Use the standard cell potential to calculate the free energy change for the cell reaction under standard conditions?

The details:
1. I have the reaction as #\sf{Zn(s)+Pb^(2+)\harrPb(s)+Zn^(2+)(aq)}#
2. The zinc reaction, being reversed, has #\tt{0.76V}#. The lead reaction is normal, and has #\tt{-0.13V}#
3. Both solutions are in #\tt{0.10M}# concentration, so #\sf{Q=\frac{[Zn^(2+)]}{[Pb^(2+)]}=1}#.

The questions:

  • Standard cell potential is #\sf{E_"cell"^o}#, which means the free energy change is signified by #\sf{\DeltaG^o}#.
    ...then, should I be calculating for #\color(tomato){\sf{E_"cell"^o}}# or #\color(lightseagreen)(\sf{E_"cell"})# when a previous question asks "Calculate the potential for the cell used..." ?
  • Also, what should the equilibrium constant be? Going ahead with the #\sf{\DeltaG^o}# equations, I got #k\approx1.84xx10^74#.
    This is very large. I don't think I got the correct answer.

1 Answer
Truong-Son N.
Aug 6, 2018

It doesn't matter. Since #Q = 1#, the current state of the reaction is that #E_(cell) = E_(cell)^@# and #DeltaG = DeltaG^@#. You should be able to prove that that is true, and it should take you less than 2 minutes.

Calculate #E_(cell)^@#, like the question suggests. As I mentioned earlier, the no-brainer way is to subtract the less positive from the more positive #E_(red)^@#.

#E_(cell)^@ = -"0.13 V" - (-"0.76 V")#

#=# #"0.63 V"#

So, the standard Gibbs' free energy change is just

#color(blue)(DeltaG^@) = -nFE_(cell)^@#

#= -("2 mol e"^(-))/("1 mol atoms") cdot "96485 C/mol e"^(-) cdot "0.63 V"#

#= -1.22 xx 10^5 "J/mol"#

#= color(blue)(-"122 kJ/mol")#

which is clearly a very spontaneous reaction. Hence, #K_c# SHOULD be huge.

At equilibrium, we know already that #DeltaG = 0# (but it does NOT mean that #DeltaG^@ = 0#!), and that #Q = K#, so that

#DeltaG^@ = -RTlnK#

and thus

#color(blue)(K_c) = e^(-DeltaG^@//RT)#

#= e^(-(-"122 kJ/mol")//("0.008314 kJ/mol"cdot "K" cdot "298.15 K"))#

#= color(blue)(1.99 xx 10^21)#

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