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 everall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Nr}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being exidized, 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. Cell Potentials under Standard Conditions (Section 20.4)

Short answer

(a) Aluminum (Al) is being oxidized, and Nickel (Ni) is being reduced. (b) Half-reactions: Oxidation: \(\mathrm{Al}(s) \longrightarrow \mathrm{Al}^{3+}(aq) + 3 e^-\) Reduction: \(\mathrm{Ni}^{2+}(aq) + 2 e^- \longrightarrow \mathrm{Ni}(s)\) (c) Anode: Aluminum half-cell (negative electrode), Cathode: Nickel half-cell (positive electrode). (d) Anode: -, Cathode: + (e) Electrons flow from the aluminum electrode to the nickel electrode. (f) Cations (Ni^2+ and Al^3+) migrate toward the nickel electrode (cathode), and Anions (NO3^- and SO4^2-) migrate toward the aluminum electrode (anode).

Step-by-step solution

(Step 1: Identify oxidized and reduced species)

First, let's determine the oxidation states for each reactant and product in the overall cell reaction: \(2 \mathrm{Al}(s) + 3 \mathrm{Ni}^{2+}(aq) \longrightarrow 2 \mathrm{Al}^{3+}(aq) + 3 \mathrm{Ni}(s)\) Oxidation states: 1. Al: from 0 to +3 2. Ni: from +2 to 0 Since Al's oxidation state increases from 0 to +3, Al is being oxidized. Since Ni's oxidation state decreases from +2 to 0, Ni is being reduced.

(Step 2: Write half-reactions)

Next, let's write the half-reactions for the oxidation and reduction processes: Oxidation half-reaction (Al): \(\mathrm{Al}(s) \longrightarrow \mathrm{Al}^{3+}(aq) + 3 e^-\) Reduction half-reaction (Ni): \(\mathrm{Ni}^{2+}(aq) + 2 e^- \longrightarrow \mathrm{Ni}(s)\)

(Step 3: Identify anode, cathode, and signs of electrodes)

In a voltaic cell, the half-cell where oxidation occurs is the anode, and the half-cell where reduction occurs is the cathode. Anode (oxidation): Aluminum half-cell (Al strip in Al(NO3)3 solution) Cathode (reduction): Nickel half-cell (Ni strip in NiSO4 solution) The anode is the negatively charged electrode (-), and the cathode is the positively charged electrode (+).

(Step 4: Determine the direction of electron flow)

In a voltaic cell, electrons flow from the anode to the cathode through the external circuit. In this case, electrons flow from the aluminum electrode to the nickel electrode.

(Step 5: Determine the direction of cation and anion migration)

In a voltaic cell, cations migrate through the solution toward the cathode, while anions migrate through the solution toward the anode. Cations (Ni^2+ and Al^3+) migrate toward the nickel electrode (cathode). Anions (NO3^- and SO4^2-) migrate toward the aluminum electrode (anode).

Key concepts

Oxidation and Reduction Reactions

Understanding oxidation and reduction reactions is fundamental to grasping electrochemistry. These reactions involve the transfer of electrons between chemical species. Oxidation is the loss of electrons, leading to an increase in oxidation state, while reduction is the gain of electrons, resulting in a decrease in oxidation state. An easy way to remember this is through the acronym 'OIL RIG': Oxidation Is Loss, Reduction Is Gain.

The reaction in the exercise illustrates a classic example where aluminum (Al) is oxidized, as it loses electrons moving from an oxidation state of 0 to +3. Conversely, nickel (Ni) is reduced as it gains electrons, transitioning from a +2 to a 0 oxidation state. This transfer of electrons is what drives the chemical reaction forward and is essential in the operation of voltaic cells.

Half-Reactions in Electrochemistry

In electrochemistry, reactions are often split into half-reactions to show the oxidation and reduction processes separately. This helps us understand how electrons are transferred in the system. For the aluminum and nickel voltaic cell, we divide the overall reaction into two half-reactions:

• The oxidation half-reaction shows aluminum metal being converted to aluminum ions: Al(s) \(\longrightarrow\) Al^3+(aq) + 3 e^-
• The reduction half-reaction involves nickel ions gaining electrons to form nickel metal: Ni^2+(aq) + 2 e^- \(\longrightarrow\) Ni(s)

These half-reactions are pivotal for identifying the reactions occurring at each electrode within a voltaic cell and are the key to understanding the electron flow and ion migration.

Anode and Cathode Identification

Identifying the anode and cathode in a voltaic cell clarifies the direction of the oxidation and reduction reactions. The anode is where oxidation occurs, and the cathode is where reduction takes place. In the given exercise, the Al strip immersed in a solution of Al(NO3)3 acts as the anode because it loses electrons (oxidation), and the Ni strip in a NiSO4 solution functions as the cathode where electrons are gained (reduction).

Remember, the anode is associated with a negative charge because it gives electrons to the external circuit. The cathode, in contrast, has a positive charge, receiving electrons from the external circuit. Recognizing this is crucial for predicting the behavior of a voltaic cell.

Electron Flow in Electrochemical Cells

Understanding the direction of electron flow in electrochemical cells is important to grasp their function. Electrons are spontaneously generated at the anode due to the oxidation reaction and then flow through the external circuit towards the cathode, where they are consumed in the reduction reaction. Therefore, in a voltaic cell, the flow of electrons always moves from anode to cathode. For our aluminum and nickel cell, this means electrons flow from the aluminum strip (anode) to the nickel strip (cathode). This directional flow of electrons also helps drive the external circuit, which could be connected to a device that uses the electrical energy produced.

Migration of Ions in Voltaic Cells

The migration of ions in voltaic cells connects the chemistry happening at the electrodes with the overall electrical circuit. When a voltaic cell operates, cations move toward the cathode to balance the charge accumulated by the reduction process, while anions migrate towards the anode, compensating for the loss of electrons due to oxidation. In the aluminum-nickel cell exercise, Al^3+ and Ni^2+ ions (cations) migrate towards the nickel electrode, and NO3^- and SO4^2- ions (anions) move toward the aluminum electrode. This migration maintains electrical neutrality in the cell and is critical for the continuous flow of electrons in the circuit.