Question

question 4 (1 point)
identify the spectator ion in the reaction:
$\ce{zn(s) + snso_{4}(aq) -> znso_{4}(aq) + sn(s)}$
$\ce{zn(s)}$
$\ce{zn^{2+}(aq)}$
$\ce{so^{2-}_{4}(aq)}$
$\ce{sn^{2+}(aq)}$
there is no spectator ion in this reaction

Explanation:

A spectator ion is an ion that remains unchanged in a chemical reaction (does not participate in the net reaction). First, we write the ionic equation. The reactants are \( \text{Zn}(s) \) and \( \text{SnSO}_4(aq) \) (which dissociates into \( \text{Sn}^{2+}(aq) \) and \( \text{SO}_4^{2 -}(aq) \)). The products are \( \text{ZnSO}_4(aq) \) (dissociates into \( \text{Zn}^{2+}(aq) \) and \( \text{SO}_4^{2 -}(aq) \)) and \( \text{Sn}(s) \). The ionic reaction is \( \text{Zn}(s)+\text{Sn}^{2+}(aq)
ightarrow\text{Zn}^{2+}(aq)+\text{Sn}(s) \). Here, \( \text{SO}_4^{2 -}(aq) \) is present on both sides (reactant and product) in the same form, so it does not participate in the redox process (Zn is oxidized, Sn²⁺ is reduced). \( \text{Zn}(s) \) and \( \text{Sn}(s) \) are solids (not ions), \( \text{Zn}^{2+} \) and \( \text{Sn}^{2+} \) are involved in the reaction (oxidation - reduction). So \( \text{SO}_4^{2 -}(aq) \) is the spectator ion.

Answer:

\( \boldsymbol{\text{SO}_4^{2 -}(aq)} \) (the option corresponding to \( \text{SO}_4^{2 -}(aq) \))