Question

Calculate the volume of H2 and O2 gases at 25°C and 1.00 atm that Will be Collected at the Cathode and anode respectively,when an aqueous solution of Naso4 is electrolyzed for 2 hours with 10 amperes current

Answer 1

To calculate the volume of H2 and O2 gases collected at the cathode and anode, you can use the ideal gas law equation. First, calculate the number of moles using Faraday's law of electrolysis. Then, use the ideal gas law equation to calculate the volumes. Make sure to convert the temperature to Kelvin by adding 273.

Step-by-step explanation:

The volume of H2 and O2 gases that will be collected at the cathode and anode respectively can be calculated using the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to calculate the number of moles of H2 and O2 gases. We can use Faraday's law of electrolysis, which states that the amount of substance produced in an electrolysis reaction is directly proportional to the amount of electric charge passed through the cell. In this case, the electric charge is given as 10 amperes for 2 hours.

Once we have the number of moles of H2 and O2 gases, we can use the ideal gas law equation to calculate their volumes. The temperature is given as 25°C, which needs to be converted to Kelvin by adding 273 to it. The pressure is given as 1.00 atm.

Answer 2

The volume of H2 gas collected at the cathode during the electrolysis is approximately 9.126 L, and the volume of O2 gas collected at the anode is approximately 4.563 L, given a current of 10 amperes for 2 hours at 25°C and 1.00 atm.

Step-by-step explanation:

To calculate the volume of H2 and O2 gases collected at the cathode and anode respectively during the electrolysis of an aqueous solution of Na2SO4, we first need to determine the amount of charge passed through the solution.

Since the current is 10 amperes and the electrolysis goes on for 2 hours, the total charge Q is given by Q = current × time = 10 A × 2 h × 3600 s/h = 72000 C.

Using Faraday's laws of electrolysis, the amount of a substance produced at an electrode is directly proportional to the quantity of electricity that passes through the electrolyte.

The Faraday constant is approximately 96500 C/mol, which is the charge required to produce one mole of a monovalent ion.

For hydrogen at the cathode (which is produced in a 2:1 ratio to electrons):

• n(H2) = Q / (2 × F) = 72000 C / (2 × 96500 C/mol) = 0.373 mol

For oxygen at the anode (which is produced in a 4:1 ratio to electrons since it takes 4 electrons to produce an O2 molecule):

• n(O2) = Q / (4 × F) = 72000 C / (4 × 96500 C/mol) = 0.1865 mol

Using the ideal gas law (PV=nRT), with P=1 atm, T = 25°C (which is 298 K), and R = 0.0821 L·atm/(K·mol), we can calculate the volume of each gas.

Volume of H2 (V = nRT/P):

• V(H2) = (0.373 mol) × (0.0821 L·atm/(K·mol)) × (298 K) / (1 atm) ≈ 9.126 L

Volume of O2 (V = nRT/P):

• V(O2) = (0.1865 mol) × (0.0821 L·atm/(K·mol)) × (298 K) / (1 atm) ≈ 4.563 L