A student set up an electrolytic cell similar to the one in the Electrolysis experiment to collect hydrogen gas over water in a burette, using a small strip of lead metal as the anode (instead of copper). Before the electrolysis, the mass of the lead strip was 9.333 g. A current of 121.1 mA was applied for a period of 3600 seconds, during which time some of the lead metal was oxidized to lead (2+). What was the mass of the lead metal strip at the end of the electrolysis?
Final mass of lead strip = ? g
(calculate your answer in grams, but do not include the units when submitting your response)
6 years ago
Answered By Leonardo F
First, we need to calculate the amount of electrical charge (Q) involved in the process. Since:
i=Q/t
In which i is the current ant t is the time, we have:
Q=i*t=(121.1e-3 A)*(3600 s)=435.96 C
The number of moles of electrons involved follows the semi-reaction of oxidation of lead:
Pb -> Pb2+ + 2e-
Since, according to Faraday's law, each mole of electron contains a charge of 96485 C, we can divide our charge found previosly by this number, to find out the number of electron moles involved:
moles of e-=(435.96 C)/(96485 C/mol)=0.004518 moles
Since we have the proportion of 1 mol of lead to 2 moles of electrons, the number of moles of lead will be:
Lead moles=(0.004518 moles)/2=0.002259 moles
If we multiply the number above by the molar mass of lead, which is 207.2 g/mol, we will have the mass consumed of lead:
Mass of lead consumed=(0.002259 moles)*(207.2 g/mol)=0.4681 g
Since it is the oxidation of lead, we subtract the above number from the original mass (9.333 g):
Final mass of lead=9.333-0.468=8.865 g
6 years ago
Answered By Harrison V
To start this problem, we need to know just what is happening. Our lead anode is losing electrons in the following reaction:
$Pb\rightarrow Pb^{2+}+2e^-$Pb→Pb2++2e−
The basic game plan is that we can figure out how much lead has been affected in this reaction if we can figure out how many electrons are involved. So let's calculate the total charge.
$Q=I\cdot t=\left(0.1211A\right)\cdot\left(3600s\right)=435.96C$Q=I·t=(0.1211A)·(3600s)=435.96C
Note that I converted the mA to A. Now that we have the charge, we need the number of electrons. By simple googling, or calculation, we know that the charge of one mole of electrons is 96,485C so we'll calculate the moles of electrons.
$\left(435.96C\right)\left(\frac{1mol}{96485C}\right)=4.518e-3mol$(435.96C)(1mol96485C )=4.518e−3mol
We're almost done! We need to divide this answer by 2 because, based on our chemical formula above, we get two electrons for every atom of lead. Finally, the molar mass of lead is 207.2g, and we subtract the spent mass of lead from the strip of lead.
$\left(\frac{0.004518mol}{2}\right)\left(\frac{207.2gPb}{1mol}\right)=0.4681gPb9.333-0.4681=8.8649gPb$(0.004518mol2 )(207.2gPb1mol )=0.4681gPb9.333−0.4681=8.8649gPb
Or accounting for significant figures, 8.865g of lead.