Chemistry Asked by isabel on October 5, 2021

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.

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.

Answered by Safdar Faisal on October 5, 2021

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.

Answered by andselisk on October 5, 2021

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