Question

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment 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 electrode compartments. (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 \(\mathrm{Al}\) is not coated with its oxide.

Short answer

In the given voltaic cell, aluminum is being oxidized at the aluminum electrode, which is the anode, and nickel is being reduced at the nickel electrode, which is the cathode. The anode has a negative charge, and the cathode has a positive charge. Electrons flow from the aluminum electrode (anode) to the nickel electrode (cathode). Al^(3+) cations move toward the nickel electrode (cathode) and NO3^(-) and SO4^(2-) anions move toward the aluminum electrode (anode).

Step-by-step solution

(a) Identifying Oxidation and Reduction

To determine which element is being oxidized and which is being reduced, look at the overall cell reaction: \[2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s)\] Here, aluminum is going from a neutral state to an ionic state with a +3 charge, which means it is losing 3 electrons. Therefore, aluminum is being oxidized. On the other hand, nickel ions are gaining 2 electrons and going from Ni^(2+) to Ni(s), so nickel is being reduced.

(b) Writing the Half-Reactions

Now that we know which element is being oxidized and which is being reduced, we can write the respective half-reactions: Oxidation half-reaction (at anode): \[\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(aq) + 3e^-\] Reduction half-reaction (at cathode): \[\mathrm{Ni}^{2+}(aq) + 2e^- \rightarrow \mathrm{Ni}(s)\]

(c) Identifying the Anode and Cathode

The anode is the electrode where oxidation occurs, and the cathode is the electrode where reduction occurs. Since aluminum is being oxidized, the aluminum electrode is the anode. Since nickel is being reduced, the nickel electrode is the cathode.

(d) Indicating the Signs of the Electrodes

In a voltaic cell, the anode is negatively charged, and the cathode is positively charged. So, the aluminum electrode (anode) has a negative charge, and the nickel electrode (cathode) has a positive charge.

(e) Determining the Direction of Electron Flow

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

(f) Determining the Direction of Ion Migration

In a voltaic cell, cations (positively charged ions) migrate towards the cathode, and anions (negatively charged ions) migrate towards the anode. For this specific cell, Al^(3+) cations move toward the nickel electrode (cathode) and NO3^(-) and SO4^(2-) anions move toward the aluminum electrode (anode).

Key concepts

Oxidation and Reduction

Oxidation and reduction are two fundamental concepts in electrochemistry and are key processes in a voltaic cell. In the given voltaic cell, we need to identify which species is oxidized and which one is reduced. Oxidation involves the loss of electrons, while reduction involves the gain of electrons.

In the overall cell reaction: \[2 \mathrm{Al}(s) + 3 \mathrm{Ni}^{2+}(aq) \longrightarrow 2 \mathrm{Al}^{3+}(aq) + 3 \mathrm{Ni}(s)\] aluminum (\(\mathrm{Al}\)) goes from a zero oxidation state to +3. This indicates oxidation as aluminum loses electrons to form \(\mathrm{Al}^{3+}\). Accordingly, nickel (\(\mathrm{Ni}^{2+}\)) ions gain electrons and are reduced from \(\mathrm{Ni}^{2+}\) to solid nickel (\(\mathrm{Ni}(s)\)). Comprehending these electron transfers is crucial for determining the half-reactions and which electrodes serve as the anode and cathode.

Electrodes Identification

In a voltaic cell, identifying which electrode acts as the anode and which one as the cathode is crucial since it determines where oxidation and reduction occur. The anode is where oxidation happens. In this cell, since aluminum is oxidized, the aluminum electrode is identified as the anode.

The cathode is where reduction takes place. Here, nickel ions are reduced, hence the nickel electrode is the cathode. It's important to remember that in a voltaic cell, the anode carries a negative charge, while the cathode carries a positive charge due to the electron flow from the anode to the cathode. Proper identification of electrodes ensures that the cell works correctly and facilitates understanding of the direction of electron flow and ion migration.

Electron Flow

Understanding electron flow within a voltaic cell is essential for grasping how electrical energy is generated. Electrons in a voltaic cell flow from the anode to the cathode.

In this voltaic setup, since aluminum is being oxidized at the aluminum electrode (anode), electrons are released here. These electrons then travel through the external circuit to the nickel electrode (cathode), where they are used to reduce nickel ions. This movement of electrons from the aluminum to the nickel electrode is what generates a current and allows the voltaic cell to produce electricity. Maintaining the correct direction of electron flow is critical for the cell's operation.

Ion Migration

Ion migration in a voltaic cell helps to balance the charge as the reaction progresses. This movement occurs in the internal part of the cell, within the electrolyte solution that separates the two electrodes.

Cations, which are positively charged ions, move towards the cathode. In this case, the aluminum ions (\(\mathrm{Al}^{3+}\)) generated at the anode after the oxidation process will move towards the cathode. Anions, which are negatively charged ions, move towards the anode. Here, the nitrate (\(\mathrm{NO}_3^-\)) and sulfate (\(\mathrm{SO}_4^{2-}\)) ions will migrate towards the aluminum electrode (anode).Understanding the direction of ion migration helps in comprehending how the cell maintains electrical neutrality and operates efficiently. Proper ion migration ensures that the electrochemical reactions can proceed without interruption, thereby facilitating continuous electron flow and electricity generation in the cell.