Question #88ec4

1 Answer
Jul 25, 2016

Answer:

Here's what I got.

Explanation:

Start by taking a look at the redox equilibrium given to you

#"Mg"_ ((s)) + "Pb"_ ((aq))^(2+) rightleftharpoons "Mg"_ ((aq))^(2+) + "Pb"_ ((s))#

Notice that magnesium metal, #"Mg"#, is being oxidized to magnesium cations, #"Mg"^(2+)#, and the lead(II) cations, #"Pb"^(2+)#, are being reduced to lead metal, #"Pb"#.

The oxidation half-reaction looks like this

#"Mg"_ ((s)) -> "Mg"_ ((aq))^(2+) + 2"e"^(-)#

The reduction half-reaction looks like this

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

Notice that #2# moles of electrons arebeing transferred in this reaction, Keep this in mind.

Now, the standard reduction potentials for these two half-cells are

#"Mg"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Mg"_ ((s)) " "E^@ = -"2.372 V"#

#"Pb"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Pb"_ ((s)) " " E^@ = -"0.126 V"#

Since the oxidation half-reaction involves the oxidation of magnesium metal to magnesium cations, you must reverse the reduction equilibrium for magnesium cations.

Consequently, the standard reduction potential must be reversed as well. You will have

#{("Mg"_ ((s)) rightleftharpoons "Mg"_ ((aq))^(2+) + 2"e"^(-)" " E_"ox"^@ = + "2.372 V"), ("Pb"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Pb"_ ((s)) " " color(white)(a)E_"red"^@ = -"0.126 V") :}#
#color(white)(aaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#
#" ""Mg"_ ((s)) + "Pb"_ ((aq))^(2+) rightleftharpoons "Mg"_ ((aq))^(2+) + "Pb"_ ((s))#

The standard cell potential, #E_"cell"^@#, for this reaction will be

#E_"cell"^@ = +"2.372 V" + (-"0.126 V") = +"2.246 V"#

Since #E_"cell"^@>0#, you know that this reaction will be spontaneous. Now, the standard cell potential is related to the standard Gibbs Free energy change, #DeltaG^@#, by the equation

#color(blue)(|bar(ul(color(white)(a/a)DeltaG^@ = -n * F * E_"cell"^@color(white)(a/a)|)))#

Here

#n# - the number of moles of electrons transferred in the reaction
#F# - the Faraday constant, equal to #"96,485 C mol"^(-1)#

In your case, #n=2#, since #2# moles of electrons are being transferred in the reaction. The change in standard Gibbs Free energy will be

#DeltaG^@ = -2 color(red)(cancel(color(black)("moles e"^(-)))) * "96,485 C" color(red)(cancel(color(black)("mol"^(-1)))) * "2.246 V"#

#DeltaG^@ = -"433,411" color(white)(a)overbrace("C" * "V")^(color(purple)("= J"))#

#DeltaG^@ = -"433,411 J"#

Finally, you can find the equilibrium constant for this reaction, #K_c#, by using the equation

#color(blue)(|bar(ul(color(white)(a/a)DeltaG^@ = - RT * ln(K_c)color(white)(a/a)|)))#

Here

#R# - the universal gas constant, useful as #"8.314 J mol"^(-1)"K"^(-1)#
#T# - the absolute temperature at which the reaction takes place

You can switch to a common log by using the fact that

#log(K_c) = (ln(K_c))/ln(10)#

This will get you

#DeltaG^@ = -ln(10) * RT * log(K_c)#

#DeltaG^@ = -2.303 * RT log(K_c)#

Rearrange to isolate the log on one side of the equation

#log(K_c) = (DeltaG^@)/(-2.303 * RT)#

Plug in your values to find

#log(K_c) = (-"433,411" color(red)(cancel(color(black)("J"))))/(-2.303 * 8.314color(red)(cancel(color(black)("J"))) "mol"^(-1)color(red)(cancel(color(black)("K"^(-1)))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))#

#log(K_c) = 75.92#

This implies that the equilibrium constant is equal to

#K_c = color(green)(|bar(ul(color(white)(a/a)color(black)(8.3 * 10^75)color(white)(a/a)|)))#

The answer is expressed in scientific notation and rounded to two sig figs.