Given the standard electrode potentials,

\( \dfrac{ K^{+}}{K}\)= –2.93V

\( \dfrac{ Ag^{+}}{Ag}\) = 0.80V

\( \dfrac{Hg^{2+}}{Hg}\) = 0.79V

\(\dfrac{ Mg^{2+}}{Mg}\) = –2.37V

\(\dfrac{Cr^{3+}}{Cr}\) = –0.74V

Arrange these metals in their increasing order of reducing power.

Asked by Pragya Singh | 1 year ago |  194

1 Answer

Solution :-

The reducing agent is stronger as the electrode potential decreases. Hence, the increasing order of the reducing power of the given metals is as given below:

Ag < Hg < Cr < Mg < K

Answered by Pragya Singh | 1 year ago

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