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 2


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.


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