Question

The cell reaction \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)+2 \mathrm{Hg}(l)\), is best represented by : (a) \(\mathrm{Cu}(s)\left|\mathrm{Cu}^{+2}(a q) \| \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\right| \mathrm{Hg}(l)\) (b) \(\mathrm{Cu}(s)\left|\mathrm{Cu}^{+2}(a q) \| \mathrm{Hg}(\mathrm{l})\right| \mathrm{HgCl}_{2}(s)\) (c) \(\mathrm{Cu}(s)\left|\mathrm{Cu}^{+2}(a q) \| \mathrm{Cl}^{-}(a q)\right| \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)|\operatorname{Hg}(l)| \operatorname{Pt}(s)\) (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\left|\mathrm{Cl}^{-}(a q) \| \mathrm{Cu}^{+2}(a q)\right| \mathrm{Cu}(s)\)

Short answer

The correct representation of the cell reaction is (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)|\mathrm{Cl}^{-}(aq) || \mathrm{Cu}^{2+}(aq) | \mathrm{Cu}(s)\).

Step-by-step solution

Understand the Cell Notation

The notation for an electrochemical cell is written in the format of anode | anode solution || cathode solution | cathode, where the single line '|' represents a phase boundary, and the double line '||' represents the salt bridge or a porous separator that allows the flow of ions.

Identify the Anode Reaction

Determine the oxidation half-reaction, which occurs at the anode. In this reaction, the oxidation occurs when mercury in \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) is oxidized to elemental mercury (Hg(l)), releasing chloride ions (Cl^-). The correct half-reaction is:\[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) \longrightarrow 2\mathrm{Hg}(l) + 2\mathrm{Cl}^{-}(aq)\]

Identify the Cathode Reaction

Determine the reduction half-reaction, which occurs at the cathode. In this reaction, the reduction occurs when copper (Cu(s)) is oxidized to \(\mathrm{Cu}^{2+}(aq)\). The correct half-reaction is:\[\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(aq) + 2e^-\]

Combine the Half-Reactions

Combine both half-reactions, ensuring to place the anode reaction on the left side and the cathode reaction on the right side in the cell notation. The correct cell notation would be:\[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)|\mathrm{Cl}^{-}(aq) || \mathrm{Cu}^{2+}(aq) | \mathrm{Cu}(s)\]

Select the Correct Representation

Looking at the options given, the correct representation of the cell reaction that matches our determined cell notation is: (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)|\mathrm{Cl}^{-}(aq) || \mathrm{Cu}^{2+}(aq) | \mathrm{Cu}(s)\).

Key concepts

Cell Reactions

Understanding cell reactions is essential when studying electrochemistry. An electrochemical cell consists of two half-cells, where chemical reactions occur. These reactions involve transfers of electrons, which move through an external circuit from one side of the cell to the other. The reaction in each half-cell involves either oxidation or reduction processes.

For the given exercise, we have the reaction \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) + \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(aq) + 2 \mathrm{Cl}^{-}(aq) + 2 \mathrm{Hg}(l)\).This represents a combination of two half-reactions occurring at respective electrodes. By carefully breaking down the reaction into these halves, we identify what is being oxidized and what is being reduced. In simplifying the reaction, we ensure students can see how electrons are transferred and the role each species plays within the cell.

Oxidation and Reduction

Oxidation and reduction are two key concepts in chemistry that are the heart of the reactions in electrochemical cells. Oxidation is the loss of electrons from a substance, while reduction is the gain of electrons. Remembering 'LEO the lion says GER'—Lose Electrons Oxidation, Gain Electrons Reduction—can be a handy mnemonic to recall this.

In the exercise, oxidized mercury from mercury(I) chloride (\(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\)) turns into liquid mercury (Hg(l)), giving up electrons (which constitutes oxidation), and these lost electrons are accepted by copper (Cu), forming copper ions (\(\mathrm{Cu}^{2+}\)), which is reduction. Thus, we have:\[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) \longrightarrow 2\mathrm{Hg}(l) + 2\mathrm{Cl}^{-}(aq)\], as the oxidation half-reaction, and\[\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(aq) + 2e^-\], as the reduction half-reaction. The flow of electrons from oxidation to reduction drives the current through the circuit attached to the cell.

Anode and Cathode Identification

Identifying the anode and cathode within a cell reaction is crucial for understanding the direction of electron flow and reaction mechanism. The anode is where oxidation occurs and cathode is where reduction takes place. These are easily confused, so it's important to note that 'An Ox' and 'Red Cat' (Anode Oxidation, Reduction Cathode) can help students to remember.

According to the step-by-step solution provided, the anode reaction involves the oxidation of mercury(I) chloride to mercury and chloride ions, and the cathode reaction is the reduction of copper to copper ions. Therefore, mercury(I) chloride represents the anode and copper represents the cathode. The correct cell notation format places the anode on the left and the cathode on the right, separated by a double vertical line (||), signifying a salt bridge or barrier that permits ionic conduction but prevents the mixing of different solutions that would lead to direct chemical reaction of the cell constituents.