Problem 73 - Chemistry The Molecular Science

Question

Assume that the anode reaction for the lithium battery is $$ \mathrm{LiC}_{6}(\mathrm{~s}) \longrightarrow \mathrm{Li}^{+}(\text {electrolyte })+\mathrm{C}_{6}(\mathrm{~s})+\mathrm{e}^{-} $$ and the anode reaction for the lead-acid storage battery is $$ \mathrm{Pb}(\mathrm{s})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) \longrightarrow \mathrm{PbSO}_{4}(\mathrm{~s})+2 \mathrm{e}^{-}+\mathrm{H}^{+}(\mathrm{aq}) $$ Compare the masses of metals consumed when each of these batteries supplies a current of \(1.0 \mathrm{~A}\) for \(10 . \mathrm{min}\).

Step-by-Step Solution

Answer

Lithium battery: 0.043 g of lithium; Lead-acid battery: 0.644 g of lead.

Step 1: Calculate Total Charge

First, calculate the total charge supplied by each battery using the formula: \( Q = I \times t \), where \( Q \) is the charge in coulombs, \( I \) is the current in amperes, and \( t \) is the time in seconds. Given \( I = 1.0 \text{ A} \) and \( t = 10 \text{ min} = 600 \text{ s} \), we find: \( Q = 1.0 \times 600 = 600 \text{ C} \).

Step 2: Calculate Moles of Electrons (Lithium Battery)

The lithium battery reaction involves the release of one mole of electrons per mole of \( \text{LiC}_6 \). Using the charge calculated and Faraday’s constant (\( F = 96485 \text{ C/mol} \)), calculate moles of electrons: Number of moles of electrons = \( \frac{600}{96485} \approx 0.00622 \text{ mol} \).

Step 3: Calculate Moles of Electrons (Lead-Acid Battery)

In the lead-acid battery, two moles of electrons are released per mole of \( \text{Pb} \). Thus, the number of moles of electrons required remains \( 0.00622 \text{ mol} \), but this corresponds to half the moles of lead: Moles of \( \text{Pb} = \frac{0.00622}{2} \approx 0.00311 \text{ mol} \).

Step 4: Calculate Mass of Lithium Consumed

For lithium, with one mole of \( \text{Li} \) consumed per mole of \( \text{LiC}_6 \), calculate the mass: Molar mass of \( \text{Li} = 6.94 \text{ g/mol} \) so the mass of lithium consumed is: \( 0.00622 \times 6.94 = 0.04316 \text{ g} \).

Step 5: Calculate Mass of Lead Consumed

Using the moles of \( \text{Pb} \) calculated in Step 3, find the mass of lead consumed. Molar mass of \( \text{Pb} = 207.2 \text{ g/mol} \) so the mass of lead consumed is: \( 0.00311 \times 207.2 = 0.64431 \text{ g} \).

Step 6: Comparison of Masses

Compare the two masses: \( 0.04316 \text{ g} \) of lithium and \( 0.64431 \text{ g} \) of lead. The lead-acid battery consumes significantly more metal than the lithium battery under the same current conditions.

Key Concepts

Lithium Battery

Lithium batteries are a popular type of rechargeable battery, often found in consumer electronics like smartphones and laptops. These batteries work through a process known as electrochemical reactions, where lithium ions move from the anode to the cathode during discharge and back again when the battery is being charged.

The chemical reactions within a lithium battery involve lithium cobalt oxide (\( \mathrm{LiCoO}_2 \)), but for this particular exercise, we're considering the case of lithium being present as lithium carbide (\( \mathrm{LiC}_6 \)). The reaction taking place at the anode is:- Lithium atoms leave the anode as lithium ions, entering the electrolyte.- Electrons are released during this process, contributing to the flow of electrical current.

This movement and transformation generate electricity, which powers electronic devices. Lithium batteries are favored for their high energy density, meaning they can store more energy in a smaller size compared to other battery types.

Lead-Acid Battery

Lead-acid batteries are one of the oldest types of rechargeable chemical batteries. They are commonly used in automobiles to provide the high surge current required by starter motors and to supply electrical energy when the engine is not running.

A lead-acid battery consists of lead dioxide (\( \mathrm{PbO}_2 \)) at the positive plate and lead (\( \mathrm{Pb} \)) at the negative plate, submerged in a sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)) electrolyte. During discharge:

• Lead at the anode reacts with the sulfuric acid to form lead sulfate (\( \mathrm{PbSO}_4 \)).
• This reaction releases electrons, which flow through the circuit.

When charged, the process is reversed, reconstituting the plates to their original form. Despite their weight and discharge inefficiency compared to newer battery technologies, lead-acid batteries are sturdy and reliable.

Electrochemistry

Electrochemistry is the field in chemistry that focuses on the interaction between electrical energy and chemical change. The fundamental principles revolve around redox (reduction and oxidation) reactions, which involve electron transfer between substances.

In the context of batteries:

• Oxidation occurs at the anode, where there is a loss of electrons.
• Reduction takes place at the cathode, where electrons are gained.

Electrochemistry enables the conversion of chemical energy into electrical energy, providing power for various devices. This interaction is at the core of battery function and makes understanding electrochemical processes vital for innovation in energy storage technologies and sustainability.

Anode Reaction

In battery chemistry, the anode reaction is a crucial concept and refers to the chemical process occurring at the anode during the discharge of a battery. It involves the oxidation of a material, leading to the release of electrons. As seen in the exercise:

• The anode reaction for a lithium battery involves lithium ions being released, alongside electrons.
• The anode reaction in a lead-acid battery results in the formation of lead sulfate on the anode, releasing two electrons per lead atom.

The anode reaction's details are important for determining the performance and efficiency of the battery. It affects the amount of energy a battery can store, its lifespan, and its environmental impact.