Question

Consider a galvanic cell consisting of a magnesium electrode in contact with $1.0\mathrm{M}\mathrm{Mg}{\left({\mathrm{NO}}_3\right)}_2$ and a cadmium electrode in contact with $1.0\mathrm{M}\mathrm{Cd}{\left({\mathrm{NO}}_3\right)}_2$. Calculate $E^{\circ }$ for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.

Short answer

The standard cell potential $E_{\text{cell}}^{\circ }$ is 1.97 V. Electrons flow from magnesium to cadmium.

Step-by-step solution

Write Half-Reactions

First, determine the standard reduction potentials for the half-reactions involved in the galvanic cell. The two half-reactions are:1. Magnesium Half-Reaction:$${\text{Mg}}_{(aq)}^{2+}+2e^{-}\to {\text{Mg}}_{(s)}{E}^{\circ }=-2.37\text{ V}$$2. Cadmium Half-Reaction:$${\text{Cd}}_{(aq)}^{2+}+2e^{-}\to {\text{Cd}}_{(s)}{E}^{\circ }=-0.40\text{ V}$$

Identify the Anode and Cathode

In a galvanic cell, the anode is where oxidation occurs and the cathode is where reduction occurs. The more negative reduction potential represents the anode, while the less negative is the cathode.- Anode (Oxidation): Magnesium (Since $-2.37\text{ V}$ is more negative than $-0.40\text{ V}$)- Cathode (Reduction): Cadmium

Calculate Standard Cell Potential, $E_{\text{cell}}^{\circ }$

The standard cell potential is calculated using the formula:$$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$$Substitute the standard reduction potentials:$$E_{\text{cell}}^{\circ }=(-0.40\text{ V})-(-2.37\text{ V})=1.97\text{ V}$$

Diagram of the Galvanic Cell

Draw a diagram to illustrate the galvanic cell layout: 1. On the left, draw the magnesium electrode (anode). 2. On the right, draw the cadmium electrode (cathode). 3. Use a line to represent the salt bridge connecting the two solutions. 4. Indicate the flow of electrons from the magnesium anode to the cadmium cathode outside the circuit.

Key concepts

Standard Cell Potential

In every galvanic cell, the most vital metric is the standard cell potential, denoted as $E_{\text{cell}}^{\circ }$. This value gives us a measure of the voltage generated by the cell under standard conditions, which usually implies concentrations of $1.0\text{ M}$ for all solutions, and a temperature of $25\text{degree}\text{C}$. By calculating the standard cell potential, we can predict the feasibility of a reaction occurring spontaneously.
To compute $E_{\text{cell}}^{\circ }$, we use the standard reduction potential values of the two half-cells. This is achieved with the formula:

• $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$

Knowing the correct values to plug into the formula is essential. For our given exercise:- Anode: Magnesium with a potential of $-2.37\text{ V}$- Cathode: Cadmium with a potential of $-0.40\text{ V}$Plugging these values into our formula yields a cell potential of $1.97\text{ V}$. A positive cell potential indicates that the galvanic cell can generate electricity spontaneously.

Anode and Cathode Identification

Identifying the anode and cathode in a galvanic cell is crucial since it determines the direction of electron flow. The anode is identified by its higher tendency to lose electrons, which is shown by its more negative reduction potential value. Conversely, the cathode is where reduction occurs, with a less negative (or more positive) potential.
In the exercise concerning magnesium and cadmium:

• Magnesium has a reduction potential of $-2.37\text{ V}$, which is more negative, signifying that it acts as the anode.
• Cadmium with a reduction potential of $-0.40\text{ V}$ plays the role of the cathode because it is less negative.

This distinction is essential as it forms the basis for understanding the cell's operation. The anode is always where oxidation happens, and the cathode is where reduction occurs.

Half-Reaction Equations

Half-reaction equations are pivotal in electrochemistry as they detail the individual steps of oxidation and reduction separately. For every galvanic cell, these half-reactions must be clearly identified to compute the overall cell reaction and potential.
For the magnesium-cadmium cell:

• Magnesium half-reaction (anode, oxidation): $${\text{Mg}}_{(s)}\to {\text{Mg}}_{(aq)}^{2+}+2e^{-}$$
• Cadmium half-reaction (cathode, reduction): $${\text{Cd}}_{(aq)}^{2+}+2e^{-}\to {\text{Cd}}_{(s)}$$

These reactions visualize the electron transfer process. At the anode, magnesium loses electrons, becoming ions, while at the cathode, cadmium ions gain electrons to form solid metal. Understanding these allows us to piece together a full cell reaction, which is the summation of these halves, excluding electrons for the balanced equation.

Electron Flow Direction

The direction of electron flow is one of the fundamental principles in the operation of a galvanic cell. Identifying this flow provides insight into how the cell generates electricity.
In our magnesium-cadmium cell:

• Electrons are released from the magnesium metal at the anode.
• These electrons travel through an external circuit to the cadmium cathode.

The path of these electrons is vital: electricity is generated as they flow from the anode to the cathode. From a practical perspective, this flow is usually shown in diagrams as moving from the left side (anode) to the right side (cathode) with a connecting wire. Thus, interpreting this electron flow helps in predicting the behavior and efficiency of the cell.