To calculate the mass of copper produced, we need to use Faraday's law of electrolysis. The formula is:
Mass of substance = (Current (A) * Time (s) * Molar mass (g/mol)) / (Faraday's constant)
The molar mass of copper is 63.55 g/mol, and Faraday's constant is 96485 C/mol. The current is given as 0.5 A and the time is 3 hours, which is equivalent to 10800 seconds.
So, the mass of copper produced is:
(0.5 A * 10800 s * 63.55 g/mol) / 96485 C/mol = 33.55 g
To calculate the volume of oxygen produced, we need to consider the stoichiometry of the reaction. From the balanced equation:
2 Cu + O2 -> 2 CuO
We can see that for every one mole of O2 produced, 2 moles of CuO are also produced. Therefore, the moles of CuO produced is equal to the moles of O2.
To convert moles of CuO to volume of O2, we need to use the ideal gas law. The formula is:
Volume (L) = (Moles * Gas constant * Temperature (K)) / Pressure (atm)
The temperature is 25 degrees Celsius, which is equivalent to 298 K. The pressure is given as 760 mmHg, which is equivalent to 1 atm.
The molar volume of a gas at STP (Standard Temperature and Pressure) is 22.4 L/mol.
So, the volume of O2 produced is:
(33.55 g / 79.55 g/mol) * (22.4 L/mol) = 9.5 L
Calculate the mass of copper and volume of oxygen (at 25 digree seltioce and 760mm t,g) that would be produce by passing current.of 0.5A through a CuSo4 solution between pt electrodes for 3 hours
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