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Posted by on Sunday, June 3, 2012 at 8:01pm.

Consider two cells, the first with Al and Ag electrodes, and the second with Zn and Ni electrodes, each in appropriate 1.00M solutions of their ions.

If 2.10g of metal is plated in the voltaic cell, how much metal is plated in the electrolytic cell?


Another plated question... I'm sorry, normally I include an explanation of what I'm confused on but I just do not know how to approach this problem at all. I know that when Ag and Ni get plated they will have a 2.98 V and when Ag and Zn get played they have a 1.98 V

Reduction Half Reaction & Standard Potential Ered°
Al3+(aq) + 3e– → Al(s) –1.66 V
Ag+(aq) + e– → Ag(s) +0.80
Zn2+(aq) + 2e– → Zn(s) –0.76
Ni2+(aq) + 2e– → Ni(s) –0.26

  • Chemistry - , Sunday, June 3, 2012 at 11:23pm

    Voltaic cell is
    Al + 3Ag^+ ==> Al^3+ + 3Ag(s)
    2.1g is how many mols of Ag?
    2.1/107.86 = 0.0195 and that will be deposited by 0.0195*96,485 coulombs = about 1878 but you can be more accurate than that.
    When in the electrolytic cell, the Al is being plated and Ag is going into soln. The Al plated will be 26.98/3 = about 9 g with one(1) coulomb; therefore,
    9g x (1878/95,485) = about 0.175g Al plated. You can arrive at the same answer by taking a shortcut.
    The equivalent weight of Al is about 9, that of Ag is about 108; therefore,
    2.1 x (9/108) = about 0.175)

  • Chemistry - , Monday, December 9, 2013 at 4:12pm

    Using this method i did not arrive at the right answer for the problem

  • Chemistry - , Tuesday, February 4, 2014 at 2:38pm

    the steps are right but its not al its zn.

  • Chemistry - , Sunday, March 9, 2014 at 9:59pm

    a)Plated Ni and Ag at 2.98 V
    b)Plated Ag and Zn at 1.98 V
    c)2.10/107.868/2*65.38=0.64g zn

    this is proven

  • Chemistry - , Friday, April 11, 2014 at 8:12pm

    STILL DOESNT WORK

  • Chemistry - , Monday, May 16, 2016 at 5:54pm

    b) Ag & Zn, 0.80 - (-0.76) = 1.56V

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