SATHEE: Redox Reactions And Electrochemistry Ques 660

Redox Reactions And Electrochemistry Ques 660

Question: Oxidation state of sulphur in anions $S O_{3}^{2 -},, S_{2} O_{4}^{2 -}$ and $S_{2} O_{6}^{2 -}$ increases in the orders:

Options:

A) $S_{2} O_{6}^{2 -} < S_{2} O_{4}^{2 -} < S O_{3}^{2 -}$

B) $S O_{6}^{2 -} < S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -}$

C) $S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$

D) $S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -} < S O_{3}^{2 -}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] In $S O_{3}^{^{2 -}}$ $x + 3 (- 2) = - 2; x = + 4$ In $S_{2} O_{6}^{2 -}$ $2 x + 4 (- 2) = - 2$ $2 x - 8 = - 2$ $2 x = 6; x = + 3$ In $S_{2} O_{6}^{2 -}$ $2 x + 6 (- 2) = - 2$ $2 x = 10; x = + 5$ hence the correct order is $S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$

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Redox Reactions And Electrochemistry Ques 660

Question: Oxidation state of sulphur in anions SO32−,,S2O42− and S2O62− increases in the orders:

Options:

Answer:

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Question: Oxidation state of sulphur in anions $S O_{3}^{2 -},, S_{2} O_{4}^{2 -}$ and $S_{2} O_{6}^{2 -}$ increases in the orders: