Which of the following chemical reactions depicts the oxidising behaviour of $H_{2} S O_{4}$ ?

2 H I + H_{2} S O_{4} \to I_{2} + S O_{2} + 2 H_{2} O

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C a (O H)_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H_{2} O

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N a C l + H_{2} S O_{4} \to N a H S O_{4} + H C l

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2 P C l_{5} + H_{2} S O_{4} \to 2 P O C l_{3} + 2 H C l + S O_{2} C l_{2}

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Solution

The correct option is C $2 H I + H_{2} S O_{4} \to I_{2} + S O_{2} + 2 H_{2} O$
In reaction A sulphur decreases its oxidation state from +6 in $H_{2} S O_{4}$ to +4 in $S O_{2}$ and iodine increases its oxidation state from -1 in $H I$ to 0 in $I_{2}$ .So $H_{2} S O_{4}$ acts as oxidising agent and $I_{2}$ acts as reducing agent.
In reaction B, C and D sulphur do not decrease its oxidation state so $H_{2} S O_{4}$ does not act as oxidizing agent.
So only reaction A depicts the oxidizing behavior of $H_{2} S O_{4}$ .

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Similar questions

H_{2} S O_{4}

Which of the following chemical reaction depicts the oxidizing behavior of $H_{2} S O_{4}$ ?

S O_{2}

95^{\circ} C

S O_{2} + C l_{2} \to S O_{2} C l_{2} (Yield = 75 %)

S O_{2} C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 80 %)

H_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + 2 H_{2} O (Yield = 100 %)

C a S O_{4}

S O_{2}

H_{2} S O_{4}

C a (O H)_{2} .

C a S O_{4}

H_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + 2 H_{2} O

S O_{2} + C l_{2} + 2 H_{2} O \to 2 H C l + H_{2} S O_{4}

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