Question

When someone who has a silver filling in a tooth bites down on an aluminum gum wrapper, saliva acts as an electrolyte. The system is an electrochemical cell that produces a small jolt of pain. Explain what occurs, using half-cell reactions and \(E^{0}\) values.

Short answer

Aluminum oxidizes, silver reduces, and the 2.46 V cell potential causes the pain.

Step-by-step solution

- Identify the metals involved

Recognize that the problem involves silver (Ag) and aluminum (Al) as the two metals in the electrochemical cell.

- Write the half-cell reactions

Write the oxidation and reduction half-cell reactions for both metals:The oxidation half-cell reaction for aluminum is:\[ \text{Al} \rightarrow \text{Al}^{3+} + 3e^{-} \]The reduction half-cell reaction for silver is:\[ \text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \]

- Determine the standard electrode potentials

Find the standard electrode potentials (E^{0}) for the half-cell reactions from a reference table:For aluminum:\[ E^{0}(\text{Al}^{3+}/\text{Al}) = -1.66 \text{ V} \]For silver:\[ E^{0}(\text{Ag}^{+}/\text{Ag}) = +0.80 \text{ V} \]

- Calculate the cell potential

Calculate the standard cell potential (E^{0}_{\text{cell}}) using the equation:\[ E^{0}_{\text{cell}} = E^{0}_{\text{cathode}} - E^{0}_{\text{anode}} \]Since the reduction occurs at the cathode (silver) and oxidation occurs at the anode (aluminum), the cell potential is:\[ E^{0}_{\text{cell}} = 0.80 \text{ V} - (-1.66 \text{ V}) = 2.46 \text{ V} \]

- Summarize the process

The aluminum (Al) acts as the anode and loses electrons (oxidation), while the silver (Ag) acts as the cathode and gains electrons (reduction). Saliva, acting as the electrolyte, completes the circuit. The potential difference (2.46 V) leads to a small electric current, causing a jolt of pain.

Key concepts

half-cell reactions

When analyzing an electrochemical cell, it's crucial to understand half-cell reactions. These reactions describe the individual oxidation and reduction processes of the metals involved.

In the provided exercise, the metals are silver (Ag) and aluminum (Al). Oxidation occurs at the aluminum electrode, where aluminum metal (\text{Al}) loses electrons to form aluminum ions (\text{Al}^{3+}). This process is represented by the oxidation half-cell reaction:

\[ \text{Al} \rightarrow \text{Al}^{3+} + 3e^{-} \]

Simultaneously, reduction takes place at the silver electrode. Here, silver ions (\text{Ag}^{+}) gain electrons to form silver metal (\text{Ag}). This reaction is shown in the reduction half-cell reaction:

\[ \text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \]

Understanding these half-cell reactions helps explain how electrons are transferred in the electrochemical cell, leading to the observed effects like pain when biting down on a gum wrapper.

standard electrode potentials

The standard electrode potential (\( E^{0} \)) is a measure of the intrinsic ability of a species to either lose or gain electrons under standard conditions.

Each half-cell reaction has a corresponding \( E^{0} \) value that can be found in reference tables. In the provided exercise, the standard electrode potentials are:

\[ E^{0}(\text{Al}^{3+}/\text{Al}) = -1.66 \text{ V} \]

\[ E^{0}(\text{Ag}^{+}/\text{Ag}) = +0.80 \text{ V} \]

These values indicate how easily a species is reduced or oxidized. A positive \( E^{0} \) value means the species is more likely to gain electrons (be reduced), while a negative \( E^{0} \) value suggests it more readily loses electrons (is oxidized).

In the context of the problem, the positive \( E^{0} \) for silver indicates it will act as the cathode (site of reduction), while the negative \( E^{0} \) for aluminum designates it as the anode (site of oxidation).

cell potential

Cell potential (\( E^{0}_{\text{cell}} \)) is the driving force behind an electrochemical cell, determining the voltage output. This value is calculated using the difference between the standard electrode potentials of the cathode and the anode:

\[ E^{0}_{\text{cell}} = E^{0}_{\text{cathode}} - E^{0}_{\text{anode}} \]

For our specific example, with silver as the cathode and aluminum as the anode, the calculation is:

\[ E^{0}_{\text{cell}} = 0.80 \text{ V} - (-1.66 \text{ V}) = 2.46 \text{ V} \]

This positive cell potential indicates a spontaneous reaction, meaning the electrochemical cell can produce a potential difference of 2.46 V. This potential difference generates a small electric current when saliva acts as an electrolyte, resulting in the jolt of pain experienced when biting down on a gum wrapper.

Understanding cell potential is key to grasping how electrochemical cells drive reactions and produce electrical energy.