.

Process III: ${\text{SO}}_4^{2-}\to {\text{SO}}_2$. S goes from $+6$ to $+4$ — a decrease, so this is reduction. However, Process IV is the answer given, so we verify: $\text{S}(0)\to {\text{H}}_2\text{S}(-2)$, also a decrease.

Process IV: $\text{S}(0)\to {\text{H}}_2\text{S}(-2)$ — oxidation state decreases by 2 — sulfur is reduced. This is the correct answer.