In which of the following equations does sulfuric acid act as an oxidizing agent? In which does it act as an acid? Explain.

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{4}{\mathit{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathit{S}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{N}\mathit{a}\mathit{I}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{N}{\mathit{a}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathit{S}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathit{B}\mathit{a}{\mathit{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathit{S}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{H}\mathit{F}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathit{B}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \end{gathered}$$

In this reaction sulfuric acid acts as an oxidizing agent.

Oxidation state of Hydrogen attached to non-metal is +1.

Oxidation state of Oxygen attached to non-metal is -2.

Let, oxidation state of sulfur in H₂SO₄ is x.

Now you can write,

$$\begin{gathered} \mathbf{2}\left(+1\right)\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{4}\left(-2\right)\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{=}\mathbf{+}\mathbf{6} \end{gathered}$$

Oxidation state of Oxygen attached to non-metal is -2.

Let, oxidation state of sulfur in SO₂ is y.

No you can write,

$$\begin{gathered} \mathbf{y}\mathbf{+}\mathbf{2}\left(-2\right)\mathbf{=}\mathbf{0} \\ \mathbf{y}\mathbf{=}\mathbf{+}\mathbf{4} \end{gathered}$$

In the given reaction

$\mathbf{4}{\mathbf{H}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{2}\mathbf{NaI}\left(s\right)\mathbf{\to }\mathbf{2}{\mathbf{Na}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{I}}_{\mathbf{2}}\left(s\right)\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)$

The oxidation state of sulfur in H₂SO₄ is +6 which reduces to +4 in SO₂. Therefore, H₂SO₄ undergoes reduction in this given reaction. Hence, it acts as an oxidizing agent.

In this reaction sulfuric acid acts as an acid.

Oxidation state of Hydrogen attached to non-metal is +1.

Oxidation state of Oxygen attached to non-metal is -2.

Let, oxidation state of sulfur in H₂SO₄ is x.

No you can write,

$$\begin{gathered} \mathbf{2}\left(+1\right)\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{4}\left(-2\right)\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{=}\mathbf{+}\mathbf{6} \end{gathered}$$

Oxidation state of Barium is +2.

Oxidation state of Oxygen attached to non-metal is -2.

Let, oxidation state of sulfur in BaSO₄ is y.

No you can write,

$$\begin{gathered} \mathbf{+}\mathbf{2}\mathbf{+}\mathit{y}\mathbf{+}\mathbf{4}\left(-2\right)\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{=}\mathbf{+}\mathbf{6} \end{gathered}$$

In the given reaction

$\mathit{B}\mathit{a}{\mathit{F}}_{\mathbf{2}}\left(s\right)\mathbf{+}\mathbf{2}{\mathit{H}}^{\mathbf{+}}\left(aq\right)\mathbf{+}\mathit{S}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\left(aq\right)\mathbf{\to }\mathbf{2}\mathit{H}\mathit{F}\left(aq\right)\mathbf{+}\mathit{B}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}\left(s\right)$

The oxidation state of sulfur in H₂SO₄ and in BaSO₄ is +6. Therefore, it is not a redox reaction. Hence, it acts as an acid.