We have a solution of $I O_{3}^{-}$ + $H^{+}$ which will show any observable changes with:

Chromium sulphate

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Ammonium sulphide

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Copper sulphate

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Potassium Iodide

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Solution

The correct options are
A Ammonium sulphide
D Potassium Iodide
(i) $I O_{3}^{-} + 5 I + 6 H^{+} ⟶ 3 I_{2} ↑ + 3 H_{2} O$
Here $I O_{3}^{-}$ in acidic medium act as oxidising agent.
(ii) $(N H_{4})_{2} S + I O_{3}^{-} + H^{+} ⟶ (N H_{4})_{2} S O_{4} + I_{2} ↑$

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Similar questions

Q. Which of the following will give any observable changes in solution with sodium peroxide?

Q. In neutral solution, one-mole periodate ion,

{I O}_{4}^{-}

, reacts with an excess of iodide to produce one mole of iodine; on acidification of the resulting solution, a further three moles of iodine is liberated.

Equations of the reactions which occur under these conditions.

I O_{4}^{-} + 2 I^{-} + H_{2} O ⟶ I O_{3}^{-} + I_{2} + 2 O H^{-}

I O_{3}^{-} + 5 I^{-} + 6 H^{+} ⟶ 3 I_{2} + 3 H_{2} O

If the statement is true enter 1, else enter 0.

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We have a solution of IO−3 + H+ which will show any observable changes with:

The correct options are A Ammonium sulphide D Potassium Iodide(i)IO−3+5I+6H+⟶3I2↑+3H2OHere IO−3 in acidic medium act as oxidising agent.(ii)(NH4)2S+IO−3+H+⟶(NH4)2SO4+I2↑

We have a solution of $I O_{3}^{-}$ + $H^{+}$ which will show any observable changes with:

The correct options are
A Ammonium sulphide
D Potassium Iodide
(i) $I O_{3}^{-} + 5 I + 6 H^{+} ⟶ 3 I_{2} ↑ + 3 H_{2} O$
Here $I O_{3}^{-}$ in acidic medium act as oxidising agent.
(ii) $(N H_{4})_{2} S + I O_{3}^{-} + H^{+} ⟶ (N H_{4})_{2} S O_{4} + I_{2} ↑$