Question

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}`.

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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.

Answer 1

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.

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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)`