Question

Consider the oxidation-reduction reaction $$ \mathrm{Al}(s)+\mathrm{Ni}^{2+}(a q) \rightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Ni}(s) $$ Sketch a galvanic cell that makes use of this reaction. Which metal ion is reduced? Which metal is oxidized? What half-reaction takes place at the anode in the cell? What half-reaction takes place at the cathode?

Short answer

In the given redox reaction, the metal ion reduced is nickel ion (Ni²⁺), and the metal oxidized is aluminum (Al). The oxidation half-reaction at the anode is \(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q) + 3e^-\), and the reduction half-reaction at the cathode is \(\mathrm{Ni}^{2+}(a q) + 2e^- \rightarrow \mathrm{Ni}(s)\). The galvanic cell can be sketched as Anode (Al) --||-- Cathode (Ni²⁺/Ni), with appropriate connections to a voltmeter and a salt bridge.

Step-by-step solution

Identify the oxidation and reduction half-reactions

In the given redox reaction: \(\mathrm{Al}(s)+\mathrm{Ni}^{2+}(a q) \rightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Ni}(s)\) Aluminum (Al) loses 3 electrons to form Al^3+: \(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q) + 3e^-\) (oxidation half-reaction) Nickel ion (Ni^2+) gains 2 electrons to form solid nickel (Ni): \(\mathrm{Ni}^{2+}(a q) + 2e^- \rightarrow \mathrm{Ni}(s)\) (reduction half-reaction)

Determine the anode and cathode

Recall that oxidation occurs at the anode and reduction occurs at the cathode. Thus: Anode: Aluminum (Al) - here, the oxidation half-reaction occurs Cathode: Nickel (Ni^2+) - here, the reduction half-reaction occurs

Sketch the galvanic cell

The galvanic cell consists of the following components: 1. Anode: Aluminum metal electrode (Al) - connected to the negative terminal of the voltmeter. 2. Cathode: Nickel metal electrode (Ni) - connected to the positive terminal of the voltmeter. 3. Electrolyte: A solution of Nickel(II) ions. 4. Salt bridge: Contains an electrolyte to allow the flow of ions between the two half-cells. Abbreviated sketch of the galvanic cell: Anode (Al) --||-- Cathode (Ni^2+/Ni) In conclusion: Reduced metal ion: Nickel ion (Ni^2+) Oxidized metal: Aluminum (Al) Half-reaction at the anode (oxidation): \(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q) + 3e^-\) Half-reaction at the cathode (reduction): \(\mathrm{Ni}^{2+}(a q) + 2e^- \rightarrow \mathrm{Ni}(s)\)

Key concepts

Oxidation-Reduction Reaction

In every chemical process known as an oxidation-reduction reaction, or redox reaction for short, there is a transfer of electrons between substances. This is pivotal in a galvanic cell, where electricity is generated through redox reactions.

Oxidation involves the loss of electrons, while reduction involves the gain. An easy way to remember this is "OIL RIG": Oxidation Is Loss, Reduction Is Gain. In our specific exercise, aluminum (Al) undergoes oxidation as it loses electrons and forms \( \mathrm{Al}^{3+} \). Meanwhile, nickel ions \( \mathrm{Ni}^{2+} \) undergo reduction by gaining electrons to transform into solid nickel \( \mathrm{Ni}(s) \).

Through this exchange, one element propels the loss of electrons (oxidation), and another element facilitates the electron gain (reduction). Together, they create a flow of electrons vital to the operation of the galvanic cell.

Anode and Cathode

In a galvanic cell, it's important to identify where oxidation and reduction happen, known respectively as the anode and cathode. Remember:

• The anode is the site of oxidation.
• The cathode is the site of reduction.

In the exercise based on our redox reaction, aluminum serves as the anode. It's here that the aluminum atoms lose electrons and become \( \mathrm{Al}^{3+} \) ions. The negative nature of the anode is crucial, as it releases electrons.

On the other hand, the nickel ions \( \mathrm{Ni}^{2+} \) act at the cathode, where they gain the electrons, becoming solid nickel \( \mathrm{Ni}(s) \). As the electrons flow from the anode to the cathode, they generate an electric current, an essential function in electrochemical cells.

Half-Reactions

Breaking down a redox reaction into half-reactions is an effective way to understand which species are oxidized or reduced. Each half-reaction gives details on electron transfer.

For the oxidation half-reaction in this galvanic cell, aluminum atoms lose three electrons:\[ \mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(aq) + 3e^- \]This process happens at the anode. Each aluminum atom transitions to its ion state, releasing electrons into the circuit.

The reduction half-reaction involves nickel ions gaining electrons:\[ \mathrm{Ni}^{2+}(aq) + 2e^- \rightarrow \mathrm{Ni}(s) \]This occurs at the cathode, where nickel ions accept electrons and are deposited as solid nickel. Both half-reactions are equally crucial since they emphasize the electron exchange that generates electrical energy.

Electrochemistry

Electrochemistry focuses on the interplay between chemical reactions and electricity, and the galvanic cell exemplifies this beautifully. These cells convert chemical energy from redox reactions into electrical energy.

By having a complete circuit, including an anode and cathode immersed in electrolytes, electrons released at the anode can travel via an external wire to the cathode. This creates an electric current. An essential component often found in these cells is the salt bridge. This connects the two half-cells, allowing ions to flow and maintain charge balance without mixing the different solutions.

In essence, electrochemistry not only allows us to harness chemical reactions to power devices but also provides insights into the fundamental behavior of the elements involved. This makes it a cornerstone of modern technology and science.