Given the standard electrode potentials:
$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$
Arrange these metals in their increasing order of reducing power.
$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$
Arrange these metals in their increasing order of reducing power.
(A) The reducing power of a metal is inversely proportional to its standard reduction potential. $A$ lower (more negative) reduction potential indicates a stronger reducing agent.
The given standard reduction potentials are:
$K^{+} / K = -2.93 \, V$
$Mg^{2+} / Mg = -2.37 \, V$
$Cr^{3+} / Cr = -0.74 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Ag^{+} / Ag = 0.80 \, V$
Arranging these in increasing order of reduction potential:
$K < Mg < Cr < Hg < Ag$
Therefore,the increasing order of reducing power is:
$Ag < Hg < Cr < Mg < K$
The given standard reduction potentials are:
$K^{+} / K = -2.93 \, V$
$Mg^{2+} / Mg = -2.37 \, V$
$Cr^{3+} / Cr = -0.74 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Ag^{+} / Ag = 0.80 \, V$
Arranging these in increasing order of reduction potential:
$K < Mg < Cr < Hg < Ag$
Therefore,the increasing order of reducing power is:
$Ag < Hg < Cr < Mg < K$
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Oxidizing power of the species will increase in the order:
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