During the electrolysis of Copper (ll) chloride solution, 3.2g of Copper was deposited at the cathode, calculate the volume of chlorine gas evolved at the anode at STP at the same time. Given cu=64, 1F=96500C. Molar volume of gas at STP=22.4/dm^3
find the moles of Cu deposited ... 3.2 g / molar mass of Cu
there will be two moles of Cl evolved for every mole of Cu
... but because Cl is diatomic in the gaseous phase
... there will be equal moles of Cu and Cl2
96,500 coulombs will deposit 64/2 = 32 g Cu; therefore,
96,500 x 3.2/32 = 9,650 coulombs
2Cl^- ==> Cl2(g) + 2e so 96,500 coulombs will displace 22.4/2 = 11.2 L Cl2 gas @ STP.
Then 11.2 x (9,650/96,500) = 11.2/10 = ? L.
Check my work.
Had the answer
To calculate the volume of chlorine gas evolved at the anode during the electrolysis of Copper (II) chloride solution, we need to use the concept of stoichiometry and the molar ratio between copper and chlorine.
Step 1: Calculate the moles of copper deposited at the cathode.
To do this, we need to convert the mass of copper (3.2g) to moles using its molar mass.
Molar mass of copper (Cu) = 64 g/mol
Moles of Cu = Mass of Cu / Molar mass of Cu
= 3.2 g / 64 g/mol
= 0.05 moles
Step 2: Determine the moles of chlorine evolved at the anode.
The balanced equation for the electrolysis of Copper (II) chloride is:
2CuCl2 -> 2Cu + Cl2
This means that for every 2 moles of Cu deposited, 1 mole of Cl2 gas is evolved.
So, moles of Cl2 = 0.05 moles / 2
= 0.025 moles
Step 3: Convert the moles of chlorine gas to volume at STP.
Using the molar volume of gas at STP (22.4 L/mol), we can calculate the volume of Cl2 gas evolved.
Volume of Cl2 gas = Moles of Cl2 * Molar volume of gas at STP
= 0.025 moles * 22.4 L/mol
= 0.56 L
Therefore, the volume of chlorine gas evolved at the anode during the electrolysis of Copper (II) chloride solution is 0.56 liters at STP.