Question

In each of the following reactions, identify which element is oxidized and which is reduced by assigning oxidation states. a. \(2 \mathrm{Cu}(s)+\mathrm{S}(s) \rightarrow \mathrm{Cu}_{2} \mathrm{S}(s)\) b. \(2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CuO}(s)\) c. \(4 \mathrm{B}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{B}_{2} \mathrm{O}_{3}(s)\) d. \(6 \mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2 \mathrm{Na}_{3} \mathrm{N}(s)\)

Short answer

a. Cu is oxidized (\(0 \rightarrow -1\)); S is reduced (\(0 \rightarrow +1\)) b. Cu in \(\mathrm{Cu}_{2} \mathrm{O}\) is oxidized (\(+1 \rightarrow +2\)); O in \(\mathrm{O}_{2}\) is reduced (\(0 \rightarrow -2\)) c. B is oxidized (\(0 \rightarrow +3\)); O in \(\mathrm{O}_{2}\) is reduced (\(0 \rightarrow -2\)) d. Na is oxidized (\(0 \rightarrow +1\)); N in \(\mathrm{N}_{2}\) is reduced (\(0 \rightarrow -3\))

Step-by-step solution

a. Identifying the oxidized and reduced elements in \({2 \mathrm{Cu}(s)+\mathrm{S}(s) \rightarrow \mathrm{Cu}_{2} \mathrm{S}(s)}\).

First, we assign oxidation states to each element in the reactants and products: \(\mathrm{Cu}\):0, \(\mathrm{S}\):0, \(\mathrm{Cu}\) in \(\mathrm{Cu}_{2} \mathrm{S}\):-1, \(\mathrm{S}\) in \(\mathrm{Cu}_{2} \mathrm{S}\):+1. Comparing the values, we see that Cu has been oxidized (loses 1 electron) and S has been reduced (gains 1 electron).

b. Identifying the oxidized and reduced elements in \({2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CuO}(s)}\).

Assigning oxidation states: \(\mathrm{Cu}\) in \(\mathrm{Cu}_{2} \mathrm{O}\):+1, \(\mathrm{O}\) in \(\mathrm{Cu}_{2} \mathrm{O}\):-1, \(\mathrm{O}\) in \(\mathrm{O}_{2}\):0, \(\mathrm{Cu}\) in \(\mathrm{CuO}\):+2, \(\mathrm{O}\) in \(\mathrm{CuO}\):-2. Cu in \(\mathrm{Cu}_{2} \mathrm{O}\) has been oxidized (loses 1 electron) and O in \(\mathrm{O}_{2}\) has been reduced (gains 2 electrons).

c. Identifying the oxidized and reduced elements in \({4 \mathrm{B}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{B}_{2} \mathrm{O}_{3}(s)}\).

Assigning oxidation states: \(\mathrm{B}\):0, \(\mathrm{O}\) in \(\mathrm{O}_{2}\):0, \(\mathrm{B}\) in \(\mathrm{B}_{2} \mathrm{O}_{3}\):+3, \(\mathrm{O}\) in \(\mathrm{B}_{2} \mathrm{O}_{3}\):-2. B has been oxidized (loses 3 electrons) and O in \(\mathrm{O}_{2}\) has been reduced (gains 2 electrons).

d. Identifying the oxidized and reduced elements in \({6 \mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2 \mathrm{Na}_{3} \mathrm{N}(s)}\).

Assigning oxidation states: \(\mathrm{Na}\):0, \(\mathrm{N}\) in \(\mathrm{N}_{2}\):0, \(\mathrm{Na}\) in \(\mathrm{Na}_{3} \mathrm{N}\):+1, \(\mathrm{N}\) in \(\mathrm{Na}_{3} \mathrm{N}\):-3. Na has been oxidized (loses 1 electron) and N in \(\mathrm{N}_{2}\) has been reduced (gains 3 electrons).

Key concepts

Oxidation States

Understanding oxidation states is crucial for identifying which elements undergo oxidation and reduction in a chemical reaction. An **oxidation state** (or oxidation number) is a theoretical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. It helps track how many electrons are lost or gained by atoms during a reaction.
Here's how you can determine oxidation states:

• For an element in its standard state, the oxidation state is always zero. For example, in reaction a, both Cu and S start with an oxidation state of 0.
• For simple monatomic ions, the oxidation state is equal to the charge of the ion. For instance, in \( \text{Na}^+ \), the oxidation state is +1.
• Oxygen usually has an oxidation state of -2 and hydrogen +1, with some exceptions.

By assigning oxidation states, you can determine which element is oxidized (oxidation state increases) and which is reduced (oxidation state decreases). This method simplifies the process of identifying electron exchange in reactions.

Oxidation

In a chemical reaction, **oxidation** refers to the process where an element loses electrons. This leads to an increase in the oxidation state of that element. Simply put, if an atom becomes more positive, it means it has undergone oxidation.
Some key points about oxidation:

• When copper transforms from \( \mathrm{Cu} \) in \( \mathrm{Cu}(s) \) to \( \mathrm{Cu}^+ \) in \( \mathrm{Cu}_{2} ext{O} \), it's losing electrons along with increasing its oxidation state.
• Oxidation is not about gaining oxygen exclusively, but it often involves reactions with oxygen, as seen in \( \text{B}(s) + \text{O}_2(g) \rightarrow \text{B}_2\text{O}_3(s) \).

Oxidation can be remembered as a loss of electrons (OIL: Oxidation Is Loss). This definition helps pinpoint which reactants are losing electrons and becoming oxidized.

Reduction

During a chemical process, **reduction** involves the gain of electrons by an element, leading to a decrease in its oxidation state. Thus, the element becomes more negative. Reduction can be seen as the opposite of oxidation.
Important facts about reduction include:

• In the transformation from \( \text{S} \) in \( \text{S}(s) \) to \( \text{S}^{2-} \) in \( \text{Cu}_2\text{S}(s) \), sulfur is gaining electrons, indicating reduction.
• Reduction is not just about losing oxygen; it's about gaining electrons. For example, oxygen is reduced in the formation of \( \text{CuO}(s) \) from \( \text{Cu}_{2}\text{O}(s) \).

Using the mnemonic RIG (Reduction Is Gain) helps remember that reduction is all about the gain of electrons. By understanding reduction, you can identify which elements are gaining electrons in any given reaction.