Redox Reactions And Electrochemistry Question 685
Question: Consider the redox reaction : $ 2SO_2O_3^{2-}+I_2\to S_4O_6^{2-}+2{{I}^{-}} $
Options:
A) $ 2SO_2O_3^{2-} $ gets oxidised to $ SO_4O_6^{2-} $
B) $ SO_2O_3^{2-} $ gets reduced to $ S_4O_6^{2-} $
C) $ I_2 $ gets reduced to $ {{I}^{-}} $
D) $ I_2 $ gets oxidised to $ {{I}^{-}} $
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Answer:
Correct Answer: D
Solution:
[d] $ 2S_2O_3^{2-}+I_2\to S_4O_6^{2-}+2{{I}^{-}} $ Oxidation half-reaction: $ \overset{+2}{\mathop{S_2}},O_3^{2-}\to \overset{+4}{\mathop{S_4}},O_6^{2-} $ Reduction half-reaction: $ I_2^{0}\to 2{{I}^{-}} $ Hence, $ S_2O_3^{2-} $ is getting oxidised to while $ I_2 $ is getting reduced to $ 2{{I}^{-}} $ . So, [d] is the correct answer.
BETA
