
Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior?
A. $C + 2\;{H_2}S{O_4} \to C{O_2} + 2S{O_2} + 2{H_2}O$
B. $Ca{F_2} + {H_2}S{O_4} \to CaS{O_4} + 2HF$
C. $Cu + 2\;{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O$
D. $S + 2\;{H_2}S{O_4} \to 3S{O_2} + 2{H_2}O$
Answer
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Hint: In redox reaction, oxidation and reduction takes place simultaneously. Oxidation and reduction in terms of oxygen transfer: oxidation is gain of oxygen and reduction is loss of oxygen. An oxidizing agent is a substance which oxidizes something else.
Complete Step by step answer: The terms oxidation and reduction can be defined in terms of adding and removing oxygen to a compound.
-Here in option A, $C + 2\;{H_2}S{O_4} \to C{O_2} + 2S{O_2} + 2{H_2}O$
Oxygen is added to carbon on going from left to right in the equation by ${H_2}S{O_4}$ , so carbon is getting oxidized by ${H_2}S{O_4}$ and hence ${H_2}S{O_4}$ is acting as oxidizing agent.
-In option B, $Ca{F_2} + {H_2}S{O_4} \to CaS{O_4} + 2HF$
In this case, $Ca{F_2}$ is not getting oxidized by ${H_2}S{O_4}$ . This reaction is a simple double displacement reaction, not redox reaction.
-In option C, $Cu + 2\;{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O$
Oxygen is getting added to copper on going from left to right in the chemical equation by ${H_2}S{O_4}$ , hence here ${H_2}S{O_4}$ is acting as an oxidizing agent.
-In option D, $S + 2\;{H_2}S{O_4} \to 3S{O_2} + 2{H_2}O$
Oxygen is added to sulphur on going from left to right by ${H_2}S{O_4}$ , in this chemical equation. Hence here also ${H_2}S{O_4}$ is acting as an oxidizing agent.
So, option B, is the correct option.
Note: Oxidation and reduction in terms of electron transfer: oxidation is loss of electron while reduction is gain of electron. The compounds which carry out the oxidation of other compounds in redox reactions are called oxidizing agents. Similarly the compound which carries out reduction of the other compound is known as reducing agent in that redox reaction.
Complete Step by step answer: The terms oxidation and reduction can be defined in terms of adding and removing oxygen to a compound.
-Here in option A, $C + 2\;{H_2}S{O_4} \to C{O_2} + 2S{O_2} + 2{H_2}O$
Oxygen is added to carbon on going from left to right in the equation by ${H_2}S{O_4}$ , so carbon is getting oxidized by ${H_2}S{O_4}$ and hence ${H_2}S{O_4}$ is acting as oxidizing agent.
-In option B, $Ca{F_2} + {H_2}S{O_4} \to CaS{O_4} + 2HF$
In this case, $Ca{F_2}$ is not getting oxidized by ${H_2}S{O_4}$ . This reaction is a simple double displacement reaction, not redox reaction.
-In option C, $Cu + 2\;{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O$
Oxygen is getting added to copper on going from left to right in the chemical equation by ${H_2}S{O_4}$ , hence here ${H_2}S{O_4}$ is acting as an oxidizing agent.
-In option D, $S + 2\;{H_2}S{O_4} \to 3S{O_2} + 2{H_2}O$
Oxygen is added to sulphur on going from left to right by ${H_2}S{O_4}$ , in this chemical equation. Hence here also ${H_2}S{O_4}$ is acting as an oxidizing agent.
So, option B, is the correct option.
Note: Oxidation and reduction in terms of electron transfer: oxidation is loss of electron while reduction is gain of electron. The compounds which carry out the oxidation of other compounds in redox reactions are called oxidizing agents. Similarly the compound which carries out reduction of the other compound is known as reducing agent in that redox reaction.
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