NCERT Class XI Chemistry Redox Reactions Solutions

Question : 29

Total: 30

Given the standard electrode potentials, K+∕K = –2.93 V, Ag+∕Ag = 0.80 V, Hg2+∕Hg = 0.79 V, Mg2+∕Mg = –2.37 V, Cr3+∕Cr = –0.74 V. Arrange these metals in increasing order of their reducing power.

Solution:

Lower the electrode potential, better is the reducing power. Since the electrode potentials increase in the order; K+∕K (– 2.93 V), Mg2+∕Mg (– 2.37 V), Cr3+∕Cr (– 0.74 V), Hg2+∕Hg (0.79 V), Ag+∕Ag (0.80 V), therefore, reducing power of metals increases in the order, i.e., Ag < Hg < Cr < Mg < K.

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