Question

Consider a battery with a lithium electrode and a silver chloride (AgCl) electrode. Assume the following chemical reactions occur in the battery, and the redox potential for each reaction is shown. \(\mathrm{AgCl}+e^{-} \rightarrow \mathrm{Ag}+\mathrm{Cl}^{-} \quad V_{r p}=0.22 \mathrm{~V}\) \(\mathrm{Li} \rightarrow \mathrm{Li}^{+}+e^{-} \quad V_{r p}=3.04 \mathrm{~V}\) (a) Which reaction is likely to occur at the cathode, and which reaction is likely to occur at the anode? Justify your answer. (b) What is the overall theoretical cell voltage? (c) If the battery is connected to a \(1 \mathrm{k} \Omega\) load, approximately what is the power delivered to that load?

Short answer

(a) AgCl at cathode, Li at anode; (b) 3.26 V; (c) 10.64 mW

Step-by-step solution

Understanding Reactions at the Electrodes

In a galvanic cell, the cathode is where the reduction takes place, and the anode is where the oxidation occurs. We notice that for the first reaction, AgCl is gaining an electron, which means reduction is happening, so this reaction likely occurs at the cathode. The lithium is losing an electron in the second reaction, indicating oxidation, hence likely occurring at the anode.

Calculate the Cell Voltage

The overall cell voltage can be calculated by adding up the reduction potential of the cathode and the oxidation potential of the anode (in reverse sign for oxidation). The cell voltage is given by the equation: \[ E_{ ext{cell}} = E_{ ext{cathode}} - E_{ ext{anode}} \] Substitute the given potentials: \[ E_{ ext{cell}} = 0.22 ext{ V} - (-3.04 ext{ V}) = 0.22 ext{ V} + 3.04 ext{ V} = 3.26 ext{ V} \]

Determine Power Delivered to the Load

Power can be calculated using the formula \[ P = \frac{V^2}{R} \] where \( V \) is the cell voltage and \( R \) is the resistance of the load. Substituting in the values: \[ P = \frac{(3.26)^2}{1000} \approx 10.6356 ext{ mW} \] This is the power delivered to the load.

Key concepts

Redox Reactions

Redox reactions are fundamental in electrochemistry, playing a crucial role in processes such as battery operation. The term "redox" is a shorthand for reduction-oxidation, which are the two-part processes where electrons are transferred between substances.
Reduction refers to the gain of electrons by a chemical species. A good way to remember this is through the mnemonic "OIL RIG"—"Oxidation Is Loss, Reduction Is Gain."
On the other hand, oxidation is the loss of electrons. In the context of a galvanic cell, these reactions are separated into half-reactions: one half-cell undergoes oxidation, while the other half undergoes reduction.

• For example, in our lithium and silver chloride battery problem, silver chloride is reduced to form silver and chloride ions, which occurs at the cathode.
• Lithium is oxidized to form lithium ions, occurring at the anode.

Understanding these reactions is key to predicting which process occurs at which electrode and calculating the overall cell voltage.

Galvanic Cells

Galvanic cells, or voltaic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. They are the backbone of traditional batteries, powering many everyday devices.
Each galvanic cell is made up of two different metals or metal compounds immersed in an electrolyte solution. These metals are called electrodes and are crucial to the cell's function.
To understand a galvanic cell:

• The anode is where oxidation occurs. Electrons are lost here.
• The cathode is where reduction takes place. Electrons are gained here.

In our example, lithium is at the anode (undergoes oxidation), and silver chloride is at the cathode (undergoes reduction). The electron movement from anode to cathode through an external circuit creates the electric current."
This simple concept is what enables a wide array of battery types, from AA batteries to modern lithium-ion cells.

Cell Voltage

Cell voltage, also known as electromotive force (EMF), is the measurement of the potential difference between the two electrodes of a galvanic cell. It represents the energy per unit charge available to drive the flow of electrons in an electric circuit.
Cell voltage is determined from the standard reduction potentials of the reactions at the cathode and anode. Here is how you calculate it:

• Identify the electrode reactions and their respective standard potentials.
• The cell voltage (E_{cell}) is found by subtracting the anode's potential from the cathode's potential.

For the given battery, the overall cell voltage is given by:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
Substituting the given values results in:
\[ E_{\text{cell}} = 0.22 \, \text{V} - (-3.04 \, \text{V}) = 3.26 \, \text{V} \]
This positive voltage confirms that the reaction is spontaneous, ensuring the flow of electrons can occur without an external stimulus. Understanding how to find and interpret cell voltage is crucial for applications in battery design and energy management.