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Question

Choose correct startement (s) regarding the following reaction. Cr2O27(aq)+3SO23(aq.)+8H+2Cr3+(aq)+3SO24(aq)+4H2O
1. Cr2O27 is oxidising agent
2. SO23 is reducing agent
3. The oxidation number of per 'S' atom in SO23 is increased by two.
4. The oxidation number of per 'Cr' atom in Cr2O27 is decreased by three.
Correct code is

A
1.2,3,4
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B
only 3,4
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C
only 1,3
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D
Only 4
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Solution

The correct option is A 1.2,3,4
The given reaction is,
Cr2O27(aq)+3SO23(aq)+8H+2Cr3+(aq)+3SO24(aq)+4H2O

Cr2O27 oxidises SO23 to SO24 Hence, it acts as an oxidizing agent.

Oppositely, SO23 reduces Cr2O27 to Cr3+. Hence, it acts as a reducing agent.

The oxidation state of S in SO23 is +4 and the oxidation state of S in SO24 is +6. The oxidation number of each S atom increases from +4 to 6 Thus, the oxidation number is increased by 2.

The oxidation number of Cr in Cr2O27 is +6 and the oxidation number of Cr in Cr3+ is +3. The oxidation number of each chromium atom decreases from +6 to +3. Thus, it decreases by 3.

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