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I. H(2)S reduces acidified KMnO(4) to Mn...

I. `H_(2)S` reduces acidified `KMnO_(4)` to `MnSO_(4)`
II. `H_(2)S` reduces acidified `K_(2)Cr_(2)O_(7)` to red colour `Cr_(2)(SO_(4))_(3)`
III. `H_(2)S` reduces nitric acid to `NO_(2)`
IV. `H_(2)S` oxidises ferrous sulphate to ferric sulphate
Then the correct statement(s) is/are:

A

only I

B

only I and III

C

only II and III

D

I,II,III and IV

Text Solution

Verified by Experts

The correct Answer is:
B
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SO_(2) reduces acidified K_(2)Cr_(2)O_(7) to

When H_(2)S is passed through acidified K_(2)Cr_(2)O_(7) solution, the solution turns :

A : H_(2)O_(2) reacts with K_(2)Cr_(2)O_(7) to give blue colour. H_(2)O_(2) can act as reducing agent.

By passing H_(2)S gas in acidified KMnO_(4) , we get

By passing H_(2)S in acidified KMnO_(4) solution we get

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