An electrolysis experiment was run for 17.5 minutes at an average current of 191 milliamps (mA). The mass lost by the copper anode was 0.0679 grams.
a) Calculate the number of coloumbs transferred. (coloumb=ampsxsec)
b) Calculate the moles of copper lost from the anode.
c) Calculate the moles of electrons transferred. The reaction at the copper anode is: Cu(s)-->Cu^2+(aq)+2e^-.
To find the number of coulombs transferred in the electrolysis experiment, we need to use the formula:
Coulombs (C) = Current (A) × Time (s)
a) Calculate the number of coulombs transferred:
Current (in amps) = 191 milliamps = 191 × 10^(-3) A
Time (in seconds) = 17.5 minutes × 60 seconds/minute = 1050 seconds
Coulombs (C) = 191 × 10^(-3) A × 1050 s
Now, let's calculate the value of Coulombs:
Coulombs (C) = 0.201 Coulombs
Therefore, the number of coulombs transferred in the experiment is 0.201 C.
b) To calculate the moles of copper lost from the anode, we need to use the formula from Faraday's laws of electrolysis:
Moles of Copper (Cu) = Mass of Copper Lost (g) / Molar Mass of Copper (g/mol)
The molar mass of copper (Cu) is 63.55 g/mol.
Moles of Copper (Cu) = 0.0679 g / 63.55 g/mol
Now, calculate the value of Moles of Copper:
Moles of Copper (Cu) = 0.001071 mol
Therefore, the moles of copper lost from the anode is 0.001071 mol.
c) To calculate the moles of electrons transferred, we can use the stoichiometry of the reaction:
1 mole of Copper (Cu) loses 2 moles of electrons (2e^-).
Since we have the moles of copper lost, we can calculate the moles of electrons:
Moles of Electrons = Moles of Copper × 2
Moles of Electrons = 0.001071 mol × 2
Now, let's calculate the value of Moles of Electrons:
Moles of Electrons = 0.002142 mol
Therefore, the moles of electrons transferred in the reaction at the copper anode are 0.002142 mol.