Problem 51 - Chemistry: The Central Science

Question

Which element is oxidized and which is reduced in the following reactions? (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow\) (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) (d) \(\mathrm{PbS}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\)

Step-by-Step Solution

Answer

In the given reactions: (a) Oxidized: Hydrogen; Reduced: Nitrogen (b) Oxidized: Aluminum; Reduced: Iron (c) Oxidized: Iodine; Reduced: Chlorine (d) Oxidized: Sulfur; Reduced: Oxygen from H2O2

Step 1: (a) N2(g) + 3 H2(g) → 2 NH3(g)

The first step for this reaction is to assign oxidation numbers to each atom. Recall that oxidation numbers represent the charges if the compound was an ionic compound. - Oxidation number of N in N2 = 0 (as it's uncombined) - Oxidation number of H in H2 = 0 (as it's uncombined) - Oxidation number of N in NH3 = -3 (since H is +1 and there are 3, to neutralize the charge N would be -3) - Oxidation number of H in NH3 = +1 Since the oxidation number of N changed from 0 to -3, N has gained electrons and has been reduced. The oxidation number of H changed from 0 to +1, indicating that H has lost electrons and has been oxidized. #a# Oxidized: Hydrogen #a# Reduced: Nitrogen

Step 2: (b) 3 Fe(NO3)2(aq) + 2 Al(s) → 3 Fe(s) + 2 Al(NO3)3(aq)

Similar to part (a), assign oxidation numbers for Fe and Al. - Oxidation number of Fe in Fe(NO3)2 = +2 (since NO3 has -1 charge and there are 2 of them) - Oxidation number of Al in Al = 0 (as it's uncombined) - Oxidation number of Fe in Fe = 0 (as it's uncombined) - Oxidation number of Al in Al(NO3)3 = +3 (since NO3 has -1 charge and there are 3 of them) Since the oxidation number of Fe changed from +2 to 0, Fe has gained electrons and has been reduced. The oxidation number of Al changed from 0 to +3, indicating that Al has lost electrons and has been oxidized. #a# Oxidized: Aluminum #a# Reduced: Iron

Step 3: (c) Cl2(aq) + 2 NaI(aq) → I2(aq) + 2 NaCl(aq)

Assign oxidation numbers for Cl and I. - Oxidation number of Cl in Cl2 = 0 (as it's uncombined) - Oxidation number of I in NaI = -1 (since Na has +1 charge) - Oxidation number of Cl in NaCl = -1 (since Na has +1 charge) - Oxidation number of I in I2 = 0 (as it's uncombined) Since the oxidation number of Cl changed from 0 to -1, Cl has gained electrons and has been reduced. The oxidation number of I changed from -1 to 0, indicating that I has lost electrons and has been oxidized. #a# Oxidized: Iodine #a# Reduced: Chlorine

Step 4: (d) PbS(s) + 4 H2O2(aq) → PbSO4(s) + 4 H2O(l)

There is a typo in the given reaction. The correct reaction should be: PbS(s) + 4 H2O2(aq) → PbSO4(s) + 4 H2O(l) Now, assign oxidation numbers for Pb and S. - Oxidation number of Pb in PbS = +2 (since S has -2 charge) - Oxidation number of S in PbS = -2 (since Pb has +2 charge) - Oxidation number of Pb in PbSO4 = +2 (as Pb is combined with -2 SO4 ion) - Oxidation number of S in SO4 = +6 (+2 for Pb, -2 for each O, 4 O* -2 = -8, +6 for S, as PbSO4 is neutral) Since the oxidation number of Pb didn't change, Pb is neither oxidized nor reduced. The oxidation number of S changed from -2 to +6, indicating that S has lost electrons and has been oxidized. Note that oxygen from the H2O2 is the reducing agent in this reaction. #a# Oxidized: Sulfur #a# Reduced: Oxygen from H2O2

Key Concepts

Oxidation Numbers

Oxidation numbers, often known as oxidation states, play a central role in understanding redox reactions—short for reduction-oxidation reactions. They represent the hypothetical charges atoms would have if the bonding were completely ionic. Assigning these numbers can help us track electron transfer between atoms.

For a neutral atom, the oxidation number is zero. For ions, it is the same as their charge. When atoms form bonds, the less electronegative element is assigned positive oxidation numbers, while the more electronegative element receives negative values. In compounds, the sum of all oxidation numbers must equal the compound's overall charge.

Why are oxidation numbers important in redox reactions? They help us detect changes in an element’s oxidation state, allowing us to identify which elements are oxidized or reduced in a chemical reaction, crucial for balancing redox equations. For example, when the oxidation number of an atom increases during a reaction, it indicates a loss of electrons, or oxidation; whereas a decrease signals a gain of electrons, or reduction.

Reducing Agent

In the theater of chemical reactions, a reducing agent is like a generous actor, giving away electrons to others. It’s a substance that donates electrons to another compound in a redox reaction, thereby being oxidized itself. As it reduces the other substance, it loses one or more electrons and increases its own oxidation number.

Diving into our textbook solution, we see that commonly, metals and hydrides act as reducing agents due to their propensity to lose electrons. For instance, in reaction (a) hydrogen, with its oxidation number going from 0 to +1, has lost electrons and is, therefore, the reducing agent. Similarly, in reaction (b), aluminum's oxidation number increases from 0 to +3, making it the reducing agent. These substances play a pivotal role in driving redox reactions as they provide the necessary electrons for the process of reduction to happen elsewhere in the chemical equation.

Oxidizing Agent

Contrasting with reducing agents, an oxidizing agent acts as the electron recipient in a redox reaction. This substance gains electrons and is reduced, which leads to a decrease in its oxidation number. Oxidizing agents are often composed of atoms with high electronegativity or molecules with high oxygen content.

In our exercise solutions, the oxidizing agents are those elements or compounds that have undergone a reduction in their oxidation number. For reaction (c), chlorine reduces its oxidation number from 0 to -1, hence, it serves as the oxidizing agent. It effectively 'oxidizes' the iodine by accepting electrons from it. Oxidizing agents are crucial for the completion of redox reactions as they facilitate the oxidation of the reducing agents, ensuring the overall conservation of charge and mass in the reactions.