SATHEE: Redox Reactions And Electrochemistry Ques 742

Redox Reactions And Electrochemistry Ques 742

Question: Consider the redox reaction : $2 S O_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}$

Options:

A) $2 S O_{2} O_{3}^{2 -}$ gets oxidised to $S O_{4} O_{6}^{2 -}$

B) $S O_{2} O_{3}^{2 -}$ gets reduced to $S_{4} O_{6}^{2 -}$

C) $I_{2}$ gets reduced to $I^{-}$

D) $I_{2}$ gets oxidised to $I^{-}$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] $2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}$ Oxidation half-reaction: $\overset{+ 2}{S_{2}}, O_{3}^{2 -} \to \overset{+ 4}{S_{4}}, O_{6}^{2 -}$ Reduction half-reaction: $I_{2}^{0} \to 2 I^{-}$ Hence, $S_{2} O_{3}^{2 -}$ is getting oxidised to while $I_{2}$ is getting reduced to $2 I^{-}$ . So, [d] is the correct answer.

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Redox Reactions And Electrochemistry Ques 742

Question: Consider the redox reaction : 2SO2O32−+I2→S4O62−+2I−

Options:

Answer:

Solution:

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Question: Consider the redox reaction : $2 S O_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}$