The reaction given below is interpreted as- H2Sg- Cl2g- 2HClg - Ss

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction to Oxidation and Reduction

The reaction ...

Question

The reaction given below is interpreted as:

$H_{2} S (g) + C l_{2} (g) \to 2 H C l (g) + S (s)$

H_{2} S

is getting oxidized and

C l_{2}

is getting reduced

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H_{2} S

is getting reduced and

C l_{2}

is getting oxidized

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Only

H_{2} S

is oxidized

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both

H_{2} S

and

C l_{2}

are reduced

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

In the reaction, H₂S + Cl₂ → 2HCl + S; hydrogen sulphide is:

2 H_{2} S + S O_{2} \to 3 S + 2 H_{2} O

H_{2} S

S O_{2}

H_{2} S

S O_{2}

H_{2} S (g) + C l_{2} \to 2 H C l (g) + S (s)

H_{2} S + C l_{2} \to 2 H C l + S

C l_{2} (a q) + H_{2} S (a q) \to S (s) + 2 H^{+} (a q) + 2 C l^{-} (a q)

= k [C l_{2}] [H_{2} S]

i s / a r e

C l_{2} + H_{2} S \to H^{+} + C l^{-} + C l^{+} + H S^{-}

C l^{+} + H S^{-} \to H^{+} + C l^{-} + S

H_{2} S \Leftrightarrow H^{+} + H S^{-}

C l_{2} + H S^{-} \to 2 C l^{-} + H^{+} + S

Join BYJU'S Learning Program