Question

A voltaic cell that uses the reaction $$ \mathrm{PdCl}_{4}{ }^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{CT}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{~V}\). (a) Write the two half-cell reactions. (b) By using data from Appendix \(E\), determine \(E_{\text {rel }}^{0}\) for the reaction involving \(\mathrm{Pd}\). (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

Short answer

(a) Oxidation half-cell reaction: \(Cd(s) → Cd^{2+}(a q) + 2e^-\). Reduction half-cell reaction: \(PdCl_4^{2-}(a q) + 4e^-→ Pd(s) + 4Cl^-(a q)\) (b) The standard reduction potential for Pd is \(E_{PdCl_4^{2-}/Pd(s)}^{0} = 1.43 V\). (c) Sketch a voltaic cell with an anode (Cd) and a cathode (Pd), connect the metals with a wire, indicate electron flow from anode to cathode, and include a salt bridge.

Step-by-step solution

Write the half-cell reactions

To find the half-cell reactions, we need to first find the species that are being reduced and oxidized in the overall cell reaction. The general form of a half-cell reaction is \(M^n+ + ne^- → M\). We notice that \(Cd(s)\) is turning into \(Cd^{2+}(a q)\) and loses 2 electrons in the process, so it's being oxidized. Therefore, we can write the oxidation half-cell reaction as: \[ Cd(s) → Cd^{2+}(a q) + 2e^-\] Next, we will write the counterpart reduction half-cell reaction. We see that \(PdCl_4^{2-}(a q)\) is turning into Pd (s) and gains 4 electrons in the process. The reduction reaction will be: \[PdCl_4^{2-}(a q) + 4e^-→ Pd(s) + 4Cl^-(a q)\]

Determine the standard reduction potential for Pd

To determine the standard reduction potential (SRP) for Pd, we can use the standard cell potential equation and the SRP for Cd from the Appendix E provided. The standard cell potential equation is: \[E_{cell}^{0} =E_{cathode}^{0} - E_{anode}^{0}\] The measured standard cell potential is \(E_{cell}^{0} = +1.03V\). From Appendix E, we find the standard reduction potential of \(Cd(s)\) to be \(E_{Cd^{2+}/Cd(s)}^{0} = -0.40V\). Since Cadmium is being oxidized, we plug this as the anode value: \[1.03V = E_{PdCl_4^{2-}/Pd(s)}^{0} - (-0.40V)\] Now we can solve for the SRP of Pd: \[E_{PdCl_4^{2-}/Pd(s)}^{0} = 1.03V + 0.40V = 1.43 V\]

Sketch the voltaic cell

To sketch the voltaic cell, you will create a representation of two half-cells under standard conditions. The anode is where the oxidation reaction occurs, and we determined that it is the half-cell containing Cd(s). The cathode is where the reduction reaction occurs, which is the half-cell containing PdCl_4^{2-}(a q) and Pd(s). 1. Start by drawing two beakers. 2. Label the left beaker "Anode: Cd(s)" and draw a piece of cadmium metal in it. In the same beaker, draw the solution containing Cd^2+ ions and label it as "Cd^2+(aq)". 3. Label the right beaker "Cathode: Pd(s)" and draw a pellet of palladium metal in it. In the same beaker, draw the solution containing PdCl_4^{2-}(aq) ions and Cl^-(aq). 4. Connect the two metals with a wire, and indicate the direction of electron flow from anode to cathode with an arrow. 5. Insert a salt bridge between the two solutions to allow for ion flow and complete the cell. In the completed voltaic cell sketch, we should see oxidation occurring at the anode (Cd(s) turning into Cd^2+(aq) and losing two electrons), and reduction occurring at the cathode (PdCl_4^{2-}(aq) gaining electrons and turning into Pd(s)). The electron flow should be from the anode (Cd) to the cathode (Pd).

Key concepts

Half-cell Reactions

In a voltaic cell, the process of conversion from chemical to electrical energy hinges on the reactions occurring at each electrode, known as half-cell reactions. For the given reaction \( \text{PdCl}_{4}^{2-}(aq) + \text{Cd}(s) \longrightarrow \text{Pd}(s) + 4 \text{Cl}^-(aq) + \text{Cd}^{2+}(aq) \), we first need to identify the species undergoing oxidation and reduction.

Oxidation occurs when a species loses electrons. Observing the reaction, we see that \( \text{Cd}(s) \) is oxidized to \( \text{Cd}^{2+}(aq) \). This results in the loss of two electrons, leading to the oxidation half-reaction:

• \( \text{Cd}(s) \rightarrow \text{Cd}^{2+}(aq) + 2e^- \)

Reduction involves the gain of electrons. Here, \( \text{PdCl}_{4}^{2-}(aq) \) is reduced to \( \text{Pd}(s) \), involving the gain of four electrons. The reduction half-reaction is therefore:

• \( \text{PdCl}_{4}^{2-}(aq) + 4e^- \rightarrow \text{Pd}(s) + 4 \text{Cl}^- \)

It is vital to balance the electron transfer in both half-reactions for maintaining charge neutrality in the cell. By understanding these processes, one gains insight into the core functioning of a voltaic cell.

Standard Cell Potential

The standard cell potential \( E_{\text{cell}}^{0} \) of a voltaic cell is a measure that indicates the potential difference between the two electrodes. It reflects the voltage generated by the cell at standard conditions (1 M concentration, 1 atm pressure, and 25°C). For the cell described, the potential is measured at \(+1.03 \text{ V}\).

Calculating the standard cell potential involves using the equation:

• \( E_{\text{cell}}^{0} = E_{\text{cathode}}^{0} - E_{\text{anode}}^{0} \)

Using this, we can calculate for unknown values. We’re given that the reduction potential for cadmium, \( E_{\text{anode}}^{0} \), is \(-0.40 \text{ V}\) because it is oxidized. To find the potential for the reaction involving \( \text{Pd} \), we rearrange the equation:
\( 1.03 \text{ V} = E_{\text{cathode}}^{0} - (-0.40 \text{ V}) \)
Solving this yields:

• \( E_{\text{cathode}}^{0} = 1.43 \text{ V} \)

This potential value is a positive indicator of a spontaneous and favorable electrochemical reaction within the cell.

Electrode Sketching

Creating a visual representation of a voltaic cell helps grasp the layout and flow of electrons. To sketch our cell, we need to distinguish roles played by the anode and cathode.

Start by envisioning two containers filled with electrolyte solutions, one for each half-cell. The left container represents the anode, where cadmium \( \text{Cd}(s) \) undergoes oxidation, producing \( \text{Cd}^{2+}(aq) \). The right container is the cathode, showcasing the reduction of \( \text{PdCl}_{4}^{2-}(aq) \) to \( \text{Pd}(s) \).

• Draw a cadmium electrode (solid piece of cadmium) in the anode's container. Label it as \( \text{Cd}(s) \), surrounded by \( \text{Cd}^{2+}(aq) \) ions.
• In the cathode's container, draw a palladium electrode, labeled \( \text{Pd}(s) \). This container should contain \( \text{PdCl}_{4}^{2-}(aq) \) and \( \text{Cl}^-(aq) \).

Connect these electrodes with a wire to establish the path for electron flow, and use an arrow to indicate the electrons moving from the anode to the cathode. There should also be a salt bridge depicted between the containers, allowing for ion exchange to sustain charge balance.

This visual aid emphasizes electron movement, with constructs like the salt bridge facilitating the continuous function of the voltaic cell by maintaining electroneutrality, which is crucial for ongoing reaction and electron flow.