Question

A battery is assembled using tin and mercury, which have the following reduction half-reactions: $$ \begin{array}{l}{\mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}} \\\ {\mathrm{Hg}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}}\end{array} $$ a. Write a balanced equation for the cell's reaction. b. What is oxidized and what is reduced? Identify the oxidizing agent and the reducing agent. c. Which reaction occurs at the anode? At the cathode? d. What is the cell potential? Use Table 20.1 e. If sodium sulfate solution is in the salt bridge, in which direction do the sulfate ions move?

Short answer

The overall balanced equation for the cell's reaction is: \( \mathrm{Sn} + \mathrm{Hg}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Hg} \) Tin (Sn) is oxidized and acts as the reducing agent. Mercury (Hg^(2+)) is reduced and acts as the oxidizing agent. The reaction at the anode is \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+}+2 \mathrm{e}^- \), and the reaction at the cathode is \( \mathrm{Hg}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg} \). The cell potential is -0.99 V. Sulfate ions (SO4^2-) move towards the anode to maintain electrical neutrality.

Step-by-step solution

Write a balanced equation for the cell's reaction

To find the overall cell reaction, we need to combine the two half-reactions in a way that the electrons cancel out: $$ \begin{array}{l}{\mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}} \\ {\mathrm{Hg}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}}\end{array} $$ First, reverse one of the half-reactions to make it an oxidation half-reaction: $$ \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+}+2 \mathrm{e}^- $$ Now, add the two half-reactions: $$ \mathrm{Sn} + \mathrm{Hg}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Hg} $$ The overall balanced chemical equation for the cell's reaction is: $$ \mathrm{Sn} + \mathrm{Hg}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Hg} $$

Identify what is oxidized and reduced, and the oxidizing and reducing agents

In the overall balanced equation: $$ \mathrm{Sn} + \mathrm{Hg}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Hg} $$ - Tin (Sn) is oxidized since it loses electrons and forms Sn^(2+). - Mercury (Hg^(2+)) is reduced since it gains electrons to form Hg. The oxidizing agent is the species that is reduced, which is Hg^(2+), and the reducing agent is the species that is oxidized, which is Sn.

Identify the reactions at the anode and cathode

Since oxidation occurs at the anode, the reaction at the anode is: $$ \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+}+2 \mathrm{e}^- $$ Since reduction occurs at the cathode, the reaction at the cathode is: $$ \mathrm{Hg}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg} $$

Calculate the cell potential

Using Table 20.1, we find the standard reduction potentials for the given half-reactions: For Sn^(2+) + 2e^(-) -> Sn, E°(cathode) = -0.14 V For Hg^(2+) + 2e^(-) -> Hg, E°(anode) = +0.85 V The cell potential (E°(cell)) is calculated as the difference between the cathode potential and the anode potential: $$ E°(\text{cell}) = E°(\text{cathode}) - E°(\text{anode}) $$ $$ E°(\text{cell}) = -0.14 - 0.85 $$ $$ E°(\text{cell}) = -0.99 \, \text{V} $$

Determine the movement of sulfate ions in the salt bridge

In the salt bridge, ions move to maintain electrical neutrality. In this case, Sn is losing two electrons, creating a positive charge (Sn^2+) in the oxidation half-cell (anode), while Hg^2+ in the reduction half-cell (cathode) gains two electrons to form neutral Hg. To maintain electrical neutrality, the negatively charged sulfate ions (SO4^2-) in the sodium sulfate salt bridge will move towards the positively charged Sn^2+ ions produced in the anode.

Key concepts

Oxidation and Reduction

In the study of electrochemical cells, the concepts of oxidation and reduction are fundamental. Oxidation is defined as the loss of electrons by a species, while reduction refers to the gain of electrons. These processes always occur simultaneously; when one species undergoes oxidation, another species undergoes reduction. This phenomenon is encapsulated in the saying, 'OIL RIG': Oxidation Is Loss, Reduction Is Gain.

For instance, consider the tin (Sn) and mercury (Hg) battery from the exercise. Tin is transformed from its metallic form to the divalent ion Sn²⁺ by losing two electrons. This is an example of oxidation. Mercury ions Hg²⁺, on the other hand, receive two electrons and are reduced to metallic mercury. Such reactions are commonly written as half-reactions, showcasing the individual oxidation or reduction processes before they are combined to illustrate the overall reaction in an electrochemical cell.

Understanding which species is oxidized and which is reduced helps identify the oxidizing and reducing agents. The oxidizing agent is the substance that accepts electrons (is reduced) and the reducing agent is the one that donates electrons (is oxidized). This is crucial for correctly solving problems related to redox reactions and electrochemical cells.

Cell Potential

The cell potential, often denoted by E°(cell), is a measure of the driving force behind the electrical current produced by an electrochemical cell. It's a quantitative expression of the cell's ability to move electrons from the oxidizing agent to the reducing agent. Cell potential is measured in volts (V), a unit that represents the potential energy difference per unit charge.

The cell potential can be calculated by taking the difference between the reduction potentials of the cathode and the anode. To find this, standard reduction potentials from reference materials are used. These potentials are measured under standard conditions, which include concentrations of 1 M for all solutions, a pressure of 1 atm for gases, and a temperature of 25°C. In the case of the tin-mercury cell, the negative cell potential (E°(cell) = -0.99 V) indicates that the reaction is non-spontaneous under standard conditions. This means external energy would be required for the reaction to proceed in the direction written.

The cell potential is pivotal in predicting the feasibility of a redox reaction and is a key component in calculating the energy that can be harnessed from electrochemical processes, whether in batteries or in broader applications such as electroplating and electrolysis.

Half-Reactions

An electrochemical cell's operation can be broken down into half-reactions. Each half-reaction represents the process occurring at one of the two electrodes: the anode or the cathode. At the anode, oxidation takes place, where electrons are released into the external circuit. At the cathode, reduction occurs as the electrons return from the external circuit to be consumed by species at the electrode.

It's important to write each half-reaction separately to understand the individual processes happening at each electrode. The breakdown into half-reactions is also essential in determining the overall cell potential, as seen in the exercise with the tin (Sn) being oxidized to Sn²⁺ and mercury (Hg²⁺) being reduced to Hg.

In an operating cell, the anode and cathode are often connected by a salt bridge that maintains electrical neutrality by allowing ions to flow. The example provided in the exercise illustrates that the sulfate ions (SO₄^2-) would move towards the anode where the positive charge is being generated due to the oxidation of tin. This movement of ions is crucial for sustaining the flow of electrons through a circuit, hence powering the device the electrochemical cell is installed in.