Problem 41 - Chemistry Principles and Practice

Question

A platinum wire is in contact with a mixture of mercury and solid mercury(I) chloride \(\left(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\right)\) in a beaker containing \(1 M\) KCl solution. A salt bridge connects this halfcell to a beaker that contains a copper electrode immersed in \(1 \mathrm{M} \mathrm{CuSO}_{4}\) solution. Voltage measurements show that the copper electrode is positive. (a) Write balanced half-reactions for the two electrodes. (b) Write the equation for the spontaneous cell reaction. (c) In which direction do electrons flow in the external circuit? (d) Would direct reaction occur if both the \(\mathrm{Hg} / \mathrm{Hg}_{2} \mathrm{Cl}_{2}\) and copper electrodes were placed in a container holding an aqueous solution that is \(1 \mathrm{M} \mathrm{CuSO}_{4}\) and \(1 \mathrm{M} \mathrm{KCl}^{2}\)

Step-by-Step Solution

Answer

Electrons flow from platinum to copper. A direct reaction is possible without the salt bridge.

Step 1: Identify Half-Reactions

The half-reaction in the beaker with platinum wire involves mercury and mercury(I) chloride. The balanced half-reaction for this is \( \mathrm{Hg}_{2}\mathrm{Cl}_{2}(s) + 2e^- \rightarrow 2\mathrm{Hg}(l) + 2\mathrm{Cl}^- \). For the copper electrode, the balanced half-reaction is \( \mathrm{Cu}^{2+}(aq) + 2e^- \rightarrow \mathrm{Cu}(s) \).

Step 2: Determine Cell Reaction

To determine the spontaneous overall cell reaction, we combine the half-reactions: \( \mathrm{Hg}_{2}\mathrm{Cl}_{2}(s) + \mathrm{Cu}(s) \rightarrow 2\mathrm{Hg}(l) + 2\mathrm{Cl}^- + \mathrm{Cu}^{2+}(aq) \).

Step 3: Determine Electron Flow

Electrons flow from the anode to the cathode in the external circuit. Since the copper electrode is positive, it is the cathode. The electrons thus flow from platinum (anode) to copper (cathode).

Step 4: Consider Direct Reaction Possibility

If both electrodes are placed in a single container, allowing direct contact between \( \mathrm{Hg} / \mathrm{Hg}_{2}\mathrm{Cl}_{2} \) and \( \mathrm{Cu}^{2+} \), a direct reaction could occur, reducing \( \mathrm{Cu}^{2+} \) to \( \mathrm{Cu} \) and oxidizing \( \mathrm{Hg} \) to form \( \mathrm{Hg}_{2}\mathrm{Cl}_{2} \). This would eliminate the need for the salt bridge.

Key Concepts

Half-Reactions

In electrochemistry, a half-reaction is an essential concept. It involves either the oxidation or the reduction process in an electrochemical cell. In this particular scenario, one of the half-reactions involves mercury and mercury(I) chloride. The balanced half-reaction for the mercury electrode is \[ \mathrm{Hg}_{2}\mathrm{Cl}_{2}(s) + 2e^- \rightarrow 2\mathrm{Hg}(l) + 2\mathrm{Cl}^- \].This reaction shows the reduction of mercury ions to liquid mercury, releasing chloride ions in the process.
On the other hand, at the copper electrode, the half-reaction involves the reduction of copper ions. It is represented as \[ \mathrm{Cu}^{2+}(aq) + 2e^- \rightarrow \mathrm{Cu}(s) \].Both reactions occur simultaneously during the electrochemical process. Each employs electron exchange, forming the basis for the electricity generation in the cell.

Spontaneous Cell Reaction

A spontaneous cell reaction is the overall reaction that occurs naturally in an electrochemical cell without any external influence. It's the sum of the two half-reactions, indicating a natural flow of electrons from the anode to the cathode.
In this case, the spontaneous cell reaction is formed by combining the two half-reactions:\[ \mathrm{Hg}_{2}\mathrm{Cl}_{2}(s) + \mathrm{Cu}(s) \rightarrow 2\mathrm{Hg}(l) + 2\mathrm{Cl}^- + \mathrm{Cu}^{2+}(aq) \].This equation simplifies the process, showing how mercury(I) chloride and copper products are involved in the spontaneous generation of electrical energy. The driving force for this reaction is the potential difference between the two electrodes, which allows the cell to operate without additional energy input.

Electron Flow

Electron flow in an electrochemical cell is a key aspect of electricity generation. Electrons move through the external circuit, creating a current, from the anode to the cathode. In this problem, the platinum wire in contact with mercury acts as the anode, where oxidation takes place. Electrons are released here due to the oxidation of mercury ions.
These electrons then travel through the external circuit to the copper electrode, which acts as the cathode because it is positively charged. At the cathode, electrons are accepted in the reduction of copper ions to solid copper. This creates a flow of electrons, which is the very essence of electricity in this setup. Thus, the flow moves from the platinum (anode) to copper (cathode), fueling the spontaneous reactions.

Electrode Potentials

Electrode potentials are crucial in determining the direction and feasibility of an electrochemical reaction. Each electrode in the cell has a specific potential, which contributes to the overall cell voltage.
In this exercise, the copper electrode demonstrates a higher potential than the mercury system. This is why the copper electrode is the cathode, exhibiting a positive charge that attracts electrons. The potential difference between the two electrodes generates the electromotive force required for the reaction.

• The potential of the mercury electrode determines its ability to be oxidized.
• The potential of the copper electrode indicates its capacity to be reduced.

The greater the difference between these potentials, the stronger the driving force for the electron flow and the more spontaneous the cell reaction. Electrode potentials are measured in volts and can be referenced to standard potentials listed in electrochemical series tables, offering significant insights into the efficiency of batteries and cells.