In which of the following equations does sulfuric acid act as an oxidizing agent? In which does it act as an acid? Explain. $$ \begin{array}{l} \text { (a) } 4 \mathrm{H}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+2 \mathrm{NaI}(s) \longrightarrow \\ \quad 2 \mathrm{Na}^{+}(a q)+\mathrm{I}_{2}(s)+\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(I) \\ \text { (b) } \mathrm{BaF}_{2}(s)+2 \mathrm{H}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{HF}(a q)+\mathrm{BaSO}_{4}(s) \end{array} $$

Sulfuric acid can act as an oxidizing agent in chemical reactions. As an oxidizing agent, it causes other substances to lose electrons while it itself is reduced. In general terms, an oxidizing agent accepts electrons from another substance. This process results in the oxidizing agent undergoing a reduction in its oxidation state.
In the given equation:
\[4 H^{+} + SO_4^{2-} + 2 NaI \rightarrow 2 Na^{+} + I_2 + SO_2 + 2 H_2O\]
The iodide ions (\text{I}^-) from the \text{NaI} are oxidized to iodine (\text{I}_2). Here, sulfuric acid (\text{H_2SO_4}) facilitates the oxidation by causing the electron loss. The sulfur in \text{SO_4^{2-}} transforms from an oxidation state of +6 to +4 in \text{SO_2}, indicating a gain of electrons by sulfuric acid, thereby reducing it. This confirms sulfuric acid's role as an oxidizing agent in this reaction.

Sulfuric acid (\text{H_2SO_4}) exhibits typical acid behavior by donating hydrogen ions (\text{H^+}) in a reaction. Acids are substances that can donate protons or \text{H^+} ions to other substances. The donation of \text{H^+} ions can lead to the formation of various products depending on the reaction partners.
In the second equation:
\[BaF_2 + 2 H^{+} + SO_4^{2-} \rightarrow 2 HF + BaSO_4\]
Sulfuric acid donates \text{H^+} ions, which react with \text{BaF_2} to form hydrofluoric acid (\text{HF}). The presence of \text{SO_4^{2-}} in the equation indicates sulfuric acid's influence. This donation of \text{H^+} ions by sulfuric acid characterizes its acid behavior in the given reaction.

A redox reaction involves the transfer of electrons between substances. The term 'redox' is derived from the words 'reduction' and 'oxidation'. Reduction refers to the gain of electrons, whereas oxidation is the loss of electrons. During a redox reaction, one substance is oxidized (loses electrons) and another is reduced (gains electrons).
In the equation:
\[4 H^{+} + SO_4^{2-} + 2 NaI \rightarrow 2 Na^{+} + I_2 + SO_2 + 2 H_2O\]
This is a redox reaction because iodide ions (\text{I}^-) are oxidized to iodine (\text{I_2}), and the sulfur in \text{SO_4^{2-}} is reduced from an oxidation state of +6 to +4 in \text{SO_2}. The transfer of electrons between iodide ions and sulfuric acid establishes this equation as a redox reaction.

Understanding oxidation states helps to identify how electrons are transferred in a reaction. The oxidation state (or oxidation number) of an element in a compound indicates the number of electrons lost or gained by an atom.
For example, in the given reaction:
\[4 H^{+} + SO_4^{2-} + 2 NaI \rightarrow 2 Na^{+} + I_2 + SO_2 + 2 H_2O\]
We see changes in oxidation states for both iodine and sulfur. Iodine in \text{NaI} has an oxidation state of -1, and in \text{I_2}, it is 0. This shows that it has lost electrons and been oxidized. For sulfur in \text{SO_4^{2-}}, the oxidation state is +6, and in \text{SO_2}, it is +4, showing a gain of electrons and reduction.
Analyzing these changes in oxidation states helps confirm the roles of substances as oxidizing agents or reducing agents in a reaction.