Final answer:
To find the volume of oxygen at NTP from electrolysis with a 5 A current for 193 seconds, Faraday's laws are applied. The calculation is based on the electric charge needed to produce oxygen gas from water. The result is approximately 56 mL, which deviates from an expected value of 965 mL in the question, possibly indicating a mistake or the need to consider real-world inefficiencies.
Step-by-step explanation:
To calculate the volume of O₂ liberated at NTP (Normal Temperature and Pressure) by passing 5 A current for 193 seconds through acidified water, we use Faraday's laws of electrolysis. According to Faraday's first law, the amount of substance liberated at an electrode is directly proportional to the quantity of electricity passing through the electrolyte. The quantity of electricity (Q) is given by the product of current (I) and time (t), so Q = I × t.
For the production of oxygen gas at an anode during water electrolysis, the reaction is:
2H₂O(I) → 4H⁺(aq) + O₂(g) + 4e⁻
Since 4 moles of electrons will produce 1 mole of O₂, we need the equivalent of 4 faradays of charge to produce 1 mole of oxygen. As 1 faraday is equal to 96500 C (coulombs), the volume of 1 mole of gas at NTP is 22.4 liters.
To determine the liberated volume of O₂,
- Calculate the quantity of electricity, Q = 5 A × 193 s = 965 C.
- Determine the number of moles of O₂, which is Q / (4 × 96500 C).
- Finally, calculate the volume of O₂ at NTP by multiplying the moles by 22.4 L.
The calculation gives us a volume of approximately 0.056 L or 56 mL of O₂. However, this value differs from the anticipated 965 mL, which might suggest an error in the problem statement or the assumption that the entire current is utilized for the production of oxygen, which in practical scenarios, might not be the case as some current could be lost or used in side reactions.