1
$\begingroup$

Considering the two equilibriums below, when the $[\ce{H+}]$ decreases, my book says that equilibrium \eqref{rxn:2} will shift to the left by a greater extent compared to equilibrium \eqref{rxn:1} because equilibrium \eqref{rxn:2} has a greater amount of $\ce{H+}.$ (4 mol compared to 2 mol)

$$ \begin{align} \ce{VO2+ + 2 H+ + e- &<=> VO^2+ + H2O}\label{rxn:1}\tag{R1}\\ \ce{VO3- + 4 H+ + e- &<=> VO^2+ + 2 H2O}\label{rxn:2}\tag{R2} \end{align} $$

However, I do not understand the link between the amount of $\ce{H+}$ and the extent of shift in position of equilibrium when faced with falling $[\ce{H+}].$ Why will equilibrium 2 shift by a greater extent?

$\endgroup$
5
  • 1
    $\begingroup$ You do not distinguish $\ce{VO2+}$, written as $\ce{VO2+}$ and $\ce{VO^2+}$ written as $\ce{VO^2+}$ // For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ). $\endgroup$
    – Poutnik
    Oct 24 '21 at 13:42
  • 1
    $\begingroup$ Try to enumerate equations properly, this skill is based on counting to 20 and on conservation of atom and charge counts. $\endgroup$
    – Poutnik
    Oct 24 '21 at 13:45
  • 1
    $\begingroup$ You can apply \ce{} to the whole equation, not for each formula, using -> as the reaction arrow. $\endgroup$
    – Poutnik
    Oct 24 '21 at 14:16
  • $\begingroup$ If you write the equilibrium expressions, the second one has hydrogen ion concentration to the 4th power in the denominator while the first one has hydrogen ion concentration only to the second power. So changing hydrogen ion concentration has more effect on the second equilibrium. So decreasing hydrogen ion concentration by a factor of 2 results in reducing product concentration by a factor of 4 for R1 and by a factor of 16 for R2. $\endgroup$
    – Ed V
    Oct 29 '21 at 13:20
  • $\begingroup$ Since you lost 50 rep with the bounty, but got good answers, I will give you 10 back. $\endgroup$
    – Ed V
    Oct 29 '21 at 21:24
2
+50
$\begingroup$

If we take the half reactions as given (so ignoring Maurice's insight about the stability of the species at different pH), the answers are in the comments.

[Ed V] If you write the equilibrium expressions, the second one has hydrogen ion concentration to the 4th power in the denominator while the first one has hydrogen ion concentration only to the second power. So changing hydrogen ion concentration has more effect on the second equilibrium.

and

[Poutnik] If you formulate Nernst equation for a particular half-reaction, you can easily derive the rate of potential change with changing pH.

Intuitive answer?

For each turnover in the forward direction of R1, two $\ce{H+}$ ions have to come together. Increasing the $\ce{H+}$ concentration by a factor 10 (lowering the pH by one unit) will make that 100-times more likely. For R2, you need four $\ce{H+}$ ions. Increasing the $\ce{H+}$ concentration by a factor 10 will make that 10,000-times more likely.

The pH dependence will be the same no matter whether you formulate an acidic or basic half reaction. You can write R2 as a basic reaction,

$$\ce{VO3^- + 2H2O + e- <=> VO^2+ + 4OH- \tag{R5}}$$

and instead of four $[\ce{H+}]$ on the left side you have four $[\ce{OH-}]$ on the right side. Lowering the pH by one unit would in turn lower $[\ce{OH-}]$ by a factor 10, making it 10,000 times less likely that four $\ce{OH-}$ ions would come together. So written as R2 or as R5, this reaction is more sensitive to changes in pH.

Reaction vs half reaction

Both R1 and R2 are half-reactions that would not attain equilibrium unless they are combined with a second half reaction. For a half-reaction, you could discuss the reduction potential and its dependence on pH. Or you could combine both half reactions with the same other half-reaction (which should be independent of pH to keep things simple).

$\endgroup$
3
  • $\begingroup$ The last sentence in my comment was sloppy: I should not have said 4 and 16, as though no reactant was being produced by the equilibrium shifts to the left. So your answer cleaned up my error nicely! $\endgroup$
    – Ed V
    Oct 29 '21 at 20:59
  • $\begingroup$ @Karsten Theis Thank for the answer! Can i clarify one thing: Does the reaction quotient tell you how likely the species in a reaction are going to come together? $\endgroup$
    – john
    Oct 30 '21 at 2:08
  • $\begingroup$ @john No, comparing the reaction quotient Q to K tells you how far from equilibrium you are, and also the Gibbs energy of reaction as $$\Delta G = R T \ln \frac{Q}{K}$$ This is related to the entropy of mixing, which most folks don't consider intuitive. However, you can use a kinetic argument for changing parameters of a reaction that has reached equilibrium like I did above. $\endgroup$ Oct 30 '21 at 2:14
1
$\begingroup$

Before studying your two redox equations (R$1$) and (R$2$), it is necessary to simply compare the ratio of the two ions dioxovanadium(V) $\ce{VO2^+}$ and vanadate $\ce{VO3^-}$ which both exist in solutions containing Vanadium at the oxidation number +$5$. Both ions occur in the left-hand side of your two equations (R$1$) and (R$2$).

Let's proceed like this. These two Vanadium containing ions are both interrelated by an equilibrium like $$\ce{VO2^+ + 2 OH- <=> VO3^- + H2O \tag{R3}}$$ which is equivalent to $$\ce{VO2^+ + H2O <=> VO3^- + 2 H+\tag{R4}}$$If sodium vanadate $\ce{NaVO3}$ solution is acidified by adding some $\ce{HCl}$ solution, the reaction (R$4$) gets inversed and produces the dioxovanadium(V) ion $\ce{VO2^+}$. This reaction is reversible. By adding some solution of $\ce{NaOH}$, the dioxovanadium(V) ion $\ce{VO2^+}$ is transformed into vanadate $\ce{VO3^-}$ like in (R$3$).

The system dioxovanadium(V)/vanadate is working like an indicator. Dioxovanadium(V) ion exists mainly in acidic solution. Vanadate ion exists mainly in basic solutions.

Now, to go back to the original questions, you should realize that the equation (R$1$) is the only one that does make sense, because you may imagine a half-cell containing both $\ce{H+}$ and $\ce{VO2^+}$ simultaneously in solution. In contrary, equation (R$2$) is purely theoretical, because the ion vanadate $\ce{VO3^-}$ does not exist significantly in acidic solution. Before taking part to the redox reaction, it is immediately transformed by $\ce{H+}$ into dioxovanadium(V) ion. This is why the reaction (R$2$) needs more $\ce{H+}$ ions than (R$1$).

$\endgroup$
6
  • $\begingroup$ You don't need to spam dots like you do with .........(3), just tag the reactions with \tag{R<number>} (I use prefix R to differentiate chemistry from math). If you refer to the reaction in text, label the reaction and use \eqref{<label>} to create a dynamic link. For the ease of reading and parsing, aligning reactions about arrows is a good idea, too (similarly as one does with multiline formulas/cases/series of equations). See my edit to the main question for the example how it can be done. Finally, chemical names should not be capitalized unless the first word in sentence. $\endgroup$
    – andselisk
    Oct 24 '21 at 14:29
  • $\begingroup$ @Putnik. You are right about the possible confusion between $\ce{VO2^+}$ and $\ce{VO^{2+}}$., I have not made this mistake. $\endgroup$
    – Maurice
    Oct 24 '21 at 14:43
  • $\begingroup$ @Maurice Hi, I don't quite understand how this allows us to compare which equilibrium has a larger shift in POE when concentration of H+ decreases? $\endgroup$
    – john
    Oct 24 '21 at 14:43
  • $\begingroup$ @john If you formulate Nernst equation for a particular half-reaction, you can easily derive the rate of potential change with changing pH. $\endgroup$
    – Poutnik
    Oct 24 '21 at 14:54
  • $\begingroup$ @Poutnik. You are right. Sorry ! I should have used the correct name, which is dioxovanadium(V) for the ion $\ce{VO2^+}$. I will edit this change. $\endgroup$
    – Maurice
    Oct 24 '21 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.