Question

Given the standard electrode potentials, $K^{+}$/K = –2.93V, $Ag^{+}$/ Ag = 0.80V, $Hg^{2+}$/ Hg = 0.79V, $Mg^{2+}$/Mg = –2.37V. $Cr^{3+}$/Cr = – 0.74V Arrange these metals in their increasing order of their reducing power.

Answer 1

Given from the question,

The electrode potential of each element is as follows:

K = -2.93

Ag = 0.80

Hg = 0.79

Mg = -2.37

Cr = 0.74

The largest negative value of standard electrode potential represents the element has highest reducing ability.

The largest positive value of standard electrode potential represents the element has highest oxidizing ability.

Such that, the increasing order of reducing power for the given elements is as follows,  Cr< Hg< Ag< Mg< K.