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 $\setminus \textsf{Z n \left(s\right) + P {b}^{2 +} \setminus \leftrightarrow P b \left(s\right) + Z {n}^{2 +} \left(a q\right)}$ 2. The zinc reaction, being reversed, has $\setminus \texttt{0.76 V}$. The lead reaction is normal, and has $\setminus \texttt{- 0.13 V}$ 3. Both solutions are in $\setminus \texttt{0.10 M}$ concentration, so $\setminus \textsf{Q = \setminus \frac{\left[Z {n}^{2 +}\right]}{\left[P {b}^{2 +}\right]} = 1}$. The questions: Standard cell potential is $\setminus \textsf{{E}_{\text{cell}}^{o}}$, which means the free energy change is signified by $\setminus \textsf{\setminus \Delta {G}^{o}}$. ...then, should I be calculating for $\setminus \textcolor{\to m a \to}{\setminus \textsf{{E}_{\text{cell}}^{o}}}$ or $\setminus \textcolor{l i g h t s e a g r e e n}{\setminus \textsf{{E}_{\text{cell}}}}$ when a previous question asks "Calculate the potential for the cell used..." ? Also, what should the equilibrium constant be? Going ahead with the $\setminus \textsf{\setminus \Delta {G}^{o}}$ equations, I got $k \setminus \approx 1.84 \times {10}^{74}$. This is very large. I don't think I got the correct answer.

Aug 6, 2018

It doesn't matter. Since $Q = 1$, the current state of the reaction is that ${E}_{c e l l} = {E}_{c e l l}^{\circ}$ and $\Delta G = \Delta {G}^{\circ}$. You should be able to prove that that is true, and it should take you less than 2 minutes.

Calculate ${E}_{c e l l}^{\circ}$, like the question suggests. As I mentioned earlier, the no-brainer way is to subtract the less positive from the more positive ${E}_{red}^{\circ}$.

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

$=$ $\text{0.63 V}$

So, the standard Gibbs' free energy change is just

$\textcolor{b l u e}{\Delta {G}^{\circ}} = - n F {E}_{c e l l}^{\circ}$

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

$= - 1.22 \times {10}^{5} \text{J/mol}$

$= \textcolor{b l u e}{- \text{122 kJ/mol}}$

which is clearly a very spontaneous reaction. Hence, ${K}_{c}$ SHOULD be huge.

At equilibrium, we know already that $\Delta G = 0$ (but it does NOT mean that $\Delta {G}^{\circ} = 0$!), and that $Q = K$, so that

$\Delta {G}^{\circ} = - R T \ln K$

and thus

$\textcolor{b l u e}{{K}_{c}} = {e}^{- \Delta {G}^{\circ} / R T}$

$= {e}^{- \left(- \text{122 kJ/mol")//("0.008314 kJ/mol"cdot "K" cdot "298.15 K}\right)}$

$= \textcolor{b l u e}{1.99 \times {10}^{21}}$