Posted by samantha on Friday, April 18, 2008 at 12:55pm.
I only read the first line or two, but isn't 6 min just 360 sec and not 3600?
6 min x (60 sec/min) = 360 seconds.
And are you sure the electrolyte is SnCl3 or is it SnCl2? And if you are plating out tin, wouldn't the half cell show Sn as the product instead of Sn^+2. Truthfully, I'm not familiar with your lab; however, check me out on the above.
yeah sorry about the 3600..in my lab i used 360 im not sure why i put 3600. And your right about the SnCl2 instead of SnCl3.
and as far as the half cell, since im using tin II chloride the half cell should be
Sn^2+ + 2e- --> Sn^+
this is the full lab..
purpose: to test the methos of stoichiometry in cells
wire lead (x2)
tin (II) chloride solution (3.25 M)
take the mass of the steel can and record your observations. 117.34g
place steel can into large beaker
place tin electrode into beaker
attach the wires to the steel can and the tin elextrode
connect the wires to the proper terminus on the power source
set timer to 6min and start timer
take the mass of the steel. 118.05g
What is the mass of the tin produced? Show all your work.
What is the theoretical mass of tin that should have been produced? Show all your work.
Determine the accuracy of your result, using a percentage difference and discuss any discrepancies. Show all your work.
If you are plating out Sn, then the half cell must be Sn^+2 + 2e = Sn.
I'll look at the remainder but give me some time.
Changing the half reaction for Sn^+2 to Sn will change all of your other calculations but there are a couple of notes here.
1. I see you have used 96,500 C per Faraday but since your balance weighings were closer than that, I wonder why you haven't used 96,485? Of course, use 96,500 if those were your instructions.
2. When I divide 1245.6 by 96,500, I get 0.01291. You probably rounded up to 0.013 and your numbers should be better than that so I wouldn't round up. Of course, if you use 96,485, you still have 0.01291 as an answer.
3.I notice you used 118 for the atomic mass of Sn but on th table I have it is listed as 118.7.
well im not all that worried about rounding off my numbers. When it comes to actually handing in the assignment i use it the way it's given to me from the book. My question was if i am doing the equation correctly, and if you could help me get started with the following question, which was "what is the theoretical mass of tin that should have been produced?". What are the equations i use to get my answer?
All you need to do is to correct for the half cell reaction Sn^+2 + 2e ==> Sn.
Step 3: Half reaction for reduction of tin
Sn^4+ + e- --> Sn^2+
amount of tin produced.
0.013mol e- x 1mol/4mol
= 0.00325mol of Sn
This step then would be 0.0129 x 1 mol e/2 mol(instead of 4) = 0.00646 mols Sn
step 4:convert amount of tin to mass
mass formed= 0.00325mol Sn x 118g
0.00646 mols Sn x (118.7 g Sn/1 mol Sn) = 0.766 g Sn. This is the theoretical yield.
You found 118.05 - 117.34 = 0.71 g Sn plated out in your experiment.
Percent error = [(correct value-your value)/correct value] * 100,
I get about 7 or 8% which isn't bad at all.
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