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:
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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