1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the current in the cell during that period?

  1. Feb 3, 2007 #1
    1. The problem statement, all variables and given/known data

    If 3.50 x 10^-3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.99 h, what is the current in the cell during that period? Assume that the gold ions carry one elementary unit of positive charge. (A)

    2. Relevant equations
    I = DeltaQ/DeltaT
    I = DeltaV/R (I don't think this is needed)

    3. The attempt at a solution
    Well, I have been reading the book and asked a friend who had taken physics before but was not able to figure this out.

    I believe I have to use the formula I = DeltaQ/DeltaT, but not sure how I would use it as in would deltaT be 2.99 h x 60? I said that because when I used it before it was always min x 60 sec.

    My main problem is I never encounter where a given is in mass form with an element such as gold. How would I convert 3.50 x 10^-3 kg into a number to fit into as equation to solve for A (current).

    Please help me or give me some tips on where to start.
  2. jcsd
  3. Feb 3, 2007 #2
    where do these guys get these questions? But you are right, you need to know the wt of gold, specically the atomic mass.Look it ip on a periodic table. Then its easy to convert that mass to the number of moles and hence the Coulombs of charge transferred which iirc is something like 96,500 coulombs/mole. But check that figure as well.
  4. Feb 3, 2007 #3
    Okay, I completely lost you when you started to mention the periodic table and Coulombs of charge.

    Anyway to explain this problem easier?
  5. Feb 3, 2007 #4


    User Avatar
    Homework Helper

    ok, you need to know a bit of chemistry to do this. (like denverdoc said)
    so one gold ions == one elementary unit of charge (simple enough)
    now you want to know how many gold ions are there in 3.5g of gold. to do this you need to know the molar mass of gold (the unit is in grams per mole)
    some definitions:
    The relative isotopic mass of carbon-12 is set at 12 exactly, and we define one mole of carbon-12 as weighing 12 g. Now, consider the mass of one mole of Chlorine atoms. Chlorine has a relative atomic mass of 35.5, so one mole of it weighs 35.5 g. In general the mass of one mole (molar mass) of an element or compound is the relative mass of the element or compound (in grams)

    So if you go to the periodic table you will find this value for gold (Au). Now, once you know this number (ie. the molar mass) then we can easily relate
    mass (m), molar mass (M) and number of mole (n) via n = m/M. And since one mole contains [tex]N_A[/tex] particles where [tex]N_A[/tex] is the Avogadro's number (6.02x10^23), so actual number can be deduced once the amount of mole of gold is known.

    once you have worked out the number of gold ions....then you know the amount of charges going through the electrode and hence can work out the average current in the time period given.
    Last edited: Feb 3, 2007
  6. Feb 3, 2007 #5
    Okay, I had to do some searching for a periodic table and got Gold atom weight is 196.97.

    So, n = 3.50/196.97 = .0178

    .0178(6.02 x 10^23) = 1.07156 x 10^22

    (1.07156 x 10^22)/2.99 = 3.584 x 10^21 (A)

    Is that the way to do it or did I make any conversion wrong?

    btw it said in the periodic table Au (gold) is Atomic weight: 196.966569 (4) not sure what that 4 is in ( )
  7. Feb 3, 2007 #6


    User Avatar
    Homework Helper

    time unit should be in seconds... observe that your current is gigantic! you forgot about the elementary charge value too
  8. Feb 3, 2007 #7
    (1.07156 x 10^22)(1.6 x 10^-19)/ 2.99 x 60


    9.56 (A)

    Can anyone confirm if this is correct or am I doing something very wrong again.
  9. Feb 4, 2007 #8
    Okay, I figure this out and realize what I did wrong.

    instead of 179.4 it was suppose to be 10764 second.

    basically it was 2.99 h x 60 minute x 60 second.

    which then give the answer of .16 (A)

    Thanks for the help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: What is the current in the cell during that period?
  1. Current in the cell (Replies: 4)