Question

A voltaic cell that uses the reaction \(\mathrm{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q)\) has a measured standard cell potential of \(+1.03 \mathrm{~V}\). (a) Write the two half-cell reactions. (b) By using data from Appendix \(\mathrm{E}\), determine \(E_{\mathrm{red}}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

Short answer

The oxidation half-reaction is $\mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+ 2 e^{-}$ and the reduction half-reaction is $\mathrm{PdCl}_{4}^{2-}(a q)+ 2 e^{-} \longrightarrow \mathrm{Pd}(s)+ 4 \mathrm{Cl}^{-}(a q)$. The reduction potential for the Pd half-reaction is $1.43 \mathrm{~V}$. The voltaic cell has the anode as solid Cd, the cathode as Pd electrode and electrons flow from Cd to Pd.

Step-by-step solution

Identify oxidation and reduction half-reactions

In the given reaction, \[ \mathrm{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q) \] We can break this down into the following half-reactions: 1. \(\mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+ 2 e^{-}\) (oxidation) 2. \( \mathrm{PdCl}_{4}^{2-}(a q)+ 2 e^{-} \longrightarrow \mathrm{Pd}(s)+ 4 \mathrm{Cl}^{-}(a q)\) (reduction)

Determine the reduction potential for the Pd half-reaction

The given standard cell potential is \(+1.03 \mathrm{~V}\). To determine the reduction potential for the Pd half-reaction, we will first find the reduction potential for the oxidation half-reaction (Cd) from Appendix E - Standard Reduction Potentials. From Appendix E, the standard reduction potential for the Cd half-reaction is: \[ \mathrm{Cd}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Cd}(s); E_{\mathrm{red}}^{\circ} = -0.40 \mathrm{~V} \] Since the Cd is actually being oxidized, we will use the opposite of the reduction potential to obtain its oxidation potential: \[ E_{\mathrm{ox}}^{\circ}(\mathrm{Cd}) = -E_{\mathrm{red}}^{\circ}(\mathrm{Cd}) = +0.40 \mathrm{~V} \] Now, using the standard cell potential given, we can determine the reduction potential for the Pd half-reaction, \(E_{\mathrm{red}}^{\circ}(\mathrm{Pd})\): \(E_{\mathrm{cell}}^{\circ}=E_{\mathrm{red}}^{\circ}(\mathrm{Pd})-E_{\mathrm{ox}}^{\circ}(\mathrm{Cd})\) Now plug the values and solve: \(1.03\mathrm{~V}=E_{\mathrm{red}}^{\circ}(\mathrm{Pd}) - 0.40 \mathrm{~V}\) \(E_{\mathrm{red}}^{\circ}(\mathrm{Pd})=1.43 \mathrm{~V}\) So, the reduction potential for the reaction involving Pd is \(1.43 \mathrm{~V}\).

Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow

To sketch the voltaic cell, we need to label the anode and the cathode, as well as indicate the direction of electron flow: 1. Anode (oxidation occurs): Solid Cd 2. Cathode (reduction occurs): PdCl2 (aq) The electron flow will be from the anode (Cd) to the cathode (Pd). In our sketch, we will have two containers (half-cells). In one container, we will have solid Cd as the anode, immersed in a solution containing Cd2+ ions. In the other container, we will have Pd electrode, immersed in a solution containing PdCl42- ions. A salt bridge will be connecting the two containers to balance the charges, while a wire connects the electrodes, indicating the direction of electron flow from Cd to Pd.

Key concepts

Electrochemical Reactions

Electrochemical reactions are chemical processes where electrons are transferred between substances, causing oxidation and reduction. These reactions are fundamental to the functioning of a voltaic cell. In such a cell, a chemical reaction creates electrical energy. This is made possible when redox reactions occur in separate compartments called half-cells. The structure of a voltaic cell ensures the movement of electrons through an external circuit, completing the electrochemical process.
Electrochemical reactions are characterized by their ability to either release or consume energy as they progress. This energy release is what allows devices like batteries to provide power. Understanding these reactions involves examining how electrons move and how this movement influences the flow of electricity, both key aspects of voltaic cells.

Standard Cell Potential

The standard cell potential is the measure of the driving force behind the cell reaction under standard conditions. It is calculated from the potentials of the two half-reactions in the voltaic cell. The cell potential is a critical value indicating the voltage output of the cell when it operates under standard conditions - typically 25°C, 1 atm pressure, and 1 M concentration for all solutions involved.
The overall standard cell potential, denoted as \(E_{\mathrm{cell}}^{\circ}\), is determined using the equation:
\[E_{\mathrm{cell}}^{\circ} = E_{\mathrm{red}}^{\circ}(cathode) - E_{\mathrm{ox}}^{\circ}(anode)\]
This relationship helps us understand the balance between reduction and oxidation at each electrode. When the standard cell potential is positive, the reaction is spontaneous and, thus, the voltaic cell can generate electricity.

Half-Cell Reactions

Half-cell reactions include separate processes of oxidation and reduction that occur at each electrode in a voltaic cell. Each half-cell reaction contributes to the overall energy conversion inside the cell.
The oxidation half-reaction is where a substance loses electrons, often releasing energy. For example, in the given voltaic cell, cadmium (Cd) is oxidized to cadmium ions ( Cd^{2+}), losing electrons in the process: \[\mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(aq) + 2e^{-}\]
Conversely, the reduction half-reaction is where a substance gains electrons, absorbing energy. In our example, palladium (\mathrm{Pd}) ions gain electrons to form solid palladium.
By understanding half-cell reactions, you can decipher the complete redox process at play and calculate critical parameters like the cell potential.

Reduction Potential

Reduction potential is a measure of the tendency of a chemical species to acquire electrons and be reduced. It is an essential concept in understanding how and why certain substances act as oxidizing agents. This potential is quantified by comparing it to the standard hydrogen electrode.
The reduction potential of a half-reaction can be found in tables like Appendix E, where each half-reaction is given a value under standard conditions. The more positive the reduction potential, the more a substance tends to gain electrons and undergo reduction.
In our reaction, the reduction potential of the Pd half-reaction is \(1.43 \mathrm{~V}\). This reveals Pd's strong ability to accept electrons, highlighting its role as the cathode in the voltaic cell setup.

Oxidation and Reduction

Oxidation and reduction, or redox reactions, are processes that occur simultaneously in an electrochemical cell. Oxidation involves the loss of electrons from one substance, while reduction involves the gain of electrons by another. These two processes together allow for electron flow, generating electrical energy in voltaic cells.
Key characteristics to remember:

• Oxidation is losing electrons (increase in oxidation number).
• Reduction is gaining electrons (decrease in oxidation number).

In a voltaic cell, the oxidation occurs at the anode, and reduction occurs at the cathode. This electron exchange across the electrodes is fundamental for the cell's ability to provide power. Understanding these concepts assists in identifying the flow of electrons which is critical for any electrochemical setup.