Video Transcript
How long would it take to reduce
one mole of Au3+ to Au solid using a current of 4.01 amps? Assume the voltage is sufficient to
perform the reduction.
In this problem, Au3+ ions are
reduced by the application of an electric current, which causes the Au3+ ions to
gain three electrons to form solid gold. We need to figure out how long
it’ll take to reduce one mole of Au3+ ions.
We can do that using this equation,
where 𝑄 is the total charge, 𝐼 is the current, and 𝑡 is the time, measured in
seconds. This problem gave us the current,
but we don’t have the total charge yet. So we’ll need to figure out how
much charge, that is, how many electrons were needed to reduce the Au3+ ions.
If we look back at our reaction
equation, we can see that every one mole of Au3+ ions reacts with three moles of
electrons. So three moles of electrons would
be required to reduce one mole of Au3+ ions, like we have in this problem. We can determine the amount of
charge delivered by three moles of electrons using this formula. We have three moles of electrons,
and Faraday’s constant is approximately equal to 96500 coulombs per mole of charged
particles. The units of moles cancel, and
we’ll find the total charge is equal to 289500 coulombs.
Now we can solve for the time. We’ll first have to isolate time so
we can solve for it. We can do this if we divide both
sides of the equation by 𝐼, which gives us this equation. Let’s just flip that around so time
is on the left-hand side. The charge we calculated was 289500
coulombs, and the current given in this problem is 4.01 amps. One amp is equivalent to one
coulomb per second. So the units of coulombs cancel,
leaving us with about 72194.514 seconds, which is a large quantity of time. So let’s convert this into
hours.
There are 3600 seconds in an
hour. So we can convert from seconds to
hours if we divide by 3600. This leaves us with about 20.0543
hours. Rounding to a sensible number of
digits gives us our final answer. So it takes 20.1 hours to reduce
one mole of Au3+ ions into solid gold using a current of 4.01 amps.
