Problem 90 - Chemistry: An Atoms-Focused Approach

Question

Determine the oxidation numbers of each of the elements in the following reactions, and identify which of them, if any, are oxidized or reduced: a. \(\mathrm{SiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{4} \mathrm{SiO}_{4}(a q)\) b. \(2 \mathrm{MnCO}_{3}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MnO}_{2}(s)+2 \mathrm{CO}_{2}(g)\) c. \(3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g)+2 \mathrm{H}^{+}(a q)\)

Step-by-Step Solution

Answer

a. SiO2 + 2H2O -> H4SiO4 b. 2MnCO3 + O2 -> 2MnO2 + 2CO2 c. 3NO2 + H2O -> 2NO3^(-) + NO + 2H^(+) Answer: a. No oxidation or reduction. b. Mn is oxidized; O2 is reduced. c. N in NO2 is both oxidized (to N in NO3^(-)) and reduced (to N in NO).

Step 1: Determine the oxidation numbers of each element in the reactants and products

In order to determine the oxidation numbers of each element, we can consider the following general rules: 1. The oxidation number of any uncombined element is 0. 2. The oxidation number of hydrogen is +1 in all its compounds, except for hydrides, where it is -1. 3. The oxidation number of oxygen is -2 in most of its compounds, except for peroxides (-1) and O in F_2O (+2). 4. The sum of the oxidation numbers of all elements in a compound is equal to zero or the charge of the ion. Now, apply these rules to find the oxidation numbers for each element in each reaction: a. SiO2 + 2H2O -> H4SiO4 - Si in SiO2: +4 (because 4+2*(-2)=0) - H in H2O: +1 - O in H2O and SiO2: -2 (follows rule 3) - H in H4SiO4: +1 - O in H4SiO4: -2 - Si in H4SiO4: +4 (because 4+4*(-2)+4*1=0) b. 2MnCO3 + O2 -> 2MnO2 + 2CO2 - Mn in MnCO3: +2 (because 2+(-2)+3=0) - C in MnCO3: +4 - O in MnCO3 and CO2: -2 - Mn in MnO2: +4 (because 4+2*(-2)=0) - C in CO2: +4 (because 4+2*(-2)=0) c. 3NO2 + H2O -> 2NO3^(-) + NO + 2H^(+) - N in NO2: +4 (because 4+2*(-2)=0) - O in NO2 and NO3^(-): -2 - H in H2O and H^+(aq): +1 - O in H2O: -2 - N in NO3^(-): +5 (because 5+3*(-2)=-1) - N in NO: +2 (because 2+1*(-2)=0)

Step 2: Identify changes in oxidation numbers and classify them as oxidation or reduction

We will now compare the oxidation numbers of each element in the reactants and products to see any changes: a. SiO2 + 2H2O -> H4SiO4 - There are no changes in oxidation numbers, so no oxidation or reduction occurs. b. 2MnCO3 + O2 -> 2MnO2 + 2CO2 - Mn: +2 to +4, indicating oxidation - O: 0 to -2 (O2 to MnO2), indicating reduction - C: no change in oxidation numbers for carbon c. 3NO2 + H2O -> 2NO3^(-) + NO + 2H^(+) - N: +4 to +5 (NO2 to NO3^(-)), indicating oxidation - N: +4 to +2 (NO2 to NO), indicating reduction - O and H: no change in oxidation numbers Now we can conclude which elements are oxidized and reduced in each reaction: a. SiO2 + 2H2O -> H4SiO4: No oxidation or reduction. b. 2MnCO3 + O2 -> 2MnO2 + 2CO2: Mn is oxidized; O2 is reduced. c. 3NO2 + H2O -> 2NO3^(-) + NO + 2H^(+): N in NO2 is both oxidized (to N in NO3^(-)) and reduced (to N in NO). The other elements have no change in oxidation state.

Key Concepts

Oxidation Numbers

Understanding oxidation numbers is crucial in identifying oxidation-reduction (redox) reactions.
They are essentially charges assigned to atoms in a molecule, which help determine how electrons are distributed.

Here are some fundamental rules for determining oxidation numbers:

• The oxidation number of any free element is 0.
• For a simple ion, it's equal to the charge on the ion (e.g., Na⁺ has an oxidation number of +1).
• Oxygen is generally assigned an oxidation number of -2, except in peroxides where it is -1.
• Hydrogen has an oxidation number of +1, except when bonded to metals in hydrides, where it is -1.
• The sum of oxidation numbers in a neutral compound must be zero, while in a polyatomic ion, it should equal the ion's charge.

By applying these rules, we can systematically assign an oxidation number to each atom throughout a chemical reaction, which will guide us in identifying any redox activity.

Redox Reactions

Redox reactions are chemical processes involving electron transfer.
In such reactions, one species loses electrons (oxidation), and another gains electrons (reduction).

To identify redox reactions, follow these steps:

• Determine oxidation numbers for each element in the reactants and products.
• Compare oxidation numbers to see which elements undergo changes.
• An increase in oxidation number represents oxidation, whereas a decrease indicates reduction.

In our example reactions: - For reaction (b), manganese (Mn) is oxidized from +2 to +4, while oxygen is reduced from 0 to -2, indicating a clear redox process. - In reaction (c), nitrogen (N) undergoes both oxidation and reduction, as it changes from +4 in NO₂ to different states in NO₃⁻ and NO. Understanding these changes allows us to pinpoint which elements are oxidized or reduced, a key aspect of analyzing redox reactions.

Chemical Reactions Analysis

Analyzing chemical reactions often begins with calculating oxidation numbers and then evaluating changes across reactants and products.
This detailed breakdown helps us pinpoint redox reactions and understand the chemical processes happening at a molecular level.

With a thorough analysis:

• We see that not all reactions involve redox changes; for instance, reaction (a) has no change in oxidation state, thus no redox activity.
• In contrast, reaction (b) involves manganese oxidation and oxygen reduction, a typical redox pair.
• Reaction (c) showcases both oxidation and reduction in nitrogen, demonstrating the intricate nature of some chemical processes.

This kind of chemical reaction analysis is fundamental not only for academia but also for practical applications like battery design and corrosion prevention. Gaining proficiency in analyzing these reactions empowers students to understand and manipulate chemical reactions effectively.