Question

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a magnesium strip placed in a solution of \(\mathrm{MgCl}_{2}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiCl}_{2}\). The overall cell reaction is $$ \mathrm{Mg}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{Mg}^{2+}(a q) $$ (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 magnesium electrode to the nickel electrode or from the nickel to the magnesium? (f) In which directions do the cations and anions migrate through the solution?

Short answer

a) Magnesium is being oxidized and Nickel ion is being reduced. b) Oxidation half-reaction: \(\mathrm{Mg}(s) \longrightarrow \mathrm{Mg^{2+}}(aq) + 2e^-\), Reduction half-reaction: \(\mathrm{Ni^{2+}}(aq) + 2e^- \longrightarrow \mathrm{Ni}(s)\) c) The Mg electrode is the anode, and the Ni electrode is the cathode. d) Anode (Mg electrode) is negative, and Cathode (Ni electrode) is positive. e) Electrons flow from the magnesium electrode to the nickel electrode. f) Cations (\(\mathrm{Mg^{2+}}\)) migrate towards the Ni electrode (cathode), and anions (\(\mathrm{Cl^-}\)) migrate towards the Mg electrode (anode).

Step-by-step solution

Identify oxidation and reduction processes

We can use the acronym "OIL RIG" (Oxidation Is Loss, Reduction Is Gain) to remember that oxidation is the loss of electrons, while reduction is the gain of electrons. The overall cell reaction is given: \[\mathrm{Mg}(s)+\mathrm{Ni^{2+}}(aq) \longrightarrow \mathrm{Ni}(s)+\mathrm{Mg^{2+}}(aq)\] From the cell reaction, we can see that \(\mathrm{Mg}\) loses two electrons and becomes \(\mathrm{Mg^{2+}}\), while \(\mathrm{Ni^{2+}}\) gains two electrons and turns into \(\mathrm{Ni}\). Therefore: a) Magnesium is being oxidized and Nickel ion is being reduced.

Write half-reactions

We will write the half-reactions for both the oxidation and reduction processes: b) Oxidation half-reaction: \[\mathrm{Mg}(s) \longrightarrow \mathrm{Mg^{2+}}(aq) + 2e^-\] Reduction half-reaction: \[ \mathrm{Ni^{2+}}(aq) + 2e^- \longrightarrow \mathrm{Ni}(s)\]

Identify Anode and Cathode

Since oxidation happens at the anode and reduction occurs at the cathode: c) The Mg electrode is the anode, and the Ni electrode is the cathode.

Determine signs of electrodes

Anode is where oxidation happens, and is the site where electrons are released, thus it's considered as the negative electrode. Cathode is where reduction occurs, and it's the site where electrons are gained, thus it is the positive electrode. d) Anode (Mg electrode) is negative, and Cathode (Ni electrode) is positive.

Determine direction of electron flow

Electrons flow from the negative electrode (anode) to the positive electrode (cathode). e) Electrons flow from the magnesium electrode to the nickel electrode.

Determine cation and anion migration

In a voltaic cell, cations (\(\mathrm{Mg^{2+}}\) in this case) migrate towards the cathode, and anions (\(\mathrm{Cl^-}\) in this case) migrate towards the anode to maintain charge neutrality. f) Cations (\(\mathrm{Mg^{2+}}\)) migrate towards the Ni electrode (cathode), and anions (\(\mathrm{Cl^-}\)) migrate towards the Mg electrode (anode).

Key concepts

Oxidation and Reduction

Understanding oxidation and reduction is key to grasping how voltaic cells operate. Let's start with the basics. Oxidation refers to the loss of electrons, while reduction means a gain of electrons. These chemical processes occur simultaneously in voltaic cells, where one substance loses electrons and is oxidized, and another gains electrons and is reduced.
In the case of a voltaic cell with magnesium and nickel, magnesium (\( \mathrm{Mg}(s) \) ) undergoes oxidation. It loses two electrons to form magnesium ions (\( \mathrm{Mg^{2+}}(aq) \) ). On the other hand, nickel ions (\( \mathrm{Ni^{2+}}(aq) \) ) undergo reduction by gaining two electrons to form nickel metal (\( \mathrm{Ni}(s) \) ). Remember "OIL RIG"—Oxidation Is Loss, Reduction Is Gain, to quickly recall these processes.
This dynamic push and pull of electrons is what generates electricity in a voltaic cell.

Half-Reaction

Half-reactions are a useful way of breaking down the chemistry in a voltaic cell into more understandable parts. Each electrode in the cell, whether it's involved in oxidation or reduction, has its own half-reaction which shows either the loss or gain of electrons.
The oxidation half-reaction at the magnesium electrode is:

• \( \mathrm{Mg}(s) \longrightarrow \mathrm{Mg^{2+}}(aq) + 2e^- \)

This equation tells us that solid magnesium loses two electrons, forming magnesium ions.
Meanwhile, the reduction half-reaction at the nickel electrode is:

• \( \mathrm{Ni^{2+}}(aq) + 2e^- \longrightarrow \mathrm{Ni}(s) \)

Here, nickel ions gain two electrons to become solid nickel.
Writing these half-reactions helps us track the electron transfer processes separately, which is crucial to analyzing and understanding the overall cell reaction.

Anode and Cathode

In any voltaic cell, you'll find two types of electrodes: the anode and the cathode. It's important to know what happens at each of these to understand the cell's chemistry.
The *anode* is where oxidation occurs. It's the electrode where electrons are lost by the reactant. In our example with magnesium, the anode is the magnesium electrode. Here, magnesium loses electrons.
On the opposite end, the *cathode* is where reduction takes place. This is the site where electrons are gained. For the nickel electrode, it is the cathode because nickel ions gain electrons to form nickel metal.
Having these roles defined helps understand which materials will undergo oxidation or reduction and why they are vital for generating electricity in the cell.

Electrode Sign and Electron Flow

Every voltaic cell has a setup where electrons flow from one electrode to the other, and the signs of these electrodes play a crucial role. At the *anode*, oxidation occurs, meaning electrons are released, making it negatively charged. With the **Mg electrode** being the anode, it holds a negative sign.
Conversely, the *cathode* is the site of reduction, where it attracts electrons. This is why the cathode generally is positively charged. Given the example, the **Ni electrode** acts as the cathode.
The path of electrons flows from the negative anode to the positive cathode. In simpler terms, electrons flow from the magnesium electrode (anode) to the nickel electrode (cathode), following this path to generate electrical energy.
These directions and signs are crucial for setting up any electrical device using voltaic or galvanic cells.

Ion Migration in Solution

An essential aspect of maintaining electrical neutrality in a voltaic cell is the migration of ions between the electrolytes in the solution. This movement ensures that the separate half-reactions at the electrodes don't lead to a buildup of charge, which would eventually stop the cell from working.
**Cations**, which are positively charged like \( \mathrm{Mg^{2+}} \) , migrate toward the cathode. This movement compensates for the electrons being gained by \( \mathrm{Ni^{2+}} \) at the cathode.
On the other hand, **anions**, like \( \mathrm{Cl^-} \) , move towards the anode. This balances the positive charges left behind when \( \mathrm{Mg} \) loses its electrons at the anode.
This migration of ions prevents charge imbalance, ensuring continuous electron flow, and thereby sustaining the operational efficiency of a voltaic cell.