To determine which ion is most likely to be reduced, we need to compare their standard reduction potentials. The standard reduction potential (E°) indicates how readily a species gains electrons and is reduced. The more positive the reduction potential, the more likely the species is to be reduced.
Here are the standard reduction potentials for the ions given:
- \( \mathrm{Zn^{2+} + 2e^- \rightarrow Zn} \): \( E^\circ = -0.76 \, \mathrm{V} \)
- \( \mathrm{Mg^{2+} + 2e^- \rightarrow Mg} \): \( E^\circ = -2.37 \, \mathrm{V} \)
- \( \mathrm{Fe^{2+} + 2e^- \rightarrow Fe} \): \( E^\circ = -0.44 \, \mathrm{V} \)
- \( \mathrm{Ni^{2+} + 2e^- \rightarrow Ni} \): \( E^\circ = -0.25 \, \mathrm{V} \)
These reduction potentials indicate how easily these ions are reduced:
- \( \mathrm{Zn^{2+}} \) has a reduction potential of -0.76 V.
- \( \mathrm{Mg^{2+}} \) has a much lower (more negative) reduction potential of -2.37 V, meaning it is much less likely to be reduced.
- \( \mathrm{Fe^{2+}} \) has a reduction potential of -0.44 V.
- \( \mathrm{Ni^{2+}} \) has a reduction potential of -0.25 V.
Comparing these values, \( \mathrm{Ni^{2+}} \) has the most positive (least negative) reduction potential among the given ions. Therefore, \( \mathrm{Ni^{2+}} \) is the most likely to be reduced.
The answer is:
D. [tex]\( \mathrm{Ni^{2+}} \)[/tex]
