Problem 17 - Chemistry The Central Science

Question

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. $$ \begin{array}{l} \text { (a) } \mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g) \\ \text { (b) } 2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow \\ \text { (c) } 3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q) \\ \text { (d) } \mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \\ \longrightarrow \mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \end{array} $$

Step-by-Step Solution

Answer

In the given balanced redox equations: (a) Iodine (I) changes its oxidation number from +5 to 0 (change of -5), and Carbon (C) changes its oxidation number from +2 to +4 (change of +2). (b) Mercury (Hg) changes its oxidation number from +2 to 0 (change of -2), and Nitrogen (N) changes its oxidation number from -2 to 0 (change of +2). (c) Sulfur (S) changes its oxidation number from -2 to 0 (change of +2), and Nitrogen (N) changes its oxidation number from +5 to 0 (change of -5). (d) Oxygen (O) in \(\mathrm{H}_{2} \mathrm{O}_{2}\) changes its oxidation number from -1 to 0 (change of +1), while Chlorine (Cl) in \(\mathrm{ClO}_{2}\) does not change its oxidation number.

Step 1: (a) Identifying the elements undergoing oxidation or reduction and calculating the change in oxidation numbers

In the provided equation: \[ \mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g) \] Here, we notice that Iodine (I) and Carbon (C) are the elements changing their oxidation states. In \(\mathrm{I}_{2} \mathrm{O}_{5}\), the oxidation number of Iodine (I) is +5, and in \(\mathrm{I}_{2}\), it is 0. In \(\mathrm{CO}\), the oxidation number of Carbon (C) is +2, and in \(\mathrm{CO}_{2}\), it is +4. Thus, the changes in oxidation numbers are as follows: Iodine (I): +5 to 0 (change of -5) Carbon (C): +2 to +4 (change of +2)

Step 2: (b) Identifying the elements undergoing oxidation or reduction and calculating the change in oxidation numbers

In the provided equation: \[ 2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2\mathrm{Hg}(s) + \mathrm{N}_{2}(g) \] In this case, Mercury (Hg) and Nitrogen (N) are changing their oxidation states. In \(\mathrm{Hg}^{2+}\), the oxidation number of Mercury (Hg) is +2, and in \(\mathrm{Hg}\), it is 0. In \(\mathrm{N}_{2} \mathrm{H}_{4}\), the oxidation number of Nitrogen (N) is -2, and in \(\mathrm{N}_{2}\), it is 0. Thus, the changes in oxidation numbers are as follows: Mercury (Hg): +2 to 0 (change of -2) Nitrogen (N): -2 to 0 (change of +2)

Step 3: (c) Identifying the elements undergoing oxidation or reduction and calculating the change in oxidation numbers

In the provided equation: \[ 3 \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)+3\mathrm{S}(s)+ 6\mathrm{H}_{2}\mathrm{O}(l) \] In this case, Sulfur (S) and Nitrogen (N) are changing their oxidation states. In \(\mathrm{H}_{2}\mathrm{S}\), the oxidation number of Sulfur (S) is -2, and in \(\mathrm{S}\), it is 0. In \(\mathrm{NO}_{3}^{-}\), the oxidation number of Nitrogen (N) is +5, and in \(\mathrm{N}_{2}\), it is 0. Thus, the changes in oxidation numbers are as follows: Sulfur (S): -2 to 0 (change of +2) Nitrogen (N): +5 to 0 (change of -5)

Step 4: (d) Identifying the elements undergoing oxidation or reduction and calculating the change in oxidation numbers

In the provided equation: \[ \mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{ O}(l)+\mathrm{O}_{2}(g) \] In this case, Oxygen (O) in \(\mathrm{H}_{2} \mathrm{O}_{2}\) and Chlorine (Cl) in \(\mathrm{ClO}_{2}\) are changing their oxidation states. In \(\mathrm{H}_{2} \mathrm{O}_{2}\), the oxidation number of Oxygen (O) is -1, and in \(\mathrm{O}_{2}\), it is 0. In \(\mathrm{ClO}_{2}\), the oxidation number of Chlorine (Cl) is +3, and in \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}\), it is +3. Here, Chlorine (Cl) does not change its oxidation state. Thus, the changes in oxidation numbers are as follows: Oxygen (O) in \(\mathrm{H}_{2} \mathrm{O}_{2}\): -1 to 0 (change of +1) Chlorine (Cl) in \(\mathrm{ClO}_{2}\): No change in the oxidation number.

Key Concepts

Oxidation Number Changes

Understanding oxidation number changes is crucial to mastering redox chemistry. Oxidation numbers, also referred to as oxidation states, help us track how electrons are transferred in chemical reactions. In redox reactions, the oxidation number of an element reflects its electron content relative to a pure element, with increases indicating loss of electrons (oxidation) and decreases indicating gain of electrons (reduction).

For example, let's consider iodine in the reaction: \[ \text{I}_{2}\text{O}_{5} + 5CO \rightarrow \text{I}_{2} + 5CO_{2} \]Initially, iodine has an oxidation number of +5 in \text{I}_{2}\text{O}_{5}, but it changes to 0 in \text{I}_{2}, signifying a gain of electrons or a reduction. Meanwhile, carbon starts with an oxidation number of +2 in CO and increases to +4 in \text{CO}_{2}, indicating that it has lost electrons or undergone oxidation. These shifts in oxidation numbers reveal which elements have been oxidized or reduced during the reaction.

Balancing Redox Reactions

Balancing redox reactions requires equalizing the electron transfer between the oxidizing and reducing agents. This balance ensures the conservation of mass and charge. A common method is the half-reaction approach, where oxidation and reduction processes are treated independently and then combined to yield the balanced equation.

For instance, in the reaction between hydrogen peroxide and chlorous acid, we write the half-reactions for the oxidation of \text{OH}^{-} from hydrogen peroxide and the reduction of \text{Cl} in chlorous acid:\[ \text{H}_{2}\text{O}_{2} \rightarrow \text{O}_{2} + 2H^{+} + 2e^{-} \]\[ 2\text{ClO}_{2} + 4e^{-} \rightarrow 2\text{ClO}_{2}^{-} \]After balancing the electrons, the half-reactions are combined to form the complete balanced equation, ensuring that all elements and charges are accounted for. This modular approach simplifies complex redox equations, allowing for a step-by-step solution.

Identifying Oxidizing and Reducing Agents

Identifying the oxidizing and reducing agents in a reaction is important to dissecting its redox nature. The oxidizing agent is the substance that gets reduced by accepting electrons, while the reducing agent is the one that gets oxidized by donating electrons.

In the reaction involving mercury(II) and hydrazine:\[ 2\text{Hg}^{2+} + \text{N}_{2}\text{H}_{4} \rightarrow 2\text{Hg} + \text{N}_{2} \]mercury(II) ions are reduced to elemental mercury and thus act as the oxidizing agent. Conversely, hydrazine is oxidized to nitrogen gas and serves as the reducing agent. Recognizing these roles is essential for understanding the driving forces behind redox processes and predicting how different substances will interact in a chemical reaction.