To determine how long a current of 1A can be drawn from a lead-acid battery as the concentration of sulfuric acid (H2SO4) decreases from 1M to 0.5M, we can use Faraday's laws of electrolysis and some basic principles of chemistry. Let's break this down step by step.
Understanding the Electrochemical Reaction
In a lead-acid battery, the discharge process involves the conversion of lead dioxide (PbO2) and sponge lead (Pb) into lead sulfate (PbSO4) while releasing sulfuric acid (H2SO4). The overall reaction can be simplified as:
- PbO2 + Pb + 2H2SO4 → 2PbSO4 + 2H2O
During this process, the concentration of H2SO4 decreases as it is consumed in the reaction.
Calculating the Moles of H2SO4
First, we need to calculate the number of moles of H2SO4 in the solution. Given that the concentration is 1M and the volume is 5 liters, we can use the formula:
Number of moles (n) = Concentration (C) × Volume (V)
Substituting the values:
n = 1M × 5L = 5 moles
Determining the Change in Moles
As the concentration decreases from 1M to 0.5M, we can calculate the moles of H2SO4 at 0.5M:
n = 0.5M × 5L = 2.5 moles
The change in moles of H2SO4 is:
Δn = 5 moles - 2.5 moles = 2.5 moles
Relating Charge to Moles of Electrons
Each mole of H2SO4 can theoretically provide 2 moles of electrons during the discharge process. Therefore, the total moles of electrons available from 2.5 moles of H2SO4 is:
Total moles of electrons = 2.5 moles × 2 = 5 moles of electrons
Using Faraday's constant (approximately 96485 C/mol), we can calculate the total charge (Q) available:
Q = moles of electrons × Faraday's constant
Q = 5 moles × 96485 C/mol = 482425 C
Calculating Time for Current Draw
Now, we can find out how long we can draw a current of 1A using the formula:
Time (t) = Charge (Q) / Current (I)
Substituting the values:
t = 482425 C / 1 A = 482425 seconds
Converting Time to More Understandable Units
To convert seconds into hours, we divide by 3600 (the number of seconds in an hour):
t = 482425 seconds / 3600 seconds/hour ≈ 134 hours
Final Thoughts
In summary, you can draw a current of 1A from the lead-acid battery for approximately 134 hours as the concentration of H2SO4 decreases from 1M to 0.5M, assuming the volume of the solution remains constant at 5 liters. This calculation highlights the relationship between concentration, charge, and time in electrochemical systems.
