Question

Consider the following redox reaction: (38) CuS + NO3 → CuSO4 + NO2 Which of the following statement is false? [1] [2] [3] Cu has an oxidation state of +2 in Cus. Cu has an oxidation state of +2 in CuSO4. S is oxidized in the above reaction. Electrons are donated from S in CuS to N in NO3". None of the above.

Answer 1

Answer : The answer is none of the above.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The given balanced chemical reaction is :

`CuS+NO_3^-\rightarrow CuSO_4+NO_2`

The oxidation state of 'Cu' in
`CuS` and
`CuSO_4` is, (+2).

In this reaction, the oxidation state of 'S' changes from (-2) to (+6) that means 'S' lost 4 electrons and it shows oxidation and the oxidation state of 'N' changes from (+5) to (+4) that means 'N' gains 1 electron and it shows reduction. That means electrons are donated from 'S' in CuS to 'N' in
`NO_3^-`.

The oxidation-reduction half reaction will be :

Oxidation :
`S^(2-)\rightarrow S^(6+)+4e^-`

Reduction :
`N^(5+)+1e^-\rightarrow N^(4+)`

All the given statements are correct.

Hence, the answer is none of the above.