Redox Reactions And Electrochemistry Question 548
Question: At 298 K, the standard reduction potentials are 1.51 V for $ MnO_4^{-}|M{n^{2+}},1.36V,\text{for }Cl_2|C{l^{-}},1.07 $ $ V,for,Br_2/B{r^{-}} $ , and 0.54 V for $ I_2/{I^{-}}.AtpH=3, $ permanganate is expected to oxidize: $ ( \frac{RT}{F}=0.059V ) $
Options:
A) $ C{l^{-}},B{r^{-}} $ and $ {I^{-}} $
B) $ B{r^{-}} $ and $ {I^{-}} $
C) $ C{l^{-}} $ and $ B{r^{-}} $
D) $ {I^{-}} $ only
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ MnO_4^{-}+8{H^{+}}+5{e^{-}}\xrightarrow{{}}M{n^{2+}}+4H_2O $ $ 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 $ Cl_2/C{l^{-}} $ is 1.36 V which is greater than that for $ MnO_4^{-}/M{n^{2+}} $
BETA
