Redox Reactions And Electrochemistry Ques 595
Question: Given:
$ E{{{}^\circ } _{\frac{1}{2}Cl_2/C{l^{-}}}}=1.36V,E{{{}^\circ } _{C{r^{3+}}/Cr}}~=-0.74V, $
$ E{{{}^\circ } _{Cr_2O_7^{2-}/C{l^{-}}}}=1.33V,E{{{}^\circ } _{MnO_4^{-}/M{n^{2+}}}}~=1.51,V $
The correct order of reducing power of the species $ ( Cr,C{r^{3+}},M{n^{2+}}andC{l^{-}} ) $ will be:
Options:
A) $ M{n^{2+}}<C{l^{-}}<C{r^{3+}}<Cr $
B) $ M{n^{2+}}<C{r^{3+}}<C{l^{-}}<Cr $
C) $ C{r^{3+}}<C{l^{-}}<M{n^{2+}}<Cr $
D) $ C{r^{3+}}<C{l^{-}}<Cr<M{n^{2+}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Lower the value of reduction potential higher will be reducing power hence the correct order will be $ M{n^{2+}}<{C^{-}}<C{r^{3+}}<Cr $
