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

##### 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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