Question

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) , and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\) . The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

Short answer

Aluminum is being oxidized, and nickel is being reduced. The oxidation half-reaction is \(2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q) +6 e^-\), and the reduction half-reaction is \(3 \mathrm{Ni}^{2+}(a q) + 6 e^- \longrightarrow 3 \mathrm{Ni}(s)\). The aluminum electrode is the anode (negative), and the nickel electrode is the cathode (positive). Electrons flow from the aluminum electrode to the nickel electrode. \(\mathrm{Ni}^{2+}\) and \(\mathrm{Al}^{3+}\) ions migrate towards the nickel electrode (cathode), while \(\mathrm{NO}_{3}^-\) and \(\mathrm{SO}_{4}^{2-}\) ions migrate towards the aluminum electrode (anode).

Step-by-step solution

(a) Identifying Oxidation and Reduction

In the overall cell reaction: $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ 1. Observe that the oxidation state of Al changes from 0 to +3, which indicates that the aluminum is losing electrons (oxidation). 2. The oxidation state of Ni changes from +2 to 0, which indicates that nickel is gaining electrons (reduction). Thus, aluminum is being oxidized, and nickel is being reduced.

(b) Writing Half-reactions

We can separate the overall cell reaction into two half-reactions: one for oxidation and one for reduction. 1. Oxidation half-reaction (Aluminum): $$ 2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q) +6 e^- $$ 2. Reduction half-reaction (Nickel): $$ 3 \mathrm{Ni}^{2+}(a q) + 6 e^- \longrightarrow 3 \mathrm{Ni}(s) $$

(c) Determining the Anode and Cathode

In a voltaic cell: 1. Anode: The electrode where oxidation occurs (loss of electrons). In this case, the aluminum electrode is the anode. 2. Cathode: The electrode where reduction occurs (gain of electrons). In this case, the nickel electrode is the cathode.

(d) Identifying Electrode Signs

In a voltaic cell: 1. Anode is negative because it releases electrons (oxidation). 2. Cathode is positive because it attracts electrons (reduction). Thus, the aluminum electrode (anode) is negative, and the nickel electrode (cathode) is positive.

(e) Determining the Direction of Electron Flow

Electrons flow from the anode to the cathode in a voltaic cell. In this case, electrons flow from the aluminum electrode (anode) to the nickel electrode (cathode).

(f) Determining the Direction of Ion Migration

In a voltaic cell solution: 1. Cations (positive ions) migrate towards the cathode. In this case, \(\mathrm{Ni}^{2+}\) and \(\mathrm{Al}^{3+}\) ions will migrate towards the nickel electrode (cathode). 2. Anions (negative ions) migrate towards the anode. In this case, \(\mathrm{NO}_{3}^-\) and \(\mathrm{SO}_{4}^{2-}\) ions will migrate towards the aluminum electrode (anode).

Key concepts

Oxidation and Reduction

In a voltaic cell like the one described, the key processes occurring are oxidation and reduction. But what do these terms mean? Oxidation refers to the loss of electrons by a substance. In this case, aluminum is oxidized. This is evidenced by the change in its oxidation state from 0 to +3, meaning it loses three electrons. Reduction, on the other hand, involves the gain of electrons. Nickel, in our voltaic cell, is reduced as it gains electrons, causing its oxidation state to change from +2 to 0.
This interplay of losing and gaining electrons between substances forms the foundation of the reaction in a voltaic cell.
Therefore,

• Aluminum undergoes oxidation.
• Nickel undergoes reduction.

These processes are crucial to the electron exchange that powers the cell.

Half-reactions

Half-reactions simplify the understanding of what happens at each electrode. In a voltaic cell, the overall chemical reaction can be split into two parts: oxidation and reduction half-reactions.

* **Oxidation Half-reaction**: This occurs at the anode, where aluminum turns into aluminum ions. The electrons released are represented in the equation as:\[2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(aq) + 6 e^- \]* **Reduction Half-reaction**: This happens at the cathode, where the nickel ions gain electrons to form solid nickel:\[3 \mathrm{Ni}^{2+}(aq) + 6 e^- \longrightarrow 3 \mathrm{Ni}(s)\]
Half-reactions are pivotal because they show the specific roles of each component in the cell, helping to track the movement and exchange of electrons. Understanding them clarifies how substances change during the reaction.

Anode and Cathode

The terms anode and cathode denote the electrodes where different reactions in a voltaic cell occur. Let's break these down:
* **Anode**: This is where oxidation takes place. For our cell, the aluminum electrode acts as the anode. It's identified by the loss of electrons from the aluminum element. * **Cathode**: This is where reduction happens. The nickel electrode serves as the cathode, where electrons are gained by the nickel ions.

It's also critical to understand the signs associated with these electrodes:

• The anode is negative, since it releases electrons into the external circuit.
• The cathode is positive, as it pulls electrons from the circuit into the cell.

Recognizing these roles helps clarify why electron flow is directed from the anode to the cathode.

Electron Flow in Galvanic Cells

Electron flow is the directional movement of electrons through the external circuit that connects the electrodes in a galvanic cell. This movement is fundamental in generating the electrical energy that a voltaic cell provides.

In our aluminum-nickel voltaic cell: - Electrons flow from the anode, which is the aluminum electrode, to the cathode, the nickel electrode. This flow happens because electrons are produced in excess at the anode during oxidation, and they move to the cathode where the reduction requires them.

This flow pattern:

• Ensures continuous movement of electrons necessary to sustain the cell reaction.
• Also influences the direction in which ions migrate, balancing the charge within each half-cell.

Understanding this electron flow is crucial, as it forms the basis of how energy is captured and used from chemical reactions in galvanic cells.