Problem 16 - Chemistry and Chemical Reactivity

Question

In the reaction $$\begin{aligned} \operatorname{CuFeS}_{2}(s)+3 \operatorname{CuCl}_{2}(\text { aq }) & \longrightarrow \operatorname{CuCl}(s)+\operatorname{FeCl}_{2}(a q)+2 S(s) \end{aligned}$$ what element is oxidized and what element is reduced? (a) Sulfur is reduced, iron is oxidized. (b) Copper is reduced, sulfur is oxidized. (c) Sulfur is reduced, copper is oxidized. (d) This is not a redox reaction.

Step-by-Step Solution

Answer

Copper is reduced, sulfur is oxidized (b).

Step 1: Identify Oxidation Numbers in Reactants

In the compound \( ext{CuFeS}_2\), copper (Cu) has an oxidation state of +1 (as it is paired with a more electronegative element), iron (Fe) has an oxidation state of +2, and sulfur (S) has a combined oxidation state of -2 because there are two sulfur atoms (each sulfur is -1). In \( ext{CuCl}_2\), copper (Cu) has an oxidation state of +2 and chlorine (Cl) has an oxidation state of -1.

Step 2: Identify Oxidation Numbers in Products

In the products, \( ext{CuCl}\) copper has an oxidation state of +1 (balancing one Cl, which is -1), and \( ext{FeCl}_2\) iron has an oxidation state of +2. The elemental sulfur \(S\) is in the elemental state, giving it an oxidation state of 0.

Step 3: Determine Changes in Oxidation Numbers

From the reactant \( ext{CuFeS}_2\) to \( ext{CuCl}\), copper changes from +1 to +1, so no change occurs here. However, sulfur in \( ext{CuFeS}_2\) is -1 and becomes 0 in the products, indicating oxidation. Iron remains at +2. Copper in \( ext{CuCl}_2\) is +2 and remains +1 in \( ext{FeCl}_2\); thus, copper undergoes a reduction.

Step 4: Identify Oxidized and Reduced Elements

Since sulfur's oxidation number increases from -1 in \( ext{CuFeS}_2\) to 0 in sulfur, sulfur is oxidized. Copper's oxidation number decreases from +2 in \( ext{CuCl}_2\) to +1 in \( ext{CuCl}\), so copper is reduced.

Key Concepts

Oxidation Numbers

Understanding oxidation numbers is crucial in identifying redox reactions. These numbers help us determine how electrons are transferred between atoms in a chemical reaction. Oxidation numbers, or oxidation states, represent the charge an atom would have if electrons were transferred completely rather than shared.

• In pure elements, the oxidation number is always zero. For example, in elemental sulfur \(S\), the oxidation state is 0.
• For simple ions, the oxidation number equals the charge of the ion. In the chloride ion \(Cl^-\), it has an oxidation state of -1.
• In compounds, certain elements have common oxidation states, such as oxygen usually being -2 and hydrogen usually being +1.

To illustrate, when we look at the compound \(CuFeS_2\), the copper (Cu) has an oxidation state of +1, while the iron (Fe) is +2, and sulfur (S) in total is -2, distributed over the two sulfur atoms. Recognizing these states helps us analyze how they change during the reaction, leading us to discover the processes of oxidation or reduction.

Oxidation and Reduction

Oxidation and reduction are processes that occur simultaneously in a redox reaction. They involve the transfer of electrons, where one element loses electrons (oxidation) and another gains them (reduction). To easily remember this, consider the mnemonic: **OIL RIG** (Oxidation Is Loss, Reduction Is Gain).In the given reaction, we can apply this concept as follows:

• Oxidation occurs when there is an increase in the oxidation number as electrons are lost. In this reaction, sulfur undergoes oxidation. Its oxidation number increases from -1 in \(CuFeS_2\) to 0 in elemental sulfur \(S\), indicating a loss of electrons.
• Reduction is identified by a decrease in the oxidation number, which happens as electrons are gained. Copper is reduced as its oxidation number decreases from +2 in \(CuCl_2\) to +1 in \(CuCl\).

Recognizing these changes ensures we accurately identify which elements undergo oxidation and reduction in any reaction, confirming our grasp of redox processes.

Copper and Sulfur Reactions

The reaction involving copper and sulfur demonstrates a fascinating redox process. When \(CuFeS_2\) reacts with \(CuCl_2\), changes in oxidation states reveal which elements are oxidized or reduced.

• The copper present in \(CuFeS_2\) does not change its oxidation state, but the copper from \(CuCl_2\) is reduced from +2 to +1, indicating that electrons are gained.
• In contrast, sulfur in \(CuFeS_2\) transforms into elemental sulfur, where its oxidation state increases from -1 to 0, showing it is oxidized and thus loses electrons.
• This indicates that there is both electron transfer and conversion of compounds, highlighting the roles of copper and sulfur within this reaction.

Reactions like this one not only deepen our understanding of elemental behavior but also underscore principles of chemistry that apply to a broad array of chemical reactions. The dynamics between copper and sulfur provide clear examples of how oxidation and reduction can occur simultaneously in the formation of new products and elements.