Question #9a203

1 Answer
May 4, 2017

Answer:

Here's what I got.

Explanation:

http://www.docbrown.info/page01/ExIndChem/electrochemistry03.htm

You know that

#2overbrace("NaCl"_ ((l)))^(color(blue)("Na"_ ((l))^(+) + "Cl"_ ((l))^(-))) -> 2"Na"_ ((s)) + "Cl"_ (2(g))# #uarr#

In this redox reaction, sodium is being reduced to sodium metal at the cathode (the negative electrode)

#"Na"_ ((l))^(+) + "e"^(-) -> "Na"_ ((s)) -># the reduction half-reaction

Chlorine is being oxidized to chlorine gas at the anode (the positive electrode)

#2"Cl"_ ((l))^(+) -> "C"_ (2(g)) + 2"e"^(-) -># the oxidation half-reaction

Now, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the redox half-reaction, which is why you have

#{(2"Na"_ ((l))^(+) + 2"e"^(-) -> 2"Na"_ ((s))), (color(white)(aaaaaa)2"Cl"_ ((l))^(+) -> "Cl"_ (2(g)) + 2"e"^(-)) :}#

So, you know that you need #1# mole of electrons to convert #1# mole of molten sodium cations to sodium metal.

Start by converting the mass of sodium metal to moles

#1.00 * 10^3 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole Na"/(22.99 color(red)(cancel(color(black)("g")))) = 4.35 * 10^4# #"moles Na"#

So, you know that you must convert #4.35 * 10^4# moles of molten sodium cations to sodium metal, so you can say that you're going to need #4.35 * 10^4# moles of electrons.

As you know, each mole of electrons is equivalent to #9.65 * 10^4# #"C"#, which means that the total charge needed to convert these many moles of molten sodium cations to sodium metal is equal to

#4.35 * 10^4 color(red)(cancel(color(black)("moles e"^(-)))) * (9.65 * 10^4color(white)(.)"C")/(1color(red)(cancel(color(black)("mole e"^(-))))) = 4.198 * 10^9# #"C"#

Now, you know that

#"1 A" = "1 C"/"1 s"#

This means that you will need

#4.198 * 10^9 color(red)(cancel(color(black)("C"))) * overbrace("1 s"/(3.00 * 10^4color(red)(cancel(color(black)("C")))))^(color(blue)(=3.00 * 10^4"A")) = 1.399 * 10^5# #"s"#

to produce the required amount of sodium metal. Convert this to hours to get

#1.399 * 10^5 color(red)(cancel(color(black)("s"))) * "1 hr"/(3600color(red)(cancel(color(black)("s")))) = color(darkgreen)(ul(color(black)("38.9 hr")))#

Now, notice that you need #2# moles of molten chloride anions and #2# moles of electrons to produce #1# mole of chlorine gas.

This means that the reaction will produce half as many moles of chlorine gas as you have moles of sodium metal, so

#4.35 * 10^4 color(red)(cancel(color(black)("moles Na"))) * "1 mole Cl"_2/(2color(red)(cancel(color(black)("moles Na")))) = 2.175 * 10^4# #"moles Cl"_2#

To find the volume occupied by the gas at STP, use the fact that STP conditions are usually defined as a temperature of #0^@"C"# and a pressure of #"1 atm"#.

Using the ideal gas law equation

#PV = nRT#

Here

  • #P# is the pressure of the gas
  • #V# is the volume it occupies
  • #n# is the number of moles of gas present in the sample
  • #R# is the universal gas constant, equal to #0.0821("atm L")/("mol K")#
  • #T# is the absolute temperature of the gas

you will have

#V = (nRT)/P#

Plug in your values to find--do not forget to convert the temperature to Kelvin!

#V = (2.175 * 10^4 color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (0 + 273.15)color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#

#color(darkgreen)(ul(color(black)(V = 4.88 * 10^5color(white)(.)"L")))#

The answers are rounded to three sig figs, the number of sig figs you have for your values.