Question

For the reaction \(16 \mathrm{Fe}(s)+3 \mathrm{S}_{8}(s) \rightarrow 8 \mathrm{Fe}_{2} \mathrm{S}_{3}(s),\) show how electrons are gained and lost by the atoms.

Short answer

In the reaction \( 16 \mathrm{Fe}(s)+3 \mathrm{S}_{8}(s) \rightarrow 8 \mathrm{Fe}_{2} \mathrm{S}_{3}(s) \), iron (Fe) atoms are oxidized, losing electrons and increasing their oxidation state from 0 to +3. Sulfur (S) atoms are reduced, gaining electrons and decreasing their oxidation state from 0 to -2.

Step-by-step solution

Write the given reaction

Write down the given reaction: \( 16 \mathrm{Fe}(s)+3 \mathrm{S}_{8}(s) \rightarrow 8 \mathrm{Fe}_{2} \mathrm{S}_{3}(s) \)

Determine the initial oxidation states of the elements

First, let's determine the oxidation state of each element before the reaction. In their elemental form, the oxidation state is always 0. For iron (Fe), the oxidation state is 0. For sulfur (S), the oxidation state is 0.

Determine the final oxidation states of the elements in the product

Next, let's determine the oxidation state of each element in the product, iron(III) sulfide (Fe₂S₃). In Fe₂S₃, the total charge must be 0, since it is a neutral compound. For iron (Fe), we can assign the oxidation state as +3, since it is iron(III) sulfide. For sulfur (S), since there are 3 sulfur atoms and 2 iron atoms with a total oxidation state of +6, the oxidation state of sulfur must be -2 to balance the charge (3 * -2 = -6).

Identify the atoms that gained and lost electrons

Now we can compare the initial and final oxidation states to determine which atoms gained and lost electrons. Iron (Fe): Initial oxidation state is 0, final oxidation state is +3. Sulfur (S): Initial oxidation state is 0, final oxidation state is -2. Since the oxidation state of iron increased from 0 to +3, it lost electrons, meaning it was oxidized. Since the oxidation state of sulfur decreased from 0 to -2, it gained electrons, meaning it was reduced.

Summarize the results

In the reaction: \( 16 \mathrm{Fe}(s)+3 \mathrm{S}_{8}(s) \rightarrow 8 \mathrm{Fe}_{2} \mathrm{S}_{3}(s) \) Iron (Fe) atoms lose electrons, which means they are oxidized. Their oxidation state increases from 0 to +3. Sulfur (S) atoms gain electrons, which means they are reduced. Their oxidation state decreases from 0 to -2.

Key concepts

Oxidation States

When dealing with redox reactions, it's crucial to understand oxidation states. These states help us track how electrons move during a chemical reaction. In a redox reaction, one element is oxidized, meaning it loses electrons, while another is reduced, gaining electrons.
In the given reaction of iron (Fe) and sulfur (S), we start by assigning oxidation states. For any element in its natural state, like Fe and S before reacting, their oxidation state is zero because they are not combined with other elements.
When Fe and S combine to form iron(III) sulfide ( Fe₂S₃ ), we readjust the oxidation states. Here, iron now has an oxidation state of +3, and sulfur has an oxidation state of -2. These changes indicate that iron loses electrons, while sulfur gains them, setting the stage for electron transfer.

Electron Transfer

Electron transfer is the heart of any redox reaction. This process involves electrons moving from one element to another, transforming reactants into products. In our example, iron is involved in oxidizing as it loses electrons, while sulfur is reduced by gaining electrons.
Iron, with its initial oxidation state of 0, gives away electrons to achieve an oxidation state of +3 in the product. This loss means each Fe atom loses three electrons. Similarly, sulfur starts at an oxidation state of 0 and gains electrons to become -2. This gain implies each sulfur atom accepts two electrons.

• Iron loses electrons, becoming oxidized
• Sulfur gains electrons, becoming reduced

Tracking these shifts in electrons is essential to understanding how substances undergo chemical changes.

Iron(III) Sulfide

Iron(III) sulfide, represented as Fe₂S₃, is a product of the reaction between iron and sulfur. It is a compound formed when iron and sulfur atoms reorganize through electron exchanges.
In Fe₂S₃, for every two iron atoms, there are three sulfur atoms. The "III" in iron(III) denotes the charge on the iron, confirming its oxidation state of +3. This notation helps chemists understand the nature of the compound, as it indicates how the compound is formed through oxidation and reduction processes.
The formation of iron(III) sulfide is a classic example of a redox reaction showing how metals like iron can combine with non-metals, such as sulfur, via electron transfers, resulting in stable, new chemical compounds.

Chemical Equations

Chemical equations display what happens during a reaction. They show the reactants and products, along with their respective amounts and states. In our scenario, the chemical equation is:
\[ 16 \, \text{Fe}(s) + 3 \, \text{S}_{8}(s) \rightarrow 8 \, \text{Fe}_{2}\text{S}_{3}(s) \]
This equation breaks down the large-scale process into simpler parts. It demonstrates how 16 iron atoms and three sulfur molecules (each consisting of 8 atoms) turn into 8 units of iron(III) sulfide.

• Shows reactants and products
• Indicates the conservation of mass
• Gives insight into electron transfer

Balancing chemical equations is vital, ensuring that all atoms present in reactants appear in equal numbers in the products, maintaining the law of conservation of mass.