Problem 10 - Chemistry and Chemical Reactivity

Question

The half-cells \(\operatorname{Sn}^{2+}(\text { aq }) | \operatorname{Sn}(s)\) and \(\operatorname{Cl}_{2}(g) | C l^{-}(\text {aq })\) 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

(a) Oxidation: \( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq)+2e^- \); Reduction: \( \text{Cl}_2(g)+2e^-\rightarrow 2\text{Cl}^-(aq) \); Overall: \( \text{Sn}(s)+\text{Cl}_2(g)\rightarrow \text{Sn}^{2+}(aq)+2\text{Cl}^-(aq) \). (b) Anode: Sn(s); Cathode: Cl₂(g). (c) Electrons flow from anode to cathode; Negative ions move from cathode to anode.

Step 1: Identify Oxidation Half-Reaction

In the voltaic cell described, the Sn metal is getting oxidized as it loses electrons. The half-reaction is \( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^- \). Here, solid tin (Sn) is oxidized to tin ions (Sn²⁺) by losing electrons.

Step 2: Identify Reduction Half-Reaction

The chlorine gas undergoes reduction as it gains electrons. The half-reaction is \( \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \). Here, chlorine gas gains electrons to form chloride ions (Cl⁻).

Step 3: Write Overall Cell Reaction

To write the overall reaction, combine the oxidation and reduction half-reactions, ensuring balance in both electrons and atoms. The overall reaction is \( \text{Sn}(s) + \text{Cl}_2(g) \rightarrow \text{Sn}^{2+}(aq) + 2\text{Cl}^-(aq) \).

Step 4: Determine Anode and Cathode Reactions

The anode compartment is where oxidation occurs, which is the Sn half-reaction \( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^- \). The cathode compartment is where reduction occurs, which is the Cl₂ half-reaction \( \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \).

Step 5: Complete Sentences on Electron Flow and Ion Movement

Electrons flow from the anode (Sn electrode) to the cathode (Cl₂ electrode) in the external circuit. Negative ions (Cl⁻) move in the salt bridge from the cathode (chlorine half-cell) to the anode (tin half-cell) to balance charge.

Key Concepts

Oxidation-Reduction Reactions

In a voltaic cell, an oxidation-reduction (or redox) reaction is the driving force that generates electrical energy. Oxidation is the process where an element loses electrons, while reduction is when an element gains electrons. Together, these reactions create a flow of electrons.
In the case of the voltaic cell made from the half-cells \(Sn^{2+}(aq) | Sn(s)\text{ and } Cl_{2}(g) | Cl^{-}(aq)\), tin metal (Sn) gets oxidized to tin ions (Sn²⁺). Concurrently, chlorine gas (Cl₂) is reduced to chloride ions (Cl⁻).
This coordination of electron loss and gain results in a systematic transfer of electrons from the oxidizing agent (tin) to the reducing agent (chlorine), which is essential for generating electricity.

Half-Reactions

When considering redox reactions in a voltaic cell, it is crucial to split the entire reaction into two "half-reactions," one for oxidation and another for reduction.
For the oxidation half-reaction:

• Tin (Sn) is oxidized as it loses two electrons.

\[Sn(s) \rightarrow Sn^{2+}(aq) + 2e^- \]In this half-reaction, the loss of electrons signifies oxidation.
On the other hand, the reduction half-reaction involves chlorine,

• Chlorine gas (Cl₂) gains two electrons to form two chloride ions (Cl⁻).

\[Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq) \]By understanding and balancing these half-reactions, we can explain the electron flow in the cell and how electric current is generated.

Electrode Compartments

In a voltaic cell, electrodes are key components where the half-reactions occur. Each electrode is housed in its own compartment, known as the "electrode compartment," where specific redox reactions take place.
The anode is the electrode where oxidation occurs, whereas the cathode is where reduction happens.

• In the tin and chlorine cell, the Sn electrode serves as the anode, hosting the oxidation half-reaction.
• The Cl₂ electrode functions as the cathode, where the reduction of chlorine occurs.

This separation into distinct compartments is vital, as it enables a directed flow of electrons through an external circuit from anode to cathode, which powers electrical devices.

Electron Flow

In a voltaic cell, electron flow is a crucial aspect of energy transfer. The pathway that electrons take can determine the efficiency of the voltaic cell in transforming chemical energy into electrical energy.
Electrons are generated at the anode, where oxidation occurs. They travel through an external circuit towards the cathode, in what essentially becomes the current.
In the provided scenario, the electrons flow from the Sn electrode (anode) to the Cl₂ electrode (cathode).

• Sn loses electrons, releasing them into the circuit.
• Cl₂ gains electrons at the cathode.

This continuous flow of electrons through the wire is harnessed to perform electrical work outside the cell.

Ion Movement

While electrons flow through an external circuit, ions move within the voltaic cell to maintain electrical neutrality. This involves both the movement of charged particles in the solution phase and through a salt bridge.
In our voltaic cell, Cl⁻ ions, formed at the cathode, travel through the salt bridge into the anode compartment.

• This movement of Cl⁻ ions ensures that negative and positive charges stay balanced in each half-cell.
• It prevents the build-up of excess charge, which would otherwise halt the operation of the cell.

This ionic movement, combined with the electron flow, ensures that the redox process can continuously operate, providing a steady stream of electrical energy.