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Step-by-Step Explanation
Step 1: Identify the given standard reduction potentials
We have the following standard reduction half-reactions and their electrode potentials:
$\displaystyle \text{Zn}^{2+} + 2 e^- \to \text{Zn}(s);\quad E^\circ = -0.76\,\text{V}$
$\displaystyle \text{Ca}^{2+} + 2 e^- \to \text{Ca}(s);\quad E^\circ = -2.87\,\text{V}$
$\displaystyle \text{Mg}^{2+} + 2 e^- \to \text{Mg}(s);\quad E^\circ = -2.36\,\text{V}$
$\displaystyle \text{Ni}^{2+} + 2 e^- \to \text{Ni}(s);\quad E^\circ = -0.25\,\text{V}$
Step 2: Recall the relationship between reduction potential and reducing power
The metal that has a more negative reduction potential is more likely to lose electrons (i.e., be oxidized) in a redox process. Therefore, it acts as a stronger reducing agent. In simpler terms:
More negative $E^\circ_{\text{red}}$ → Stronger tendency to get oxidized → Stronger reducing agent.
Step 3: Arrange the metals in order of increasing reducing power
Examine and compare the given reduction potentials from the least negative to the most negative:
$\displaystyle E^\circ(\text{Ni}^{2+}/\text{Ni}) = -0.25\,\text{V}$ (least negative)
$\displaystyle E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\,\text{V}$
$\displaystyle E^\circ(\text{Mg}^{2+}/\text{Mg}) = -2.36\,\text{V}$
$\displaystyle E^\circ(\text{Ca}^{2+}/\text{Ca}) = -2.87\,\text{V}$ (most negative)
This means Ni has the smallest (least negative) magnitude of reduction potential, thus it is the weakest reducing agent, while Ca with the most negative reduction potential is the strongest reducing agent.
Step 4: Final order of reducing power
From weakest to strongest reducing agent, the order is:
Ni < Zn < Mg < Ca
This matches the correct answer given in the question.
Conclusion
The reducing power of the metals increases as their standard reduction potential becomes more negative. Therefore, the correct order is Ni < Zn < Mg < Ca.
