Effect of H+ concentration on equilibria

Question

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?

Answer 1

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).

Answer 2

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$1$).