For the redox reaction given below, answer the following questions.
$$
2 \mathrm{HNO}_{3}+\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \rightarrow 2 \mathrm{NO}_{2}+\mathrm{S}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}
$$
a. What is the oxidation state of $\mathrm{S}$ in $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ ? $\qquad$
b. What is the oxidation state of $\mathrm{N}$ in $\mathrm{HNO}_{3}$ ? $\qquad$
c. What is the element that is oxidized? $\qquad$
d. What is the oxidizing agent? $\qquad$

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the average oxidation state of sulfur in Na2S2O3**: To find the oxidation state of sulfur in Na2S2O3, we first consider the oxidation states of sodium (Na) and oxygen (O). Sodium, being an alkali metal, has an oxidation state of +1, and oxygen typically has an oxidation state of -2. The formula Na2S2O3 can be broken down into 2 Na⁺ ions and one S2O3²⁻ ion. Since there are two sodium ions, their total oxidation state is +2. The overall charge of the S2O3²⁻ ion must be -2 to balance the charge of the sodium ions. ‖ ‖ 2 ⋮ **Apply the oxidation state rules**: The sum of the oxidation states of all atoms in a neutral compound is zero. For the S2O3²⁻ ion, the sum of the oxidation states of sulfur (which we will call x) and oxygen must equal -2: 2x + 3(-2) = -2. Solving for x gives us the average oxidation state of sulfur in the ion. ‖ ‖ 3 ⋮ **Solve for the oxidation state of sulfur**: From the equation 2x + 3(-2) = -2, we find that 2x - 6 = -2, which simplifies to 2x = 4, and therefore x = 2. This is the average oxidation state of the two sulfur atoms in the S2O3²⁻ ion. ‖ ∻a Answer∻ ⚹ +2 (average oxidation state of sulfur in Na2S2O3) ⚹ ∻Key Concept∻ ⚹ The oxidation state of an element in a compound can be calculated by using the known oxidation states of other elements and the overall charge of the compound or ion. ⚹ ∻Explanation∻ ⚹ The average oxidation state of sulfur in Na2S2O3 is +2, which is found by balancing the oxidation states of all elements in the S2O3²⁻ ion and ensuring the total charge is -2. ⚹ ∻Solution∻ ‖ 1 ⋮ **Determine the oxidation state of nitrogen in HNO3**: Nitrogen's oxidation state in HNO3 can be found by considering the known oxidation states of hydrogen (H) and oxygen (O). Hydrogen has an oxidation state of +1, and oxygen has an oxidation state of -2. ‖ ‖ 2 ⋮ **Apply the oxidation state rules**: In a neutral molecule like HNO3, the sum of the oxidation states must equal zero. The formula HNO3 can be broken down into 1 H⁺ ion and one NO3⁻ ion. Since the hydrogen ion has an oxidation state of +1, and there are three oxygen atoms each with an oxidation state of -2, the sum of the oxidation states is 1 + x + 3(-2) = 0, where x is the oxidation state of nitrogen. ‖ ‖ 3 ⋮ **Solve for the oxidation state of nitrogen**: From the equation 1 + x - 6 = 0, we find that x - 5 = 0, which simplifies to x = +5. This is the oxidation state of nitrogen in HNO3. ‖ ∻b Answer∻ ⚹ +5 (oxidation state of nitrogen in HNO3) ⚹ ∻Key Concept∻ ⚹ The oxidation state of an element in a compound can be determined by using the known oxidation states of other elements in the compound and the requirement that the sum of the oxidation states equals the net charge of the compound. ⚹ ∻Explanation∻ ⚹ The oxidation state of nitrogen in HNO3 is +5, which is found by balancing the oxidation states of hydrogen and oxygen and ensuring the total is zero for a neutral molecule. ⚹ ∻Solution∻ ‖ 1 ⋮ **Identify the element that is oxidized**: To determine which element is oxidized, we compare the oxidation states of all elements in the reactants and products. An element is oxidized if its oxidation state increases during the reaction. ‖ ‖ 2 ⋮ **Compare oxidation states**: In the reactant Na2S2O3, we previously determined the average oxidation state of sulfur to be +2. In the products, sulfur appears as elemental S with an oxidation state of 0 and as part of Na2SO4, where it has an oxidation state of +6. Since the oxidation state of sulfur increases from +2 to 0 and +6, sulfur is the element that is oxidized. ‖ ∻c Answer∻ ⚹ Sulfur (S) ⚹ ∻Key Concept∻ ⚹ Oxidation involves an increase in the oxidation state of an element during a chemical reaction. ⚹ ∻Explanation∻ ⚹ Sulfur is oxidized because its oxidation state increases from +2 in Na2S2O3 to 0 and +6 in the products. ⚹ ∻Solution∻ ‖ 1 ⋮ **Identify the oxidizing agent**: The oxidizing agent is the substance that gains electrons and is reduced in the process. To find the oxidizing agent, we look for the element that undergoes a decrease in oxidation state during the reaction. ‖ ‖ 2 ⋮ **Compare oxidation states**: In the reactant HNO3, we previously determined the oxidation state of nitrogen to be +5. In the product NO2, the oxidation state of nitrogen is +4. Since the oxidation state of nitrogen decreases from +5 to +4, nitrogen is reduced, and HNO3, which contains the nitrogen, is the oxidizing agent. ‖ ∻d Answer∻ ⚹ HNO3 (nitric acid) ⚹ ∻Key Concept∻ ⚹ The oxidizing agent is the substance that is reduced (decrease in oxidation state) and gains electrons in a redox reaction. ⚹ ∻Explanation∻ ⚹ HNO3 is the oxidizing agent because it contains nitrogen, which is reduced from an oxidation state of +5 to +4 during the reaction.◊What is the role of the oxidizing agent in a redox reaction?⍭ Generate me a similar question◊⚹

Sia