Question #88ec4
1 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,
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
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"^@ = +"2.372 V" + (-"0.126 V") = +"2.246 V"#
Since
#color(blue)(|bar(ul(color(white)(a/a)DeltaG^@ = -n * F * E_"cell"^@color(white)(a/a)|)))#
Here
In your case,
#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,
#color(blue)(|bar(ul(color(white)(a/a)DeltaG^@ = - RT * ln(K_c)color(white)(a/a)|)))#
Here
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.