Which reactions involve iodometric titrations?

2 N a_{2} S_{2} O_{3} + I_{2} ⟶ N a_{2} S_{4} O_{4} + 2 N a I

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2 {M n O}_{4}^{-} + 2 I^{-} + 16 H^{+} ⟶ 2 M n^{2 +} + 8 H_{2} O + 5 I_{2}

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N a_{3} A s O_{3} + I_{2} + H_{2} O ⟶ N a_{2} A s O_{4} + 2 H I

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C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} ⟶ 2 C r^{3 +} + 7 H_{2} O + 3 I_{2}

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Solution

The correct options are
A $2 N a_{2} S_{2} O_{3} + I_{2} ⟶ N a_{2} S_{4} O_{4} + 2 N a I$
C $N a_{3} A s O_{3} + I_{2} + H_{2} O ⟶ N a_{2} A s O_{4} + 2 H I$
Iodometry, also known as iodometric titration, is a method of volumetric chemical analysis, a redox titration where the appearance or disappearance of elementary iodine indicates the end point.
It involves indirect titration of iodine liberated by reaction with the analyte (redusing agent) whereas iodimetry involves direct titration using iodine as the titrant.
To a known volume of sample, an excess but known amount of iodide is added which the oxidizing agents oxidizes iodide to iodine. Iodine dissolves in the iodide-containing solution to give triodide ions which have a dark brown color.
So, in options A and C, iodine reacts with oxidising agent and produces iodide. Hence, options A and C are correct options.
Options B and D are iodimetry reactions. Hence, options B and D are wrong.

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Similar questions

(i) C r_{2} O_{7}^{2 -} + 3 S O_{2} + 2 H^{+} \to 2 C r^{3 +} +

+ H_{2} O

(i i) 2 M n O_{4}^{-} + 5 S O_{3}^{2 -} + 6 H^{+} \to 2 M N^{2 +} +

3 H_{2} O

(i i i) C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +}

7 H_{2} O

{C r_{2} O_{7}}^{2 -}

M

{C r_{2} O_{7}}^{2 -} + 6 I^{-} + 14 H^{+} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O

M n O_{4}^{-}

4.56 \times 10^{- 3} M s^{- 1}

2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 M n^{2 +} + 5 I_{2} + 8 H_{2} O

I_{2}

2 N a_{2} S_{2} O_{3} + I_{2} = N a_{2} S_{4} O_{6} + 2 N a I, I_{2}

n

C r_{2} O_{7}^{2 -} + n e^{-} + 14 H^{+} \to 2 C r^{3 +} + 7 H_{2} O

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