A 1 mol sample of zinc can reduce the greatest number of moles of which of the following ions? — Electrochemistry Chemistry Question

STEPS:

1. Identify the Oxidation Half^-Reaction for Zinc: Zinc ($Zn$) is a metal that typically undergoes oxidation to the $+2$ state: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$. According to this stoichiometry, 1 mole of $Zn$ atoms will release 2 moles of electrons.
2. Understand the Reduction Process: This question tests the concept of redox stoichiometry. To "reduce" an ion, those 2 moles of electrons provided by the zinc must be accepted by the target ions. The number of moles of an ion that can be reduced depends on the charge of the ion; ions with a lower positive charge require fewer electrons per mole and thus can be reduced in greater quantities.
3. Analyze the Electron Requirements for the Cations:
* $Ag^+$ (Choice C): Each silver ion requires 1 electron to be reduced to solid silver ($Ag^+ + e^- \rightarrow Ag$).
* $Pb^{2+}$ (Choice B): Each lead(II) ion requires 2 electrons to be reduced ($Pb^{2+} + 2e^- \rightarrow Pb$).
* $Al^{3+}$ (Choice A): Each aluminum ion requires 3 electrons to be reduced ($Al^{3+} + 3e^- \rightarrow Al$).
4. Calculate the Total Moles Reduced by 1 Mol of Zinc: Using the 2 moles of electrons available from 1 mole of $Zn$:
* Moles of $Ag^+$ = $2\text{ mol } e^- / 1\text{ mol } e^- \text{ per ion} = \mathbf{2.0\text{ mol}}$
* Moles of $Pb^{2+}$ = $2\text{ mol } e^- / 2\text{ mol } e^- \text{ per ion} = \mathbf{1.0\text{ mol}}$
* Moles of $Al^{3+}$ = $2\text{ mol } e^- / 3\text{ mol } e^- \text{ per ion} = \mathbf{0.67\text{ mol}}$
5. Conclusion: Because $Ag^+$ has the lowest positive charge among the cations, the 2 moles of electrons from zinc can reduce a larger molar quantity of it compared to $Pb^{2+}$ or $Al^{3+}$. Therefore, zinc can reduce 2 moles of $Ag^+$, which is the greatest number of moles among the options.

WHY_OTHERS_WRONG:

• $Al^{3+}$ and $Pb^{2+}$: These cations have higher positive charges than $Ag^+$. Because they require more electrons per mole (3 and 2 respectively) to reach their elemental state, the fixed supply of electrons from 1 mole of zinc is exhausted more quickly, resulting in fewer total moles reduced.
• $Cl^-$ and $N^{3-}$: These are anions already in their reduced states (negative oxidation numbers). In standard aqueous chemistry, they cannot accept more electrons to be reduced further; rather, they would typically act as spectator ions or reducing agents themselves.