# Math Help - chemistry - faraday and electrolysis

1. ## 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?

2. Originally Posted by applejelly
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).

$\mbox{Cost} = \frac{\0.04}{kWh}\times 1.17\ kW\times 158.6\ h = \displaystyle{\boxed{\7.42}}$

3. oh woah i didn't know about standard voltage.

Thanks!