Problem 10 - Chemistry and Chemical Reactivity

Question

The half-cells \(\operatorname{Sn}^{2+}(\text { aq }) | \operatorname{Sn}(s)\) and \(C l_{2}(g) | C l^{-}(a q)\) are linked to create a voltaic cell. (a) Write equations for the oxidation and reduction half-reactions and for the overall (cell) reaction. (b) Which half-reaction occurs in the anode compartment, and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the electrode to the electrode. Negative ions move in the salt bridge from the \(-\) half-cell to the half-cell.

Step-by-Step Solution

Answer

Oxidation: \( \operatorname{Sn}(s) \rightarrow \operatorname{Sn}^{2+}(aq) + 2e^- \). Reduction: \( Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq) \). Anode: \( \operatorname{Sn}^{2+}(aq) | \operatorname{Sn}(s) \), Cathode: \( Cl_2(g) | Cl^-(aq) \). Electrons flow from \( \operatorname{Sn} \) to \( Cl_2 \).

Step 1: Identifying Oxidation and Reduction

Start by identifying which species undergo oxidation and reduction. The half-cell \( \operatorname{Sn}^{2+}(aq) | \operatorname{Sn}(s) \) indicates that \( \operatorname{Sn}(s) \) is oxidized to \( \operatorname{Sn}^{2+}(aq) \). The half-cell \( Cl_2(g) | Cl^-(aq) \) suggests that \( Cl_2(g) \) is reduced to \( Cl^-(aq) \).

Step 2: Writing Oxidation Half-Reaction

The oxidation reaction occurs at the anode, where \( \operatorname{Sn}(s) \) is oxidized to \( \operatorname{Sn}^{2+}(aq) \). The half-reaction is: \[ \operatorname{Sn}(s) \rightarrow \operatorname{Sn}^{2+}(aq) + 2e^- \]

Step 3: Writing Reduction Half-Reaction

Now, write the reduction half-reaction occurring at the cathode, where \( Cl_2(g) \) gains electrons to form \( Cl^-(aq) \). The half-reaction is: \[ Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq) \]

Step 4: Writing Overall Cell Reaction

Combine the oxidation and reduction half-reactions to obtain the overall cell reaction. Make sure electrons are canceled out: \[ \operatorname{Sn}(s) + Cl_2(g) \rightarrow \operatorname{Sn}^{2+}(aq) + 2Cl^-(aq) \]

Step 5: Identifying Anode and Cathode

The anode is where oxidation occurs, which in this case is the \( \operatorname{Sn}^{2+}(aq) | \operatorname{Sn}(s) \) half-cell. The cathode, where reduction occurs, is the \( Cl_2(g) | Cl^-(aq) \) half-cell.

Step 6: Describing Electron Flow

Electrons flow from the anode to the cathode in the external circuit. Thus, electrons flow from the \( \operatorname{Sn} \) electrode to the \( Cl_2 \) electrode.

Step 7: Describing Ion Movement in Salt Bridge

Negative ions in the salt bridge flow from the cathode to the anode half-cell to maintain charge balance. Thus, negative ions move from the \( Cl^- \) half-cell to the \( \operatorname{Sn}^{2+} \) half-cell.

Key Concepts

Oxidation Half-Reaction

In a voltaic cell, the oxidation half-reaction is the process where a substance loses electrons. Specifically, oxidation occurs at the anode. In our example, solid tin, represented as \( \operatorname{Sn}(s) \), undergoes oxidation. This means it loses two electrons and is transformed into aqueous \( \operatorname{Sn}^{2+}(aq) \).
This process can be represented by the equation: \[ \operatorname{Sn}(s) \rightarrow \operatorname{Sn}^{2+}(aq) + 2e^- \] - **Location**: Anode - **Electron Loss**: 2 electrons are lost - **Starting Substance**: \( \operatorname{Sn}(s) \) - **Ending Substance**: \( \operatorname{Sn}^{2+}(aq) \) The oxidation half-reaction is crucial because it provides electrons for the cell's electrical current. Without this electron transfer, the voltaic cell wouldn't function.

Reduction Half-Reaction

The reduction half-reaction happens at the cathode. It's all about gaining electrons. In the given voltaic cell, chlorine gas \( Cl_2(g) \) is reduced, meaning it gains two electrons to form chloride ions \( Cl^-(aq) \). The reduction process can be expressed as: \[ Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq) \] - **Location**: Cathode - **Electron Gain**: Two electrons are gained - **Starting Substance**: \( Cl_2(g) \) - **Ending Substance**: \( 2Cl^-(aq) \) Reduction is vital because it completes the electron flow circuit in the voltaic cell. By accepting electrons, it allows the cell to produce an electrical current that can do useful work.

Electrode Compartment

Each half-reaction in a voltaic cell occurs in separate electrode compartments. One compartment hosts the anode while the other houses the cathode.
The two compartments make up the entire cell but serve distinct roles. This separation is necessary to harness energy from spontaneous chemical reactions and convert it into electrical energy. - **Anode Compartment**: Here, oxidation occurs. In the given example, this is the \( \operatorname{Sn}^{2+}(aq) | \operatorname{Sn}(s) \) half-cell.
- **Cathode Compartment**: Here, reduction occurs. For our cell, it's the \( Cl_2(g) | Cl^-(aq) \) half-cell.
These compartments are the heart of the voltaic cell, allowing controlled electron transfer from the anode to the cathode.

Electron Flow

Electron flow is the key to electricity in a voltaic cell. It is the movement of electrons from one electrode to the other through an external circuit. Electrons must always flow from the anode to the cathode.
In the case of our voltaic cell, electrons move from the tin electrode at the anode to the chlorine electrode at the cathode. This flow generates an electric current, which is the primary purpose of voltaic cells. - **Direction**: From Anode (Sn) to Cathode (Cl) - **Pathway**: External circuit The electrons provide not only the current but also the necessary charges to sustain the reduction and oxidation reactions.

Salt Bridge

The salt bridge plays a crucial role in maintaining charge balance within the voltaic cell. It connects the two half-cells, allowing ions to flow without allowing the solutions to mix.
This movement of ions is essential because it ensures that each compartment maintains electroneutrality as electrons are transferred.
- **Purpose**: Maintains charge balance - **Ion Movement**: Negative ions travel from the cathode compartment to the anode compartment Without the salt bridge, the build-up of charges would quickly stop the cell from working, as the electron transfer would halt. This component is vital for the continuous flow of electrons and therefore, the continuous production of electric current.