This a question that offers an answer with electrochemical equilibrium. Our experiment is as follows:
- We have a typical beaker of volume $V = \pu{200 mL}$ of water. On top of it we have a volume $V_\mathrm{a} = \pu{1 dm3}$ of air, at ambient conditions of $p = \pu{1 bar}$ and $T = \pu{25 °C}$.
- We add a mass $m = \pu{1 g}$ of zinc.
- Determine the final state after a long period of time.
First attempt
One of the most important issues with these type of problems is to determine which types and how many reactions will take place. Let's consider for simplicity that only the oxidation of zinc takes place:
$$\ce{Zn(s) -> Zn^2+(aq) + 2e^- \tag{1}}$$
However, if only this happens we have the following problems:
- The electrolyte solution becomes positively charged with $\ce{Zn^2+}$ without any counterbalance of anions. This is a violation of charge conservation taking place at the macroscopic scale in the liquid phase.
- The solid zinc becomes negatively charge with $\ce{e^-}$. This is a violation of charge conservation taking place at the macroscopic scale in the electronic phase.
Thus, we reject that only Eq. (1) takes place.
A more rapid way to show that this is impossible is the known phrase taught at at general chemistry: 'an oxidation needs a reduction or vice versa', so we cannot have a single oxidation.
Second attempt
We know that self-ionisation of pure water brings negligible but some amount of protons and hydroxides to the solution. This raises the interesting question: can the two electrons from Eq. (1) be grabbed by the hydrogen ions, and produce hydrogen gas? This situation will not violate any law of charge conservation, so we see whether this is possible by thermodynamics.
The system has two electrochemical reactions and one chemical reaction
\begin{align}
\ce{Zn(s) &-> Zn^2+(aq) + 2e^-} \tag2 \\
\ce{2H+(aq) + 2e^- &-> H2(g)} \tag3 \\
\ce{H2O(l) &-> H+(aq) + OH^-(aq)} \tag4 \\
\end{align}
Eqs. (2) and (3) lead to two electrochemical equilibrium, while Eq. (4) to a chemical equilibrium
\begin{align}
E &= E^\circ(\ce{Zn^2+/Zn}) - \frac{RT}{2F}\ln ([\ce{Zn^2+}]) \tag5 \\
E &= E^\circ(\ce{H+/H2}) - \frac{RT}{2F}\ln
\left(\frac{p_\ce{H2}}{[\ce{H+}]^2}\right)
\tag6 \\
K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] \tag7
\end{align}
and we count 5 unknowns: $E$, $[\ce{Zn^2+}]$, $p_\ce{H2}$, $[\ce{H+}]$, and $[\ce{OH-}]$. We need two more equations. The first one is charge conservation
\begin{align}
[\ce{H+}] + 2[\ce{Zn^2+}] &= [\ce{OH^-}] \tag8 \\
\end{align}
The other one is a relation between the partial pressure of hydrogen and other species. Let $\xi_2$ and $\xi_3$ be the extents of reactions 2 and 3. Then, the amount of hydrogen, protons, and hydroxides in terms of these extents are
\begin{align}
n_\ce{H2} &= \xi_2 \tag{9} \\
n_\ce{H+} &= n_\ce{H+}^0 - 2\xi_2 + \xi_3 \tag{10} \\
n_\ce{OH-} &= n_\ce{OH-}^0 + \xi_3 \tag{11}
\end{align}
By Eq. (11) we can write $\xi_3 = n_\ce{OH-} - n_\ce{OH-}^0$. Introducing this expression and Eq. (9) in Eq. (10) gives
\begin{align}
n_\ce{H+} &= n_\ce{H+}^0 + 2n_\ce{H2} + n_\ce{OH-} - n_\ce{OH-}^0 \\
2n_\ce{H2} &= n_\ce{H+}^0 - n_\ce{H+} + n_\ce{OH-} - n_\ce{OH-}^0 \\
2n_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0)V \\
p_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0)
V\left(\frac{RT}{2V_\mathrm{a}}\right) \tag{12} \\
\end{align}
and we note that the volume $V$ refers to the species in the liquid phase, which is not the same as the volume $V_\mathrm{a}$ for the gas phase.
And we have a set of 5 non-linear equations for the system, which we repeat again
\begin{align}
E &= E^\circ(\ce{Zn^2+/Zn}) - \frac{RT}{2F}\ln ([\ce{Zn^2+}]) \tag{13} \\
E &= E^\circ(\ce{H+/H2}) - \frac{RT}{2F}\ln
\left(\frac{p_\ce{H2}}{[\ce{H+}]^2}\right) \tag{14} \\
K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] \tag{15} \\
[\ce{H+}] + 2[\ce{Zn^2+}] &= [\ce{OH^-}] \tag{16} \\
p_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0)
V\left(\frac{RT}{2V_\mathrm{a}}\right) \tag{17} \\
\end{align}
It is possible to combine this equations to a single unknown, but this will unecessary extend the answer. We tabulate the solutions:
\begin{array}{|c|c|}
[\ce{H+}] \; \pu{(mol dm^-3)} & \pu{9.0632E-13} \\
[\ce{OH-}] \; \pu{(mol dm^-3)} & \pu{5.5168E-2} \\
[\ce{Zn^2+}] \; \pu{(mol dm^-3)} & \pu{1.1034E-3} \\
p_\ce{H2} \; \pu{(bar)} & \pu{0.27352} \\
E \; (\pu{V}) & \pu{-0.69598} \\
\end{array}
The results indicate:
- Indeed, the hydrogen ions were consumed and a bit of zinc entered the solution.
- The electrode potential $E = \phi^\mathrm{M} - \phi^\mathrm{S} < 0$, thus the electric potential of the liquid phase became more negative.
By mass conservation, we can finally see how much the mass of zinc solution changed (remember we put $\pu{1 g}$ initially)
\begin{equation}
m_\ce{Zn} = m_\ce{Zn}^0 - [\ce{Zn^2+}]VM(\ce{Zn}) \to
\boxed{m_\ce{Zn} = \pu{0.99278 g}}
\end{equation}
where $M(\ce{Zn}) = \pu{65.41 g mol^-1}$ is the molar mass of zinc.
