Consider the reaction: H$$_2$$SO$$_{3(aq)}$$ + Sn$$^{4+}_{(aq)}$$ + H$$_2$$O$$_{(l)}$$ → Sn$$^{2+}_{(aq)}$$ + HSO$$^-_{4(aq)}$$ + 3H$$^+_{(aq)}$$. Which of the following statements is correct?

First, we need to understand the reaction: $$H_{2}SO_{3}(aq) + Sn^{4+(aq) + H2O(l) -> Sn^{2+}(aq) + HSO4^{-}(aq) + 3H^{+}(aq)}$$. To determine which statement is correct, we must identify the oxidizing and reducing agents by analyzing the changes in oxidation numbers.

Recall the rules for oxidation numbers: Hydrogen (H) is usually +1, Oxygen (O) is usually -2, and the sum of oxidation numbers in a neutral compound is zero, while in an ion, it equals the charge. Let's assign oxidation numbers to each element in the reactants and products.

Starting with the reactants:

• In $$H_{2}SO_{3}$$: Hydrogen (H) is +1 (since it's combined with non-metals). Let the oxidation number of Sulfur (S) be $$x$$. Oxygen (O) is -2. So, $$2 \times (+1) + x + 3 \times (-2) = 0$$. Simplifying: $$2 + x - 6 = 0$$ → $$x - 4 = 0$$ → $$x = +4$$. Thus, S in $$H_{2}SO_{3}$$ has an oxidation number of +4.
• In $$Sn^{4+}$$: The ion has a +4 charge, so Sn has an oxidation number of +4.
• In $$H_{2}O$$: Hydrogen is +1, Oxygen is -2.

Now for the products:

• In $$Sn^{2+}$$: The ion has a +2 charge, so Sn has an oxidation number of +2.
• In $$HSO_{4}^{-}$$: Hydrogen (H) is +1. Oxygen (O) is -2. Let the oxidation number of Sulfur (S) be $$y$$. The ion has a -1 charge. So, $$+1 + y + 4 \times (-2) = -1$$. Simplifying: $$1 + y - 8 = -1$$ → $$y - 7 = -1$$ → $$y = +6$$. Thus, S in $$HSO_{4}^{-}$$ has an oxidation number of +6.
• In $$H^{+}$$: Hydrogen has an oxidation number of +1.

Now, compare the oxidation numbers:

• Sulfur (S): From +4 in $$H_{2}SO_{3}$$ to +6 in $$HSO_{4}^{-}$$. This is an increase of +2, indicating oxidation (loss of electrons).
• Tin (Sn): From +4 in $$Sn^{4+}$$ to +2 in $$Sn^{2+}$$. This is a decrease of -2, indicating reduction (gain of electrons).

Since Sulfur in $$H_{2}SO_{3}$$ is oxidized, $$H_{2}SO_{3}$$ is the reducing agent (it causes reduction by being oxidized). Since Tin in $$Sn^{4+}$$ is reduced, $$Sn^{4+}$$ is the oxidizing agent (it causes oxidation by being reduced).

Now, evaluate the options:

• Option A: "Sn^{4+} is the oxidizing agent because it undergoes oxidation" → Incorrect, because Sn^{4+} undergoes reduction, not oxidation.
• Option B: "Sn^{4+} is the reducing agent because it undergoes oxidation" → Incorrect, because Sn^{4+} undergoes reduction, and it is not the reducing agent.
• Option C: "H2SO3 is the reducing agent because it undergoes reduction" → Incorrect, because H2SO3 undergoes oxidation, not reduction.
• Option D: "H2SO3 is the reducing agent because it undergoes oxidation" → Correct, as H2SO3 undergoes oxidation and is therefore the reducing agent.

Hence, the correct answer is Option D.