Problem 22 - Chemistry: The Central Science

Question

The element tin is generally recovered from deposits of the ore cassiterite \(\left(\mathrm{SnO}_{2}\right)\). The oxide is reduced with carbon, and the crude metal is purified by electrolysis. Write balanced chemical equations for the reduction process and for the electrode reactions in the electrolysis. (Assume that an acidic solution of \(\mathrm{SnSO}_{4}\) is employed as an electrolyte in the electrolysis.)

Step-by-Step Solution

Answer

In summary, the balanced chemical equations are: 1. Reduction of tin oxide with carbon: \(SnO_2 + C \rightarrow Sn + CO_2\) 2. Electrode reactions in the electrolysis of tin sulfate: - Anode (oxidation): \(Sn \rightarrow Sn^{2+} + 2e^-\) - Cathode (reduction): \(2H^+ + 2e^- \rightarrow H_2\) - Overall balanced equation: \(Sn + 2H^+ \rightarrow Sn^{2+} + H_2\)

Step 1: Write the unbalanced chemical equation for the reduction of tin oxide with carbon

The given information tells us that the reduction of tin oxide (SnO2) occurs using carbon (C). We can write the general (unbalanced) equation for this reaction: SnO2 + C → Sn + CO2

Step 2: Balance the chemical equation for the reduction of tin oxide with carbon

To balance the chemical equation, we must ensure that the number of atoms of each element is equal on both sides of the equation. In this case, there are already 1 Sn atom, 2 O atoms, and 1 C atom on each side. Therefore, the equation is already balanced: SnO2 + C → Sn + CO2

Step 3: Write the unbalanced half-reactions for the electrolysis of tin sulfate

We are given that an acidic solution of tin sulfate (SnSO4) is used as the electrolyte for the electrolysis. In an electrolytic cell, oxidation occurs at the anode, and reduction occurs at the cathode. We can write the unbalanced half-reactions for each electrode: Anode (oxidation): Sn → Sn^{2+} + 2e^- Cathode (reduction): H^+ + e^- → 1/2 H2

Step 4: Balance the half-reactions for the electrolysis of tin sulfate (SnSO4)

The half-reactions are already balanced with respect to the number of Sn and H atoms. However, we need to balance the overall charge in each reaction: Anode (oxidation): Sn → Sn^{2+} + 2e^- Cathode (reduction): 2H^+ + 2e^- → H2

Step 5: Write the overall balanced equation for the electrolysis of tin sulfate

Now that each half-reaction is balanced, we can find the overall balanced equation for the electrolysis of tin sulfate. As the number of electrons involved in both half-reactions is equal (2e^-), we can combine the half-reactions to find the overall reaction: Sn + 2H^+ → Sn^{2+} + H2 In conclusion: The balanced chemical equation for the reduction of tin oxide with carbon is: SnO2 + C → Sn + CO2 The balanced chemical equations for the electrode reactions in the electrolysis of tin sulfate are: Anode (oxidation): Sn → Sn^{2+} + 2e^- Cathode (reduction): 2H^+ + 2e^- → H2 Overall balanced equation: Sn + 2H^+ → Sn^{2+} + H2

Key Concepts

Chemical Equation Balancing

Balancing chemical equations is a crucial step in understanding how chemical reactions occur and comply with the law of conservation of mass. A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the equation.

For instance, in the reduction of tin oxide with carbon, the equation starts as unbalanced: \[ \text{SnO}_2 + \text{C} \rightarrow \text{Sn} + \text{CO}_2 \].

However, a closer inspection reveals that there is already an equal number of atoms for each element on both sides, suggesting the reaction is balanced as is. For equations that aren't initially balanced, you would typically adjust coefficients, the numbers in front of the compounds or elements, to achieve balance without changing the chemical identity of the substances involved.

In more complex equations, balancing may require a systematic approach involving the half-reaction method which separates the oxidation and reduction components of a reaction.

Reduction of Tin Oxide

The reduction of tin oxide to tin is a crucial step in the process of obtaining pure tin from its ore, cassiterite. In this reduction process, carbon acts as a reducing agent, meaning it gives electrons to the tin oxide:

\[ \text{SnO}_2 + \text{C} \rightarrow \text{Sn} + \text{CO}_2 \].

In this simplified example, carbon monoxide (CO) or carbon dioxide (CO2) can be formed, depending on the conditions. The reaction also demonstrates the redox principle, where one species is oxidized (carbon) and the other is reduced (tin oxide). This equation represents a direct reduction method that is distinct from electrolysis, which often occurs at lower temperatures.

Electrode Reactions

During the electrolysis of tin sulfate, two distinct electrode reactions take place: oxidation at the anode and reduction at the cathode. Each electrode facilitates a half-reaction that contributes to the overall change in the electrolytic cell.

For tin, the oxidation reaction at the anode can be written as:\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2\text{e}^- \],which shows tin atoms losing electrons (oxidation). At the cathode, the reduction reaction usually involves hydrogen ions (protons) from the acid accepting electrons to form hydrogen gas:\[ 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \].

This distinction between the reactions at the two electrodes is essential for understanding the role of each electrode in driving the electrolysis process and for troubleshooting if there are problems with the reaction.

Half-Reaction Method

The half-reaction method is an effective way to balance complex redox reactions. It breaks these reactions down into two separate parts: oxidation (loss of electrons) and reduction (gain of electrons). Each half-reaction is balanced individually for atoms and charge, applying the principle that the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction.

For example, during the electrolysis of tin sulfate, the half-reactions are balanced as follows:\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2\text{e}^- \] (anode),\[ 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \] (cathode).

The electrons cancel when the half-reactions are combined to give the overall reaction:\[ \text{Sn} + 2\text{H}^+ \rightarrow \text{Sn}^{2+} + \text{H}_2 \].

Employing the half-reaction method can simplify the process of balancing reactions in an electrochemical context and is broadly used to understand various electrolysis experiments.