To determine the reduction half-reaction for the given reaction, we must first identify the species that are being reduced. Remember that reduction involves the gain of electrons and a decrease in oxidation state.
Given reaction:
[tex]\[ \operatorname{Mo}(0) + \operatorname{ZnCl}_2(\operatorname{aq}) + \operatorname{MgCl}_2(\operatorname{aq}) + \operatorname{Zn}(s) \][/tex]
Among the options provided, let's analyze and identify the proper reduction half-reaction.
- Option A: [tex]\(\operatorname{Mg}^{2+} + 2 e^{-} \rightarrow \operatorname{Mg}(0)\)[/tex]
This involves the reduction of magnesium ions to magnesium metal. However, in the context of the given reaction, it does not make sense as magnesium is not directly involved in the main reaction sequence provided.
- Option B: [tex]\(\operatorname{Zn}(0) + 2 n^{24} + 2 e^{-}\)[/tex]
This appears to be an incorrectly written half-reaction. Moreover, it doesn't represent a standard half-reaction and contains inconsistencies.
- Option C: [tex]\(\operatorname{Mo}(0) \rightarrow Mo^{2+} + 2 e^{-}\)[/tex]
This represents an oxidation half-reaction, where molybdenum metal is losing electrons and being oxidized, not reduced.
- Option D: [tex]\(\operatorname{Zn}^{2+} + 2 e^{-} \rightarrow \operatorname{Zn}(0)\)[/tex]
This half-reaction involves the reduction of zinc ions to zinc metal. Here, zinc ions are gaining electrons, which is consistent with the definition of a reduction half-reaction.
After careful analysis, we can conclude that the correct reduction half-reaction is:
[tex]\[ \operatorname{Zn}^{2+} + 2 e^{-} \rightarrow \operatorname{Zn}(0) \][/tex]
Therefore, the answer is:
D. [tex]\(\operatorname{Zn}^{2+} + 2 e^{-} \rightarrow \operatorname{Zn}(0)\)[/tex]
