Asked by Masre
Two cells are connected in series . One contains AlCl3, and the other contains AgNO3 as the electrolytes. What mass of Ag is deposited when 18g of Al is deposited at cathode?
Answers
Answered by
DrBob222
96,485 coulombs will deposit 27/3 = 9 g Al
This cell deposited 18 g Al so you must have had 96,485 x 18/2 = 48,242 C flowing through the cell.
96,485 coulombs will deposit 107.9/1 = about 108 but you need to do the math on all of these estimates.
So Ag deposited will be 107.9 x (48,242/96, 485) = about 107.9/2 = ? g Ag.
Post your work if you have further questions.
This cell deposited 18 g Al so you must have had 96,485 x 18/2 = 48,242 C flowing through the cell.
96,485 coulombs will deposit 107.9/1 = about 108 but you need to do the math on all of these estimates.
So Ag deposited will be 107.9 x (48,242/96, 485) = about 107.9/2 = ? g Ag.
Post your work if you have further questions.
Answered by
Masre
MAl/EAl=MAg/MAg
18g/9=MAg/108g
MAg=18gx108/9
=1994/9
=216g/mol
I think this is my answer for this question
18g/9=MAg/108g
MAg=18gx108/9
=1994/9
=216g/mol
I think this is my answer for this question
Answered by
DrBob222
Your answer of 216 grams Ag (not grams/mol) is correct and mine is wrong. I goofed in my first calculations. Here is what I should have written.
<b>96,485 coulombs will deposit 27/3 = 9 g Al
This cell deposited 18 g Al so you must have had 96,485 x 18/9 = 192,970 C flowing through the cell.
96,485 coulombs will deposit 107.9/1 = about 108 g Ag but you need to do the math on all of these estimates.
So Ag deposited will be 107.9 x (192,970/96, 485) = about 107.9*2 = ? g which is 215.8 g Ag.
<b>96,485 coulombs will deposit 27/3 = 9 g Al
This cell deposited 18 g Al so you must have had 96,485 x 18/9 = 192,970 C flowing through the cell.
96,485 coulombs will deposit 107.9/1 = about 108 g Ag but you need to do the math on all of these estimates.
So Ag deposited will be 107.9 x (192,970/96, 485) = about 107.9*2 = ? g which is 215.8 g Ag.
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