# Disproportionation of silver(I) in aqueous solution

Use the following equations to predict whether or not $$\ce{Ag+}$$ ions will disproportionate in solution:

\begin{align} \ce{Ag+(aq) + e- &-> Ag} &\qquad E^\circ = \pu{+0.80 V}\\ \ce{Ag^2+(aq) + e- &-> Ag+} &\qquad E^\circ = \pu{+2.00 V} \end{align}

I used the method $$E_\mathrm{cell} = E_\mathrm{red} - E_\mathrm{ox}$$ and thought that $$\pu{0.80 V}$$ would be more negative. Therefore, it is oxidised and the reaction would be thermodynamically feasible.

However, my textbook and another webpage has a different answer. I am a bit stuck on how to get the right answer. Please someone explain me.

## 2 Answers

It is convenient to solve problems like that with a Latimer diagram, which is a great tool for predicting conditions for the reactions of disproportionation and synproportionation.

A generic Latimer diagram

$$\ce{A ->[E_1] B ->[E_2] C}$$

posesses the following properties:

• If $$E_2 > E_1,$$ then $$\ce{B}$$ is thermodynamically unstable and disproportionates to $$\ce{A}$$ and $$\ce{C}.$$
• If $$E_2 < E_1,$$ then the mixture of $$\ce{A}$$ and $$\ce{C}$$ is thermodynamically unstable and synproportionates to $$\ce{B}.$$

Now we can visualize your problem using a Latimer diagram

$$\ce{Ag^2+ ->[\pu{+2.00 V}] Ag+ ->[\pu{+0.80 V}] Ag}$$

and the condition of disproportionation resulting from the application of Nernst equation:

if the potential to the right of the species is higher than the potential on the left, it will disproportionate.

Since $$E^\circ(\ce{Ag+(aq)/Ag}) < E^\circ(\ce{Ag^2+(aq)/Ag+(aq)}),$$ disproportionation is thermodynamically unfavorable, and silver(I) can be considered stable in solution.

You can get the same result from a linear combination of both equations written for the disproportionation:

\begin{align} \ce{Ag+(aq) + e- &-> Ag} &\quad E^\circ_1 = \pu{+0.80 V} & \tag{1}\\ \ce{Ag^2+(aq) + e- &-> Ag+(aq)} &\quad E^\circ_2 = \pu{+2.00 V} &\quad|\cdot (-1)\tag{2}\\ \hline \ce{2 Ag+(aq) &-> Ag + Ag^2+(aq)} &\quad E^\circ = \pu{-1.20 V} \end{align}

Since resulting $$E^\circ = \pu{-1.20 V} < 0,$$ free Gibbs energy $$Δ_\mathrm{r}G^\circ = -nFE^\circ > 0,$$ and the disproportionation of silver(I) in solution can be considered a thermodynamically unfavorable process.

In this question, the equations given are: \begin{align} \tag{1} \ce{Ag+(aq) + e- &-> Ag} &\qquad E^\circ = \pu{+0.80 V}\\ \tag{2} \ce{Ag^2+(aq) + e- &-> Ag+} &\qquad E^\circ = \pu{+2.00 V} \end{align}

Now, since we know that $$\Delta G$$ is additive, we can use this property to proceed. (This is also the proof for why $$E_\mathrm{cell} = E_\mathrm{red} - E_\mathrm{ox}$$)

As the first step, we first find the value of $$\Delta G$$ using the formula $$\Delta G = -nFE$$

So, for the first reaction, we see that $$\Delta G = -0.8 \times 96500 = \pu{-77,200 J}$$

For the second reaction, similarly we get $$\Delta G = \pu{-193,000 J}$$

Now, rewriting the two equations using $$\Delta G$$ instead of $$E$$ in ($$1$$) and ($$2$$), we get: \begin{align} \tag{3} \ce{Ag+(aq) + e- &-> Ag} &\qquad \Delta G = \pu{-77,200 J}\\ \tag{4} \ce{Ag^2+(aq) + e- &-> Ag+} &\qquad \Delta G = \pu{-193,000 J} \end{align}

Now, the final reaction that we need is $$\tag{5} \ce{2Ag+ -> Ag + Ag^{2+}}$$

This can be achieved by subtracting ($$2$$) from ($$1$$). Now when we subtract the two, due to the additive property of $$\Delta G$$ we can simply subtract the $$\Delta G$$ of ($$4$$) from the $$\Delta G$$ of ($$3$$)

Doing so, we get: \begin{align} \ce{2Ag+ -> Ag + Ag^{2+}} &\qquad \Delta G = \pu{115,800 J} \end{align}

So, we could end the question here, since we can see that the value of $$\Delta G$$ is positive and so the reaction is not spontaneous. However, since the question was asked in terms of $$E_\mathrm{cell}$$, we can convert this into $$E_\mathrm{cell}$$ using the given formula relating $$\Delta G$$ and $$E_\mathrm{cell}$$. We get:

\begin{align} \ce{2Ag+ -> Ag + Ag^{2+}} &\qquad E_\mathrm{cell}= \pu{-1.20 V} \end{align}

As you can see the cell potential is also negative. So, there was a mistake made in finding the right manipulation of the chemical reactions given.