| \[E{{{}^\circ }_{\frac{1}{2}C{{l}_{2}}/C{{l}^{-}}}}=1.36V,\text{ }E{{{}^\circ }_{C{{r}^{3+}}/Cr}}~=-0.74V,\] |
| \[E{{{}^\circ }_{C{{r}_{2}}O_{7}^{2-}/C{{l}^{-}}}}=1.33V,\text{ }E{{{}^\circ }_{MnO_{4}^{-}/M{{n}^{2+}}}}~=1.51\,V\] |
| The correct order of reducing power of the species \[\left( Cr,C{{r}^{3+}},\text{ }M{{n}^{2+}}and\text{ }C{{l}^{-}} \right)\] will be: |
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) \[C{{r}^{3+}}<C{{l}^{-}}<Cr<M{{n}^{2+}}\]
Correct Answer: A
Solution :
[a] Lower the value of reduction potential higher will be reducing power hence the correct order will be \[M{{n}^{2+}}<{{C}^{-}}<C{{r}^{3+}}<Cr\]
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