Iodine oxidises $S_{2} O_{3}^{- 2}$ ion into:

S O_{3}^{- 2}

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S O_{4}^{- 2}

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S_{4} O_{6}^{- 2}

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S^{- 2}

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Solution

The correct option is C $S_{4} O_{6}^{- 2}$
$+ 2$ $+ 2.5$
$I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{-}$
Iodine oxidises $S_{2} O_{3}^{2} -$ ion to $S_{4} O_{6}^{2} -$

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Similar questions

Q. The order of increasing oxidation number ofS in

S_{8}, S_{2} O_{8}^{- 2}, S_{2} O_{3}^{- 2}, S_{4} O_{6}^{- 2}

is given below :

Identify the monoatomic and polyatomic ions from the following equation-
$P b (N O_{3})_{2}$ + $K_{2} S O_{4}$ $\to$ $2 K N O_{3}$ + $P b S O_{4}$

Q. Oxidation of thiosulphate

(S_{2} O_{3}^{2 -})

ion by iodine gives

Two half reactions for the following equation are:
$4 H_{2} S O_{4} + K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} ⟶ C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 N a_{2} S O_{4} + 4 H_{2} O$

Q. Balance the following chemical equation:

C r_{2} O_{7}^{- 2} (a q) + S O_{3}^{- 2} (a q) \to C r^{+ 3} (a q) + S O_{4}^{- 2} (a q)

(Acidic medium)

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Iodine oxidises S2O−23 ion into:

The correct option is C S4O−26 +2 +2.5I2+2S2O2−3→S4O2−6+2I−Iodine oxidises S2O23− ion to S4O26−

Iodine oxidises $S_{2} O_{3}^{- 2}$ ion into:

The correct option is C $S_{4} O_{6}^{- 2}$
$+ 2$ $+ 2.5$
$I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{-}$
Iodine oxidises $S_{2} O_{3}^{2} -$ ion to $S_{4} O_{6}^{2} -$