When the gases; sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction $S O_{2} + 2 H_{2} S \to 2 H_{2} O + 3 S$ occurs.
Here hydrogen sulphide is acting as :

an oxidizing agent

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a reducing agent

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a dehydrating agent

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a catalyst

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Solution

The correct option is C a reducing agent
The oxidation state of sulphur in $H_{2} S$ is $- 2$ whereas after the reaction the oxidation state of sulphur is changing to 0. Hence it is acting like a reducing agent.

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Similar questions

When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water the reaction is :

SO₂+2H₂S=2H₂O+3S

Here hydrogen sulphide is acting as :

A. Oxidizing agent

B. Reducing agent

Q. When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction is as shown. Here, hydrogen sulphide is acting as:

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

Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:
$2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$
(i). calculate the volume of hydrogen suulphide at STP.
Also, (ii). calculate the volume of oxygen required at STP. Which will complete the combustion of hydrogen sulphide determined in litres.

Q. An aqueous solution of hydrogen sulphide and sulphur dioxide when mixed together yield:

Q. In the reaction of sulphur dioxide and hydrogen sulphide, identify the reducing agent and choose it.

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When the gases; sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction SO2+2H2S→2H2O+3S occurs.Here hydrogen sulphide is acting as :

The correct option is C a reducing agentThe oxidation state of sulphur in H2S is −2 whereas after the reaction the oxidation state of sulphur is changing to 0. Hence it is acting like a reducing agent.

When the gases; sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction $S O_{2} + 2 H_{2} S \to 2 H_{2} O + 3 S$ occurs.
Here hydrogen sulphide is acting as :

The correct option is C a reducing agent
The oxidation state of sulphur in $H_{2} S$ is $- 2$ whereas after the reaction the oxidation state of sulphur is changing to 0. Hence it is acting like a reducing agent.