Exercises
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 their increasing order of reducing power.
Reducing power of a metal increases as its standard reduction potential (E°) becomes more negative — a lower (more negative) E° means the metal is oxidised more readily, i.e., it is a stronger reducing agent.
Given E° values:
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
Arranging in increasing order of reducing power means arranging in increasing order of negative E° (or decreasing E° value):
Ag (E° = 0.80 V) < Hg (E° = 0.79 V) < Cr (E° = -0.74 V) < Mg (E° = -2.37 V) < K (E° = -2.93 V)
Answer: Increasing order of reducing power: Ag < Hg < Cr < Mg < K
