Calculating Gibbs free energy change when the state of ions is unknown

### Problem



> 
> Calculate $\Delta G^\circ$ for the following reaction:  
> 
> $$\ce{Zn(s) + Ag2O(s) + H2O(l) -> Zn^{+2}(aq) + 2Ag(s) + 2OH-(aq)}$$  
> 
> Given $E^\circ\_{\ce{Ag+/Ag}} = \pu{0.80 V}$, $E^\circ\_{\ce{Zn^{+2}/Zn}} = \pu{-0.76 V}$, $K\_\mathrm{sp}$ of $\ce{AgOH} = \pu{2\times 10^{-8} M^2}$
> 
> 
> **A**. $\pu{-301 kJ/mol}$  
> 
> **B**. $\pu{-213 kJ/mol}$
> 
> 
> 


### Answer



> 
>  **B**
> 
> 
> 


### Question


This reaction can be split into two redox reactions:  

$\ce{Zn -> Zn^{+2} + 2e-}$  

$\ce{Ag+ + e- -> Ag}$


Accordingly, my initial approach was to take the $E^\circ$ of the entire cell, which comes out to be $\pu{1.56 V}$. Then using the formula $\Delta G = -nFE$, We get option **A** as the answer.


What am I missing?




The key thing to note here is that $E\_\mathrm{cell} = E^\circ$ only when the ions involved are all in their standard states, i.e. concentrations of all ions is $\pu{1 M}$. But here we need to find $\Delta G^\circ$ for the reaction:  

$$\ce{Zn(s) + Ag2O(s) + H2O(l) -> Zn^{+2}(aq) + 2Ag(s) + 2OH-(aq)},$$


whereas the redox reactions involved in the cell would be:  

\begin{align}
\ce{Zn &-> Zn^{+2} + 2e-},\\
\ce{Ag+ + e- &-> Ag},
\end{align}


which requires $\ce{Ag+}$ to be present in its standard state. However, the first reaction given only tells us that $\ce{Ag2O}$ and $\ce{H2O}$ on the side of reactants are in their standard states. *We hence don't know anything yet about the concentration of $\ce{Ag+}$, required in the cell*.


Using the expression of $K\_\mathrm{sp}$, $\ce{[Ag+][OH-]} = \pu{2E-8 M2}$:  

From the products side of the first reaction, we know that $\ce{OH-}$ is in its standard state, so $\ce{[OH-] = 1 M}$, which gives $\ce{[Ag+]} = \pu{2E-8 M}$.


Now we use Nernst equation for this cell:


\begin{align}
 \Delta G^\circ &= -nFE\_\mathrm{cell}\\
 \Delta G^\circ &= -nF\left(E^\circ\_\mathrm{cell} 
 - \frac{0.059}{n}\log\frac{\ce{Zn^{+2}}}{\ce{Ag+}}\right)
\end{align}


$\ce{[Zn^{+2}]} = \pu{1 M}$ as it is in its standard state. Using $\ce{[Ag+]} = \pu{2E-8 M}$, we get:


$$ \Delta G^\circ = -2 \times 96.5 \times (1.56 - 0.059(8-\log2)) = \pu{-213 kJ/mol}$$


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