A) \[C{{l}^{-}},B{{r}^{-}}\] and \[{{I}^{-}}\]
B) \[B{{r}^{-}}\] and \[{{I}^{-}}\]
C) \[C{{l}^{-}}\] and \[B{{r}^{-}}\]
D) \[{{I}^{-}}\] only
Correct Answer: B
Solution :
[b] \[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}}+4{{H}_{2}}O\] \[E=1.51-\frac{0.059}{5}\log \frac{[M{{n}^{2+}}]}{[MnO_{4}^{-}]{{[{{H}^{+}}]}^{8}}}\] Taking \[M{{n}^{2+}}\] and \[MnO_{4}^{-}\] in standard state i.e. 1 M, \[E=1.51-\frac{0.059}{5}\times 8\log \frac{1}{[{{H}^{+}}]}\] \[E=1.51-\frac{0.059}{5}\times 8\times 3=1.2268V\] Hence at this pH, \[MnO_{4}^{-}\] will oxidise only \[B{{r}^{-}}\] and \[{{I}^{-}}\] as SRP of \[C{{l}_{2}}/C{{l}^{-}}\] is 1.36 V which is greater than that for \[MnO_{4}^{-}/M{{n}^{2+}}\]
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