Redox Reactions And Electrochemistry Ques 727
Question: Given
(i) $ MnO_4^{-}+8{H^{+}}+5{e^{-}}\xrightarrow{{}} $ $ M{n^{2+}}+4H_2,O $ (ii) $ MnO_2+4{H^{+}}+2{e^{-}}\xrightarrow{{}} $ $ M{n^{2+}}+2H_2,OE^{0}=x_2V $ Find $ E^{0} $ for the following reaction $ MnO_4^{-}+4{H^{+}}+3{e^{-}}\xrightarrow{{}}MnO_2+2H_2,O $
Options:
A) $ x_2-x_1 $
B) $ x_1-x_2 $
C) $ \frac{5x_1-2x_2}{3} $
D) $ \frac{2x_1-5x_2}{3} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \Delta G_1^{0}=-5x_1F $ ; $ \Delta G_2^{0}=-2x_2F $ $ \Delta G_3^{0}=\Delta G_1^{0}-\Delta G_2^{0} $ ; $ \Delta G_3^{0}=2x_2F-5x_1F $
$ \therefore $ $ x=\frac{5x_1-2x_2}{3} $
