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. (a) \(\mathrm{I}_{2} \mathrm{O}_{\mathrm{s}}(s)+5 \mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow 3 \mathrm{~S}(s)+\) \(2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

In the given redox reactions: (a) Iodine has a change in oxidation number of -5 and Carbon has a change of +2. (b) Mercury has a change in oxidation number of -2 and Nitrogen has a change of +2. (c) Sulfur has a change in oxidation number of +2 and Nitrogen has a change of -3.

Step 1: Determine the initial oxidation numbers

We will first determine the oxidation numbers of each element before the reaction occurs: (a) \(\mathrm{I}_{2} \mathrm{O}_{\mathrm{s}}(s)+5 \mathrm{CO}(\mathrm{g})\) - Iodine (I) in \(\mathrm{I}_{2} \mathrm{O}_{\mathrm{s}}\) has an oxidation number of +5 - Oxygen (O) in \(\mathrm{I}_{2} \mathrm{O}_{\mathrm{s}}\) has an oxidation number of -2 - Carbon (C) in CO has an oxidation number of +2 - Oxygen (O) in CO has an oxidation number of -2 (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q)\) - Mercury (Hg) in \(\mathrm{Hg}^{2+}\) has an oxidation number of +2 - Nitrogen (N) in \(\mathrm{N}_{2} \mathrm{H}_{4}\) has an oxidation number of -2 - Hydrogen (H) in \(\mathrm{N}_{2} \mathrm{H}_{4}\) has an oxidation number of +1 (c) \(3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)\) - Hydrogen (H) in \(\mathrm{H}_{2} \mathrm{S}\) has an oxidation number of +1 - Sulfur (S) in \(\mathrm{H}_{2} \mathrm{S}\) has an oxidation number of -2 - Hydrogen (H) in \(\mathrm{H}^{+}\) has an oxidation number of +1 - Nitrogen (N) in \(\mathrm{NO}_{3}^{-}\) has an oxidation number of +5 - Oxygen (O) in \(\mathrm{NO}_{3}^{-}\) has an oxidation number of -2

Step 2: Determine the final oxidation numbers

Now we will determine the oxidation numbers of each element after the reaction occurs: (a) \(\mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(\mathrm{~g})\) - Iodine (I) in \(\mathrm{I}_{2}\) has an oxidation number of 0 - Carbon (C) in \(\mathrm{CO}_{2}\) has an oxidation number of +4 - Oxygen (O) in \(\mathrm{CO}_{2}\) has an oxidation number of -2 (b) \(2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) - Mercury (Hg) in \(\mathrm{Hg}\) has an oxidation number of 0 - Nitrogen (N) in \(\mathrm{N}_{2}\) has an oxidation number of 0 - Hydrogen (H) in \(\mathrm{H}^{+}\) has an oxidation number of +1 (c) \(3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) - Sulfur (S) in S has an oxidation number of 0 - Nitrogen (N) in \(\mathrm{NO}\) has an oxidation number of +2 - Oxygen (O) in \(\mathrm{NO}\) has an oxidation number of -2 - Hydrogen (H) in \(\mathrm{H}_{2} \mathrm{O}\) has an oxidation number of +1 - Oxygen (O) in \(\mathrm{H}_{2} \mathrm{O}\) has an oxidation number of -2

Step 3: Calculate the change in oxidation numbers

Now we will calculate the change in oxidation numbers for each element involved in the redox reactions: (a) - Iodine (I) changes from +5 to 0, so the change is -5 - Carbon (C) changes from +2 to +4, so the change is +2 (b) - Mercury (Hg) changes from +2 to 0, so the change is -2 - Nitrogen (N) changes from -2 to 0, so the change is +2 (c) - Sulfur (S) changes from -2 to 0, so the change is +2 - Nitrogen (N) changes from +5 to +2, so the change is -3

Step 4: Results

From the calculations, we can conclude that: (a) In the redox reaction, Iodine has a change in oxidation number of -5 and Carbon has a change of +2. (b) In the redox reaction, Mercury has a change in oxidation number of -2 and Nitrogen has a change of +2. (c) In the redox reaction, Sulfur has a change in oxidation number of +2 and Nitrogen has a change of -3.

Key Concepts

Oxidation Numbers

Oxidation numbers are like a bookkeeping tool for chemists to keep track of electron transfers in chemical reactions. They help us understand which elements gain or lose electrons. Each element in a compound has an oxidation number, representing its charge or partial charge. A change in an element’s oxidation number during a reaction indicates a redox reaction.

In the provided reactions:

• Iodine in different compounds goes from +5 to 0, indicating it gains electrons (reduction).
• Carbon goes from +2 to +4, indicating it loses electrons (oxidation).
• Mercury transitions from +2 to 0, suggesting it gains electrons as well.
• Nitrogen increases from -2 to 0, representing a loss of electrons.
• In the sulfur cases, sulfur goes from -2 to 0, meaning electron loss.
• Nitrogen in a nitrate moves from +5 to +2, indicating a gain.

Tracking these numbers is essential to identify oxidation-reduction (redox) reactions and to understand the electron movements taking place.

Redox Reactions

Redox reactions are central to chemistry, involving the transfer of electrons between molecules. A mnemonic to remember is: "LEO the lion says GER" where LEO stands for Loss of Electrons is Oxidation, and GER means Gain of Electrons is Reduction.

Every redox reaction is made up of two half-reactions:

• Oxidation: The process where an element loses electrons, increasing its oxidation state.
• Reduction: The process where an element gains electrons, reducing its oxidation state.

In these reactions, the substance that gets oxidized loses electrons (acting as a reducing agent) while the substance reduced gains electrons (acting as an oxidizing agent).

Understanding these roles is crucial. For instance, in one reaction, iodine is reduced from +5 to 0, while carbon increases from +2 to +4, hence getting oxidized.

Chemical Equations

Chemical equations represent what happens in a chemical reaction. They depict reactants turning into products, balanced to show that matter is conserved. It’s like a recipe telling what you end with from what you start.

In oxidation-reduction reactions, the equations balance not only atoms but also charges. This ensures the overall charge stays the same before and after the reaction.

• Each side of the equation must have equal numbers of atoms for each element involved.
• The total charge must be the same on both sides of the equation.

For the original exercise:

• In reaction (a), five moles of carbon monoxide react to balance the carbon numbers.
• In reaction (b), mercury changes from aqueous ions to liquid form.
• In reaction (c), the balance of hydrogen and sulfur is maintained across reactants and products.

This balance in chemical reactions is crucial to accurately represent the conservation of mass and charge.