A current of 4.71 A is passed through a #Pb(NO_3)_2# solution for 1.80 hours. How much lead is plated out of the solution?

Your starting point here will be to use the definition of an ampere to find the charge that passed through the solution in that period of time.

This will allow you to find how many moles of electrons took part in the reaction.

As you know, an ampere is simply a measure of the rate of flow of electric charge. More specifically, an ampere is defined as an electric current one coulomb per second.

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 A" = "1 C"/"1 s")color(white)(a/a)|)))#

In your case, a current of #"4.71 A"# tells you that you get #"4.71 C"# of charge per second.

This means that you can use the current as a conversion factor to determine that total charge that passed through the solution in #1.80# hours. Convert the time from hours to second first

#1.80 color(red)(cancel(color(black)("h"))) * (60color(red)(cancel(color(black)("min"))))/(1color(red)(cancel(color(black)("h")))) * "60 s"/(1color(red)(cancel(color(black)("min")))) = "6480 s"#

You will thus have

#6480color(red)(cancel(color(black)("s"))) * overbrace("4.71 C"/(1color(red)(cancel(color(black)("s")))))^(color(purple)("= 4.71 A")) = "30520.8 C"#

Now, a coulomb is equivalent to the charge of #6.242 * 10^(18)# electrons. This means that the total charge that passes through solution will be equivalent to

#30520.8 color(red)(cancel(color(black)("C"))) * (6.242 * 10^(18)"e"^(-1))/(1color(red)(cancel(color(black)("C")))) = 1.905 * 10^(23)"e"^(-)#

To convert this to moles of electrons, use Avogadro's number

#1.905 * 10^(23)color(red)(cancel(color(black)("e"^(-)))) * overbrace("1 mole e"^(-)/(6.022 * 10^(23)color(red)(cancel(color(black)("e"^(-))))))^(color(purple)("Avogadro's number")) = "0.3163 moles e"^(-)#

Now, your solution contains lead(II) cations, #"Pb"^(2+)#, which result from the dissociation of the lead(II) nitrate, #"Pb"("NO"_3)_2#. In order to plate the lead out of solution, you must reduce the lead(II) cations to lead metal.

The reduction half-reaction looks like this

#"Pb"_ ((aq))^(2+) + color(red)(2)"e"^(-) -> "Pb"_ ((s))#

Notice that you need #color(red)(2)# moles of electrons in order to produce #1# mole of lead metal. This means that you will have

#0.3163 color(red)(cancel(color(black)("moles e"^(-)))) * "1 mole Pb"/(color(red)(2)color(red)(cancel(color(black)("moles e"^(-))))) = "0.15815 moles Pb"#

Finally, to convert this to grams of lead, use the element's molar mass

#0.15815 color(red)(cancel(color(black)("moles Pb"))) * "207.2 g"/(1color(red)(cancel(color(black)("mole Pb")))) = color(green)(|bar(ul(color(white)(a/a)"32.8 g"color(white)(a/a)|)))#

The answer is rounded to three sig figs.

It can be easily solved by applying Faraday's law of electrolysis as follows. According to this law we have the following equation.

• #W=(Exxcxxt)/F ......(1)#
  where:

• #W#=Mass (in g ) of an ion discharged at any electrode

• #E#=Equivalent mass of that ion.
• #c# = Current Passed in Ampere
• #t#= time (in sec) during which electricity is passed.
• #F#= 1 faraday = 96500C.

Here the equivalent mass of #Pb^(2+)#

#E=(" molar mass of" Pb^(2+)) / "Its valency"=207.2/2=103.6 g/"equivalent"#

#c=4.71A#

#t=1.8hrs=1.8xx60xx60s#

Inserting these values in equation (1) we have
the mass of lead plated out of the solution

#W=(103.6xx4.71xx1.8xx60xx80)/96500g=32.76g#