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 \(\mathrm{NaCl}\) 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?

Short answer

(a) Anode: Cadmium electrode; Cathode: Platinum electrode. (b) The Cd electrode loses mass as the cell reaction proceeds. (c) The overall cell reaction: \(Cd + Cl_2 \rightarrow Cd^{2+} + 2Cl^-\). (d) The emf generated by the cell under standard conditions is 1.763 V.

Step-by-step solution

Identify the half-reactions

Considering the given information, we can identify two half-reactions: (i) Cd in the cadmium electrode reacts with nitrate ions: \[Cd \rightarrow Cd^{2+} + 2e^-\] (ii) Cl2 gas reacts with chloride ions at the platinum electrode: \[Cl_2 + 2e^- \rightarrow 2Cl^-\]

Determine the direction of electron flow

To determine the direction of electron flow, we need to find the standard electrode potentials (E°) for each half-reaction. Consulting a table of standard reduction potentials, we get: (i) \(Cd^{2+} + 2e^- \rightarrow Cd\), \(E°_{Cd} = -0.403 V\) (ii) \(Cl_2 + 2e^- \rightarrow 2Cl^-\), \(E°_{Cl} = +1.36 V\) Since Cl2 has a higher reduction potential, the electrons will flow towards the chlorine half-cell, and thus the direction of the electron flow is from the cadmium electrode to the platinum electrode.

Write the net cell equation

Combine the two balanced half-reactions and cancel out the electrons to write the net cell equation: \[Cd + Cl_2 \rightarrow Cd^{2+} + 2Cl^-\]

Determine the anode and cathode

As the electrons flow from the cadmium electrode, it is the anode (oxidation takes place). The platinum electrode, where the reduction of Cl2 takes place, is the cathode. (a) Anode: Cadmium electrode; Cathode: Platinum electrode

Determine if the Cd electrode gains or loses mass

As the cell reaction proceeds, Cd metal is oxidized to Cd2+ ions, thereby losing mass. (b) The Cd electrode loses mass as the cell reaction proceeds.

Calculate the overall cell potential

Calculate the emf generated by the cell under standard conditions using the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\) \(E°_{cell} = 1.36 V - (-0.403 V) = 1.763 V\) (d) The emf generated by the cell under standard conditions is 1.763 V.

Key concepts

Electrode Potentials

Electrode potentials are crucial in understanding voltaic cells since they determine the direction of electron flow. In a voltaic cell, each half-reaction has its own potential, commonly referred to as its standard electrode potential ( E° ). This potential indicates a substance's tendency to gain or lose electrons, termed as reduction or oxidation, respectively.
To evaluate these potentials, we typically consult a standard electrode potential table. Each entry in the table provides the potential for a half-reaction under standardized conditions. For the given voltaic cell, the reaction involving cadmium ( Cd^{2+} + 2e^- ightarrow Cd ) has a standard electrode potential of -0.403 V. This negative value suggests that cadmium is more prone to oxidation.
Conversely, the electrode potential for the reaction involving chlorine ( Cl_2 + 2e^- ightarrow 2Cl^- ) is +1.36 V, showing a strong tendency for reduction. Based on the more positive potential, electrons will naturally move towards the half-cell with the higher positive potential—here, the chlorine half-cell.

Half-Reactions

Voltaic cells operate through two separate half-reactions happening simultaneously. Each half-reaction occurs in its respective half-cell. Understanding these half-reactions helps predict the overall chemical change in the voltaic cell.
In our example, the first half-reaction is the oxidation of cadmium: - Cd transforms to Cd^{2+} ions while releasing electrons: Cd ightarrow Cd^{2+} + 2e^- This signifies the anode where oxidation happens.
- The second half-reaction involves chlorine gas gaining electrons to form chloride ions: Cl_2 + 2e^- ightarrow 2Cl^- This represents the cathode where reduction takes place.
The combination of these half-reactions leads to the overall cell reaction. It's vital to balance the electrons in the half-reactions to ensure that the total number of electrons gained equals the number lost. When combined appropriately, as in our example, they yield: - Cd + Cl_2 ightarrow Cd^{2+} + 2Cl^-

Salt Bridge

The salt bridge is a vital component of voltaic cells, serving to maintain electrical neutrality. In our cadmium-chlorine cell, the salt bridge connects the two half-cells, allowing the voltaic process to continue smoothly.
A salt bridge typically contains a gel or liquid with salt ions, which doesn't affect the chemical reactions occurring in either half-cell. Its main role is to allow the ions to migrate. This migration ensures that as the reaction proceeds, charge balance is maintained in both half-cells.
Without the salt bridge, the solutions would eventually become charged, halting the electrical circuit. Essential ions from the salt bridge move to counterbalance this, supporting uninterrupted electron flow across the electrode and maintaining the cell's operation.

Standard Conditions

Standard conditions provide a baseline for comparing cell potential values, crucial for consistency in electrochemical measurements. Standard conditions are defined as 1 M concentration for aqueous solutions, a pressure of 1 atmosphere for gases, and a temperature of 25°C (298 K).
These parameters allow for a uniform comparison of cells' standard electrode potentials found in literature.
For our voltaic cell involving cadmium and chlorine, operating under these standard conditions provides an expected cell potential of 1.763 V.
This calculated potential ( E°_{cell} ) is derived using the formula: - E°_{cell} = E°_{cathode} - E°_{anode} Plugging in the values from our reaction: - E°_{cell} = 1.36 V - (-0.403 V) = 1.763 V This helps to identify a cell's feasibility to do work and further predict the cell's behavior under different conditions.