Question

A lead storage battery delivers a current of \(6.00 \mathrm{~A}\) for one hour and 22 minutes at a voltage of \(12.0 \mathrm{~V}\). (a) How many grams of lead are converted to \(\mathrm{PbSO}_{4}\) ? (b) How much electrical energy is produced in kilowatt hours?

Short answer

Question: A lead-acid battery delivers a current of 6.00 A for 1 hour and 22 minutes at 12.0 V. Calculate (a) the mass of lead converted to lead sulfate during the discharge process and (b) the total electrical energy produced by the battery. Answer: (a) To calculate the mass of lead converted to lead sulfate, first calculate the number of electrons transferred during discharge, then the quantity of lead involved in the reaction, and finally the mass of lead using the molar mass of lead. (b) To find the total electrical energy produced by the battery, multiply the voltage, current, and time, then convert the energy to kilowatt-hours.

Step-by-step solution

(a) Step 1: Write down the related chemical equations

The half-reactions for the discharge of a lead-acid battery are as follows: \[ \mathrm{Pb(s) + SO_{4}^{2-}(aq)} \to \mathrm{PbSO_{4}(s)} + 2 e^- \] This reaction takes place at the anode (negative electrode). Each lead atom releases two electrons. For the purpose of the exercise, we can focus on this half-reaction.

(a) Step 2: Calculate the number of electrons transferred during discharge

The battery delivers a current of 6.00 A for 1 hour and 22 minutes or (1 + 22/60) hours. Since current represents the flow of electrons per unit time, we can calculate the total charge transferred during this period using the formula: Charge (Q) = Current (I) × Time (t). The charge of an electron is given as \(1.60\times10^{-19} \mathrm{~C}\). Therefore, the number of electrons transferred (n) can be calculated using the formula n = Q / \(1.60\times10^{-19} \mathrm{~C}\). From this, we derive: n = (6.00 A × (1 + 22/60) hours) / \(1.60\times10^{-19} \mathrm{~C}\)

(a) Step 3: Calculate the quantity of lead involved in the reaction

Since 2 electrons are released for every lead atom that is converted to lead sulfate, the total number of lead atoms converted can be calculated as: Number of Pb atoms = Number of electrons transferred / 2 = n / 2

(a) Step 4: Compute the mass of lead converted

To calculate the mass of lead converted, we need to know the molar mass of lead, which is approximately 207.2 g/mol. Multiply the molar mass of lead by the number of Pb atoms obtained in step 3, and then divide by Avogadro's number (approximately \(6.022\times10^{23} \mathrm{~atoms/mol}\)) to find the mass of lead (m): m = (Number of Pb atoms × 207.2 g/mol) / \(6.022\times10^{23} \mathrm{~atoms/mol}\)

(b) Step 1: Calculate the electrical energy produced

The electrical energy produced by the battery can be calculated by multiplying the voltage, current, and time. Using the given voltage of 12.0 V and the already calculated time in hours, we have: Electrical Energy (E) = Voltage (V) × Current (I) × Time (t) = 12.0 V × 6.00 A × (1 + 22/60) hours

(b) Step 2: Convert Electrical Energy to Kilowatt-hours

To convert the electrical energy to kilowatt-hours (kWh), we need to divide by 1000 (since 1 kW = 1000 W): E (kWh) = E (W) / 1000

Key concepts

Electrochemical Reactions

Electrochemical reactions are the foundation of how batteries work, allowing for the storage and release of energy through chemical processes. In the context of the lead storage battery, the chemical equation at the anode during discharge is:
\[ \mathrm{Pb(s) + SO_{4}^{2-}(aq)} \to \mathrm{PbSO_{4}(s)} + 2 e^- \]
This reaction is an example of an oxidation process where lead (Pb) loses electrons to form lead sulfate (\(\mathrm{PbSO}_{4}\)). These reactions occur at the electrodes and are responsible for the conversion of chemical energy into electrical energy.

Electrical Energy Conversion

Converting chemical energy into electrical energy is a key function of a lead storage battery. When a battery is discharging, the electrochemical reactions within release electrons that flow through an external circuit, generating an electric current. This current multiplied by the voltage and time gives us the amount of electrical energy produced. For instance, the lead storage battery in the exercise produces:
\[ \text{Electrical Energy (E)} = \text{Voltage (V)} \times \text{Current (I)} \times \text{Time (t)} \]
The voltage and current are directly related to the energy output, where a higher voltage or current results in a greater amount of energy released.

Stoichiometry in Electrochemistry

Stoichiometry in electrochemistry involves the quantitative relationship between the elements in an electrochemical reaction. In the lead storage battery, for each lead atom that oxidizes, two moles of electrons are released. Therefore, by knowing the amount of charge (current times time) and the charge of one electron, we can calculate the number of electrons transferred, and subsequently, using the reaction's stoichiometry, determine the amount of lead that reacted. This process highlights the predictability of chemical transformations in an electrochemical system.

Molar Mass Calculation

Molar mass calculation is crucial for converting between the mass of a substance and the number of atoms or molecules it contains. In this exercise, we're interested in the mass of lead transformed during battery operation. Knowing the molar mass of lead (207.2 g/mol) allows us to convert the number of lead atoms to grams by multiplying it with the number of moles, which will then be divided by Avogadro's number to obtain the final mass in grams.
\[ \text{Mass of lead (m)} = \left(\frac{\text{Number of Pb atoms} \times \text{Molar mass of Pb}}{\text{Avogadro's number}}\right) \]

Avogadro's Number Application

Avogadro's number (approximately \(6.022\times10^{23}\)) is key in the realm of chemistry for converting between atoms, molecules, or ions and moles. In the calculation of the mass of lead turned into lead sulfate in the battery, Avogadro's number is utilized to translate the number of lead atoms into moles since the molar mass provides the mass for one mole of a substance. Hence, we divide the product of the number of lead atoms and the molar mass of lead by Avogadro's number to obtain the mass in grams.