Lower the value of reduction potential higher will be reducing power hence the correct order will be Mn2+<Cl−<Cr3+<Cr
Given : E21Cl2/Cl−o=1.36 V,ECr3+/Cro=−0.74 VECr2O72−/Cr3+o=1.33 V,EMnO4−/Mn2+o=1.51 V The correct order of reducing power of the species (Cr,Cr3+,Mn2+ and Cl−)will be:
Held on 23 Apr 2013 · Verified 6 Jul 2026.
Mn2+<Cl−<Cr3+<Cr
Mn2+<Cl3+<Cl−<Cr
Cr3+<Cl−<Mn2+<Cr
Cr3+<Cl−<Cr<Mn2+
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