Results 1 to 3 of 3

Math Help - chemistry - faraday and electrolysis

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    Exclamation chemistry - faraday and electrolysis

    You want to cover a motor bike with chromium(3+). You need about 1000g. The ampere is 9.75. How much would you have to pay if it costs 4 cents per KW per hour (KWH)?

    I used Q = nF [ n = moles of electrons, F = faraday] to solve for Q. Then, I plugged it in to Q = It to find the time. Is that the right way to do it? But after that, I got stuck. How do I find the price?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539
    Quote Originally Posted by applejelly View Post
    You want to cover a motor bike with chromium(3+). You need about 1000g. The ampere is 9.75. How much would you have to pay if it costs 4 cents per KW per hour (KWH)?

    I used Q = nF [ n = moles of electrons, F = faraday] to solve for Q. Then, I plugged it in to Q = It to find the time. Is that the right way to do it? But after that, I got stuck. How do I find the price?

    Since you need 1000 grams of chromium, we need to determine how many moles this represents. Since the mass of one mole of chromium is 52.00 g:

    1000\ \mbox{g}\times \frac{1\ \mbox{mole}}{(52.00\ g)} = 19.23\ \mbox{moles chromium}

    Since the appropriate half-reaction is:
    Cr^{+3} + 3e^{-}\longrightarrow Cr, each mole of chromium requires three moles of electrons.

    19.23\ \mbox{moles Cr} \times \frac{3\ \mbox{moles}\ e^{-}}{1\ \mbox{mole Cr}} = 57.69\ \mbox{moles of electrons}

    Since, 1\ \mbox{mole of}\ e^{-}\ \mbox{possess}\ 96485\ C\ \mbox{of charge}

    57.69\ \mbox{moles}\ e^{-} \times \frac{96485\ C}{1\ \mbox{mole}\ e^{-}} = 5.566 \times 10^6\ C

    Electric power is the product of current and voltage. If we assume that this is being carried out at a standard 120 volts,

    \mbox{Power} = 9.75\ A\times 120\ V = 1170\ W = 1.17\ kW

    The time required for this process can be found by taking the quotient of the charge and the current,

    \mbox{Time} = \frac{5.566 \times 10^6\ C}{9.75\ A} = 5.709 \times 10^5\ \mbox{seconds} = 158.6\ \mbox{hours}

    The cost for the process can be found by taking the product of the price per kWh, the power (in kW), and the time (in hours).

    \mbox{Cost} = \frac{\$0.04}{kWh}\times 1.17\ kW\times 158.6\ h = \displaystyle{\boxed{\$7.42}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    2
    oh woah i didn't know about standard voltage.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. chemistry help
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: October 10th 2010, 07:46 AM
  2. chemistry help please
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 19th 2009, 02:56 PM
  3. chemistry
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: December 22nd 2008, 09:28 AM
  4. Help me in chemistry
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 2nd 2008, 07:09 AM
  5. Replies: 10
    Last Post: April 28th 2007, 05:00 PM

Search Tags


/mathhelpforum @mathhelpforum