Calculate the volume of gas liberated at the anode at STP during the electrolysis of a CuSO4 solution by a current of 1 A passed for 16 minutes and 5 seconds.
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Cu is deposited at the negative electrode (the cathode) and oxygen is liberated at the positive electrode (anode in an electrolysis cell) by the equation 2H2O(l) ===> 4H+(aq) + O2(g) + 4e–
coulombs = amperes x seconds
C = 1 A x 16 min x (60 sec/min) + 5 sec = 965
96,485 C will liberate (22.4/4 = 5.6 ) L O2 gas @ STP; therefore, 965 C will liberate.
5.6 L x (965/96,485) = ? L
NOTE: I suspect the author of the problem used 96,500 instead of the more exact 96,485 so the final calculation comes out easier as
5.6 L x (965/96,500) = ? L
To calculate the volume of gas liberated at the anode during the electrolysis of a CuSO4 solution, we need to use Faraday's laws of electrolysis.
For copper sulfate (CuSO4), the electrolysis reaction at the anode is as follows:
2H2O(l) → O2(g) + 4H+(aq) + 4e-
From the reaction, we can see that 4 moles of electrons (4e-) are required to liberate 1 mole of gas (O2).
Given:
Current (I) = 1 A
Time (t) = 16 minutes and 5 seconds = 16 × 60 + 5 = 965 seconds
To calculate the amount of charge (Q) passed in coulombs, we can use the equation:
Q = I × t
Q = 1 A × 965 seconds
Q = 965 coulombs
Next, we need to calculate the number of moles of electrons (n) using Faraday's constant (F):
n = Q / F
Faraday's constant (F) is approximately 96485 C/mol.
n = 965 C / 96485 C/mol
n ≈ 0.01 mol
Since 4 moles of electrons are required to liberate 1 mole of gas, we can determine the number of moles of gas (m) liberated as follows:
m = n / 4
m = 0.01 mol / 4
m = 0.0025 mol
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of gas liberated (V) can be calculated as:
V = m × 22.4 liters/mol
V = 0.0025 mol × 22.4 liters/mol
V = 0.056 liters
Therefore, the volume of gas liberated at the anode during the electrolysis of a CuSO4 solution, with a current of 1 A passed for 16 minutes and 5 seconds at STP, is approximately 0.056 liters.
To calculate the volume of gas liberated at the anode during electrolysis, we need to use Faraday's laws of electrolysis. These laws state that the amount of substance produced or consumed during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.
To calculate the volume of gas, we need to determine the number of moles of gas produced using Faraday's laws and then use the ideal gas law to calculate the volume.
Step 1: Calculate the total charge passed
The total charge passed (Q) can be calculated using the formula:
Q = I × t
where I is the current in amperes and t is the time in seconds.
In this case, I = 1 A, t = 16 minutes and 5 seconds = 16 × 60 + 5 = 965 seconds.
Q = 1 A × 965 s = 965 coulombs.
Step 2: Convert charge to moles of electrons
To find the moles of electrons (n) transferred, we need to divide the total charge passed by the Faraday constant (F).
Faraday constant (F) = 96,485 C/mol e⁻.
n = Q / F
n = 965 C / 96,485 C/mol e⁻ = 0.010 mol e⁻
Step 3: Determine the moles of gas produced
The stoichiometry of the electrolysis reaction would determine the number of moles of gas produced. In this case, copper sulfate (CuSO4) is being electrolyzed, so the balanced equation for the reaction at the anode is:
2H₂O → O₂ + 4H⁺ + 4e⁻
Since 4 electrons are required to produce 1 mole of O₂ gas, we multiply the moles of electrons (n) by 1/4 to get the moles of O₂ gas produced.
moles of O₂ gas produced = (0.010 mol e⁻) × (1 mol O₂/4 mol e⁻) = 0.0025 mol O₂
Step 4: Convert moles of gas to volume using the ideal gas law
The ideal gas law equation is given by:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
At STP (Standard Temperature and Pressure), we can use the following values:
P = 1 atm
T = 273 K
R = 0.0821 atm·L/mol·K
V = (nRT) / P
V = (0.0025 mol) x (0.0821 atm·L/mol·K) x (273 K) / (1 atm)
V = 0.56 L
Therefore, the volume of gas liberated at the anode during the electrolysis of a CuSO4 solution is 0.56 liters at STP.