Given the standard electrode potentials, K +/K = −2.93V, Ag+/Ag = 0

1.	Given the standard electrode potentials, K +/K = −2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = −2.37 V, Cr 3+/Cr = − 0.74V Arrange these metals in their increasing order of reducing power.
Answer» The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K+/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag. Hence, the reducing power of the given metals increases in the following order: Ag < Hg < Cr < Mg < K

Given the standard electrode potentials, K +/K = −2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = −2.37 V, Cr 3+/Cr = − 0.74V Arrange these metals in their increasing order of reducing power.

Answer»

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K+/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag.

Hence, the reducing power of the given metals increases in the following order: Ag < Hg < Cr < Mg < K

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Given the standard electrode potentials, K +/K = −2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = −2.37 V, Cr 3+/Cr = − 0.74V Arrange these metals in their increasing order of reducing power.

Given the standard electrode potentials, K +/K = −2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = −2.37 V, Cr 3+/Cr = − 0.74V Arrange these metals in their increasing order of reducing power.

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