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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 :

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 :

[tex]CuS+NO_3^-\rightarrow CuSO_4+NO_2[/tex]

The oxidation state of 'Cu' in [tex]CuS[/tex] and [tex]CuSO_4[/tex] 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 [tex]NO_3^-[/tex].

The oxidation-reduction half reaction will be :

Oxidation : [tex]S^{2-}\rightarrow S^{6+}+4e^-[/tex]

Reduction : [tex]N^{5+}+1e^-\rightarrow N^{4+}[/tex]

All the given statements are correct.

Hence, the answer is none of the above.

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