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. $$\pu{-213 kJ/mol}$$

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?

• Please note that this question is currently self-answered. I found the correct numerical value for the answer, but I'm still unsure if I have solved it correctly (sometimes wrong methods give the right end-answers too). Hence I've posted it here so that other users can verify if my method is correct or wrong. If wrong, the problem still stands, and of course others' insights and answers are welcomed!
– TRC
Commented Aug 21, 2021 at 4:24
• Very good way of using this site @TRC. Commented Aug 22, 2021 at 13:33
• @Mathew For future reference, you can make that even shorter (if you like): \pu{2\times 10^{-8} M^2} -> \pu{2E-8 M2} $\pu{2E-8 M2}$ Commented Aug 22, 2021 at 18:03

1 Answer

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:

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

• standard state oh- is 1? Commented Aug 22, 2021 at 13:37
• @Buraian From what I've learnt standard state of an ion is defined to be a concentration 1 M at STP. If you feel there's a mistake in that, please let me know.
– TRC
Commented Aug 22, 2021 at 14:29
• @TRC. The concentration of $\ce{Zn^{2+}}$ cannot be $1$ M, as you write. In the presence of $\ce{OH^-}$ ions, zinc makes a rather insoluble precipitate of $\ce{Zn(OH)2}$. The solubility product of $\ce{Zn(OH)2}$ is equal to $1.24 ·10^{-17}$. So the solubility of $\ce{Zn(OH)2}$ cannot be higher than $\ce{10^{-6}}$ M. Commented Aug 22, 2021 at 20:29
• @Maurice You could run it in a cell with acidic pH in in zinc half cell and basic pH in the silver half cell.
– Karsten
Commented Aug 22, 2021 at 22:20
• @Karsten Theis. You are right. I have not thought of this possibility. Commented Aug 24, 2021 at 16:39