Problem 48
Question
During the discharge of a lead storage battery, the density of sulphuric acid fell from \(1.294\) to \(1.139 \mathrm{~g} / \mathrm{mL}\). Sulphuric acid of density \(1.294\) \(\mathrm{g} / \mathrm{mL}\) is \(39 \%\) by weight and that of \(1.139 \mathrm{~g} / \mathrm{mL}\) is \(20 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) by weight. The battery holds \(3.5\) litres of the acid and the volume remained practically constant during the discharge. Calculate the number of ampere- hours for which the battery must have been used. The charging and discharging reactions are : [1986-5 Marks] Anode : \(\mathrm{Pb}+\mathrm{SO}_{4}^{2-}=\mathrm{PbSO}_{4}+2 \mathrm{e}^{-}\)(discharging) Cathode : \(\mathrm{PbO}_{2}+4 \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}+2 \mathrm{e}^{-}=\mathrm{PbSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) (discharging) Note : Both the reactions take place at the anode and cathode respectively during discharge. Both reaction get reverse during charging.
Step-by-Step Solution
Verified Answer
The battery was used for approximately 529.9 ampere-hours.
1Step 1: Understand the Problem
The problem requires calculating the ampere-hours for which a lead storage battery has been used. This involves determining how much sulphuric acid was used during the discharge of the battery.
2Step 2: Determine Initial and Final Mass of \\(\mathrm{H}_{2}\mathrm{SO}_{4}\\)
Initially, the sulphuric acid is 39% by weight at a density of 1.294 g/mL. The battery contains 3.5 L (or 3500 mL) of this acid. Thus, the initial mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is calculated as follows:\[ \text{Initial mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)} = 0.39 \times 1.294 \times 3500 = 1768.49 \, \text{g} \]After discharge, the acid is 20% by weight with a density of 1.139 g/mL:\[ \text{Final mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)} = 0.20 \times 1.139 \times 3500 = 797.3 \, \text{g} \]
3Step 3: Calculate Mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) Used
Subtract the final mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) from the initial mass:\[ \text{Mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) used} = 1768.49 - 797.3 = 971.19 \, \text{g} \]The molar mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is 98.08 g/mol, so:\[ \text{Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) used} = \frac{971.19}{98.08} = 9.9 \text{ mol} \]
4Step 4: Determine Ampere-Hours
Each mole of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) provides 2 moles of electrons during discharge. Thus, the total moles of electrons is 19.8 mol (since 9.9 mol of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) are used).Using Faraday's constant, 1 mole of electrons corresponds to 96500 coulombs. Therefore, we find:\[ \text{Total charge} = 19.8 \times 96500 = 1907700 \, \text{C} \]Since 1 Ah = 3600 C, the number of ampere-hours is:\[ \text{Ampere-hours} = \frac{1907700}{3600} \approx 529.9 \text{ Ah} \]
Key Concepts
Sulphuric Acid ConcentrationElectrochemical ReactionAmpere-Hour Calculation
Sulphuric Acid Concentration
In a lead storage battery, sulphuric acid plays a critical role as the electrolyte. As the battery discharges, the concentration of sulphuric acid decreases. Initially, it may have a density like 1.294 g/mL, with a high percentage of sulphuric acid (39% by weight in this exercise). During discharge, the lead and lead dioxide plates react with the sulphuric acid, transforming it into lead sulfate. This decreases the acid's concentration. In our example, after discharge, the concentration reduces to 20% by weight, corresponding to a lower density of 1.139 g/mL.
These changes in concentration help us determine the extent to which the battery has been utilized. By comparing initial and final concentrations, one can calculate how much sulphuric acid has reacted, which is critical to understanding the overall battery discharge process.
The consistency in volume (like the 3.5 liters here) ensures that changes in density directly reflect the change in concentration, simplifying calculations and leading us to figure out the consumed charge.
These changes in concentration help us determine the extent to which the battery has been utilized. By comparing initial and final concentrations, one can calculate how much sulphuric acid has reacted, which is critical to understanding the overall battery discharge process.
The consistency in volume (like the 3.5 liters here) ensures that changes in density directly reflect the change in concentration, simplifying calculations and leading us to figure out the consumed charge.
Electrochemical Reaction
The functioning of a lead storage battery hinges on electrochemical reactions. During discharge, specific chemical changes occur at both the anode and cathode that transform chemical energy into electrical energy. At the anode, lead (Pb) reacts with sulphate ions to form lead sulfate (PbSO₄) and releases electrons. This reaction is given by:
\[ \mathrm{Pb} + \mathrm{SO}_{4}^{2-} = \mathrm{PbSO}_{4} + 2 \mathrm{e}^{-} \]
At the cathode, lead dioxide (PbO₂) combines with hydrogen ions and sulphate ions, consuming electrons to form water and more lead sulfate:
\[ \mathrm{PbO}_{2} + 4 \mathrm{H}^{+} + \mathrm{SO}_{4}^{2-} + 2 \mathrm{e}^{-} = \mathrm{PbSO}_{4} + 2 \mathrm{H}_{2} \mathrm{O} \]
Both these reactions contribute to the depletion of sulphuric acid and the simultaneous formation of lead sulfate. These transformations release electrons that travel through the circuit, delivering the electrical energy needed to power devices. The electrons generated at the anode move to the cathode through an external circuit, which creates an electron flow—the current of the battery.
Understanding these reactions is vital not only for calculating energy usage but also for optimizing battery design and efficiency.
\[ \mathrm{Pb} + \mathrm{SO}_{4}^{2-} = \mathrm{PbSO}_{4} + 2 \mathrm{e}^{-} \]
At the cathode, lead dioxide (PbO₂) combines with hydrogen ions and sulphate ions, consuming electrons to form water and more lead sulfate:
\[ \mathrm{PbO}_{2} + 4 \mathrm{H}^{+} + \mathrm{SO}_{4}^{2-} + 2 \mathrm{e}^{-} = \mathrm{PbSO}_{4} + 2 \mathrm{H}_{2} \mathrm{O} \]
Both these reactions contribute to the depletion of sulphuric acid and the simultaneous formation of lead sulfate. These transformations release electrons that travel through the circuit, delivering the electrical energy needed to power devices. The electrons generated at the anode move to the cathode through an external circuit, which creates an electron flow—the current of the battery.
Understanding these reactions is vital not only for calculating energy usage but also for optimizing battery design and efficiency.
Ampere-Hour Calculation
An ampere-hour (Ah) measures the battery's charge capacity, essentially capturing how much current a battery can provide over time. When calculating the ampere-hours during a battery's discharging phase, one must consider the total charge transferred, which originates from the electrochemical reactions involving sulphuric acid.
From the exercise, we see a decrease in the concentration of sulphuric acid, allowing us to calculate how many grams of acid reacted. Knowing the molar mass of sulphuric acid, find the moles used. In our case this was about 9.9 moles. Since each mole of sulphuric acid involved corresponds to the transfer of 2 moles of electrons, the total electron moles is double that amount—19.8 moles of electrons.
Using Faraday's constant (96500 C/mol), one translates the moles of electrons into total charge, calculating the number of coulombs. Then, since an ampere-hour is 3600 coulombs, divide the total charge by 3600 to find how many ampere-hours the battery discharged over its use.
This example resulted in approximately 529.9 Ah, providing valuable insights into the battery's performance during its operational period.
From the exercise, we see a decrease in the concentration of sulphuric acid, allowing us to calculate how many grams of acid reacted. Knowing the molar mass of sulphuric acid, find the moles used. In our case this was about 9.9 moles. Since each mole of sulphuric acid involved corresponds to the transfer of 2 moles of electrons, the total electron moles is double that amount—19.8 moles of electrons.
Using Faraday's constant (96500 C/mol), one translates the moles of electrons into total charge, calculating the number of coulombs. Then, since an ampere-hour is 3600 coulombs, divide the total charge by 3600 to find how many ampere-hours the battery discharged over its use.
This example resulted in approximately 529.9 Ah, providing valuable insights into the battery's performance during its operational period.
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