Question

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

Short answer

To summarize the solution: (a) The two half-cell reactions are: 1. \(\mathrm{PdCl}_{4}^{2-}(a q) \longrightarrow\mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+2 e^{-}\) (Pd half-reaction) 2. \(\mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+2 e^{-}\) (Cd half-reaction) (b) The standard reduction potential for the Pd-containing reaction is: \(E_{\text {Pd-red }}^{\circ} = 1.43\ \text{V}\) (c) In the sketched voltaic cell, the anode is Cd (where oxidation occurs), and the cathode is Pd (where reduction occurs). The electrons flow from the anode to the cathode through the external wire. The salt bridge connects the two half-cells, balancing the charges and ensuring electrical neutrality.

Step-by-step solution

Write the two half-cell reactions

First, we need to determine the two half-cell reactions. One of them involves Cd and the other one involves Pd. From the overall reaction, we can break them into two half-reactions: 1. Pd-containing half-reaction: \[ \mathrm{PdCl}_{4}^{2-}(a q) \longrightarrow\mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+2 e^{-} \] Here, Pd is being reduced as it gains 2 electrons. 2. Cd-containing half-reaction: \[ \mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+2 e^{-} \] In this half-reaction, Cd is being oxidized as it loses 2 electrons.

Determine the standard reduction potential (E°) for the Pd-containing reaction

From Appendix E, we can find the standard reduction potential (E°) for the Cd-containing reaction as: \[ \mathrm{Cd^{2+}(a q)+2 e^{-}} \longrightarrow \mathrm{Cd}(s);\quad E_{\text { red }}^{\circ} = -0.40\ \text{V} \] We are also given the standard cell potential (E°cell) for the voltaic cell: \[ E_{\text { cell }}^{\circ} = +1.03\ \text{V} \] Using the Nernst equation for standard cell potential, \[ E_{\text { cell }}^{\circ} = E_{\text { cathode }}^{\circ} -E_{\text { anode }}^{\circ} \] Since the standard cell potential is positive, the Pd-containing half-reaction must have the higher reduction potential and therefore must be the cathode. The Cd-containing half-reaction is then the anode. Therefore, we can write the Nernst equation as: \[ E_{\text { cell }}^{\circ} = E_{\text {Pd-red }}^{\circ} - (-0.40\ \text{V}) \] Now solve for the standard reduction potential (E°) for the Pd-containing reaction: \[ E_{\text {Pd-red }}^{\circ} = E_{\text { cell }}^{\circ} +0.40\ \text{V} = 1.03\ \text{V} + 0.40\ \text{V} = 1.43\ \text{V} \]

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

To sketch the voltaic cell, you can follow these steps: 1. Draw two containers, one for the anode half-cell and the other for the cathode half-cell. 2. In the anode half-cell (left side), place a Cd electrode, and label it "Anode" along with the Cd half-reaction: \[ \mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+2 e^{-} \] 3. In the cathode half-cell (right side), place a Pd electrode, and label it "Cathode" along with the Pd half-reaction: \[ \mathrm{PdCl}_{4}^{2-}(a q) \longrightarrow\mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+2 e^{-} \] 4. Connect the two electrodes with an external wire to allow electron flow from the anode (Cd) to the cathode (Pd), and place the voltmeter across the wire. (Don't forget to draw an arrow showing the electron flow direction from anode to cathode). 5. Add a salt bridge between the two cells to ensure electrical neutrality. The salt bridge contains ions that can flow between the two half-cells, balancing the charges. In the end, your sketched voltaic cell should show two half-cells with their respective electrodes, the external wire with an arrow indicating the electron flow, and the salt bridge.

Key concepts

Standard Cell Potential

When studying electrochemistry and voltaic cells, the standard cell potential (E°cell) is a crucial concept representing the voltage difference between two half-cells. It determines how much electrical energy is produced by the complete redox reaction occurring within the cell. The ability to harness this energy is what makes batteries and similar devices function.

The standard cell potential is calculated under standard conditions, which include a temperature of 298 K, a 1 M concentration for all aqueous solutions, and a pressure of 1 atm for any gases involved. The E°cell is the difference between the potential at the cathode and the anode. If the E°cell is positive, as in the exercise, it indicates a spontaneous reaction, meaning the voltaic cell will naturally produce electric current.

Half-Cell Reactions

A voltaic cell works through half-cell reactions, which are simply the division of the overall redox reaction into two parts—one part that involves oxidation and the other involving reduction. In the given problem, cadmium (Cd) is oxidized, losing electrons, whereas palladium (Pd) is reduced, gaining electrons.

The division into half-cells allows us to analyze each electrode's activity separately. The electrode at which oxidation occurs is called the anode and the electrode at which reduction occurs is termed the cathode. Electrons flow from the anode to the cathode through an external circuit, generating electricity. Each half-cell's activity is essential for the proper functioning of the cell as a whole.

Standard Reduction Potential

The standard reduction potential (E°red) is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Each half-reaction has a standard reduction potential, which is the potential of a reduction reaction measured relative to the standard hydrogen electrode (SHE), assumed to be zero volts by convention.

The numerical value of E°red is crucial in determining the direction of electron flow in a cell. The substance with the higher (more positive) E°red becomes the cathode, while the lower (more negative) becomes the anode. This is precisely the reason the Pd-containing reaction in our problem is the cathode since it has the higher E°red compared to the Cd-containing reaction, which in turn becomes the anode.

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that allows us to calculate the cell potential under non-standard conditions. While the standard cell potential gives us a snapshot at the ideal scenario, the Nernst equation accounts for actual concentrations of reactants and products, temperature, and other conditions that deviate from the standard.

Mathematically expressed as:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is the Faraday's constant, and Q is the reaction quotient. For the example provided, the Nernst equation is used to establish the relationships between cell potential and half-cell reactions to determine the Pd-containing reaction's standard reduction potential.