Problem 42 - Chemistry: The Central Science in SI Units, Global Edition

Question

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Step-by-Step Solution

Answer

In this voltaic cell, the anode is the Cd electrode, and the cathode is the Pt electrode. The Cd electrode loses mass as the cell reaction proceeds. The overall cell reaction is: \( \mathrm{Cd} + \mathrm{Cl}_{2} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{Cl}^{-} \). Under standard conditions, the emf generated by the cell is 1.76 V.

Step 1: Identify anode and cathode

A voltaic cell contains two half-cells. In each half-cell, a redox reaction occurs. The electrode where oxidation occurs is called the anode, and the electrode where reduction occurs is called the cathode. In this cell, we have a Cd electrode in a Cd(NO3)2 solution and a Pt electrode in a NaCl solution with Cl2 gas. Cadmium half-cell (Cd to Cd2+): \[ \mathrm{Cd} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{e}^{-} \] Chlorine half-cell (Cl2 to 2Cl-): \[ \mathrm{Cl}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-} \] Here, Cd is being oxidized and Cl2 is being reduced. Thus: - Anode: Cd electrode - Cathode: Pt electrode

Step 2: Determine mass change in Cd electrode

Since the Cd electrode serves as the anode, it undergoes oxidation, losing electrons and converting into Cd2+. Therefore, the Cd electrode loses mass as the cell reaction proceeds.

Step 3: Write the overall cell reaction

Next, we combine the half-reactions to write the balanced overall cell reaction. \[ \mathrm{Cd} + \mathrm{Cl}_{2} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{Cl}^{-} \]

Step 4: Calculate the emf under standard conditions

To calculate the emf under standard conditions, we use the following equation: \[ E_{cell} = E_{cathode} - E_{anode} \] Using the standard reduction potentials for the two half-reactions: - Cd2+ + 2e → Cd: \( E^{\circ} = -0.40V \) (anode) - Cl2 + 2e → 2Cl-: \( E^{\circ} = +1.36V \) (cathode) By substituting the values, \[ E_{cell} = (1.36) - (-0.40) = 1.76V \] The emf generated by the voltaic cell under standard conditions is 1.76 V.

Key Concepts

Redox Reactions

Voltaic cells operate through redox reactions, which are chemical processes where electrons are transferred between species. In a redox reaction, one element loses electrons, which is known as oxidation, and another gains electrons, which is referred to as reduction. These reactions are vital not only for the function of voltaic cells but also in various natural and industrial processes.
In a typical voltaic cell, you'll encounter two half-reactions: one for oxidation and one for reduction. For example:

• The cadmium half-cell: \[ \mathrm{Cd} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{e}^{-} \]
• The chlorine half-cell: \[ \mathrm{Cl}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-} \]

These half-reactions showcase how cadmium loses electrons to form Cd²⁺ ions while chlorine gains electrons to form Cl⁻ ions. Together, they complete the cell's redox reaction, which enables the voltaic cell to produce electrical energy.

Anode and Cathode

In a voltaic cell, electrodes are crucial components – specifically, the anode and cathode. Identifying their roles in a cell gives insight into the overall function and direction of electron flow.
The **anode** is where oxidation occurs. Here, the substance loses electrons. In our example, the cadmium electrode serves as the anode, demonstrated by the reaction:\[ \mathrm{Cd} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{e}^{-} \].
Electrons flow from the anode, as it undergoes oxidation, causing it to gradually lose mass.
The **cathode**, on the other hand, is where reduction happens. Here, the substance gains electrons. In our voltaic cell, the platinum electrode acts as the cathode via the reaction:\[ \mathrm{Cl}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-} \].
The electrons are accepted here, driving the reduction process.
Understanding these concepts is essential to comprehend the working of electrochemical cells, where the flow of electrons from anode to cathode generates electric current.

Standard Electrode Potentials

Standard electrode potentials, abbreviated as E°, are measured in volts (V) and indicate the tendency of a substance to lose or gain electrons under standard conditions. These potentials are essential for determining the voltage produced by voltaic cells.
In the context of our example, the standard electrode potential for each half-reaction provides insight into the overall reaction's feasibility. We see the standard potentials represented as:

• Cd²⁺ + 2e⁻ → Cd: \( E^{\circ} = -0.40V \)
• Cl₂ + 2e⁻ → 2Cl⁻: \( E^{\circ} = +1.36V \)

The negative value for cadmium implies it more readily loses electrons (serving as the anode), while the positive value for chlorine shows a stronger propensity to gain electrons (making it the cathode).
To find the emf of the cell, we apply the formula:
\[ E_{cell} = E_{cathode} - E_{anode} \]
By substituting our values:
\[ E_{cell} = (1.36V) - (-0.40V) = 1.76V \]
Hence, this measurement of 1.76 V represents the maximum potential difference a cell can achieve under standard conditions, indicating its capacity to perform electrical work.