Question:
which option the order of reducing
power is correct.
EᶱCr2O72-/Cr3+ = 1.33V
EᶱMnO4–/Mn2+ = 1.51V
EᶱCl2/Cl– = 1.36V
EᶱCr3+/Cr = -0.74V
(i) Cr3+ < Cl–< Mn2+ < Cr
(ii) Mn2+ < Cl–< Cr3+ < Cr
(iii) Cr3+ < Cl–< Cr2O72– < MnO4–\
(iv) Mn2+ < Cr3+ < Cl–< Cr
Solution:
Option (ii) is the answer.
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