Chemistry

Chemistry

Problem

Given the standard electrode potentials, K+/K=−2.93V\text{K}^+/\text{K} = -2.93\text{V}, Ag+/Ag=0.80V\text{Ag}^+/\text{Ag} = 0.80\text{V}, Hg2+/Hg=0.79V\text{Hg}^{2+}/\text{Hg} = 0.79\text{V}, Mg2+/Mg=−2.37V\text{Mg}^{2+}/\text{Mg} = -2.37\text{V}, Cr3+/Cr=−0.74V\text{Cr}^{3+}/\text{Cr} = -0.74\text{V}, arrange these metals in their increasing order of reducing power.
Intuitive explanation

Reducing power measures how readily a metal donates electrons—a stronger reducer loses electrons more easily. The key insight is that more negative electrode potentials correspond to stronger reducing agents: a metal with a very negative E°E° sits far below the hydrogen reference point, making it energetically favorable for it to be oxidized. Simply rank these metals from most negative to most positive potential, and you've ordered them from strongest to weakest reducer.

Intuitive explanation

Reducing power measures how readily a metal donates electrons—a stronger reducer loses electrons more easily. The key insight is that more negative electrode potentials correspond to stronger reducing agents: a metal with a very negative E°E° sits far below the hydrogen reference point, making it energetically favorable for it to be oxidized. Simply rank these metals from most negative to most positive potential, and you've ordered them from strongest to weakest reducer.

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