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

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**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.

**The details:**

1. I have the reaction as

2. The zinc reaction, being reversed, has

3. Both solutions are in

**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

It doesn't matter. Since

Calculate

#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^@ = -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)#