Chemistry

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?

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  1. 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.

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    DrBob222
  2. MAl/EAl=MAg/MAg

    18g/9=MAg/108g
    MAg=18gx108/9
    =1994/9
    =216g/mol
    I think this is my answer for this question

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  3. 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.
    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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    DrBob222

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