Given that the standard electrode potentials (E°) of metals are:
K/K = −2.93 V, Ag/Ag = 0.80 V, Cu/Cu = 0.34 V, Mg/Mg = −2.37 V, Cr/Cr = −0.74 V, Fe/Fe = −0.44 V.
Arrange these metals in an increasing order of their reducing power.
K/K = −2.93 V, Ag/Ag = 0.80 V, Cu/Cu = 0.34 V, Mg/Mg = −2.37 V, Cr/Cr = −0.74 V, Fe/Fe = −0.44 V.
Arrange these metals in an increasing order of their reducing power.
Answer: Ag < Cu < Fe < Cr < Mg < K
Solution
The more negative the standard electrode potential, the greater the tendency to be oxidised, i.e. the stronger the metal as a reducing agent.
Arranging by E° from highest to lowest gives the increasing order of reducing power:
Arranging by E° from highest to lowest gives the increasing order of reducing power:
CBSE Class 12 Board · 2010 · All India · Subjective