Respuesta :
Answer : The oxidation state of 'S' in reactant is (-2) and product side is (0) and the oxidation state of 'N' in reactant side is (+5) and product side is (+2).
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 chemical reaction is:
[tex]3H_2S(aq)+2H^+(aq)+2NO_3^{-}(aq)\rightarrow 3S(s)+2NO(g)+4H_2O(l)[/tex]
In this reaction,
The oxidation state of 'S' in [tex]H_2S[/tex] is, (-2)
The oxidation state of 'S' in [tex]S[/tex] is, (0)
The oxidation state of 'N' in [tex]NO_3^-[/tex] is, (+5)
The oxidation state of 'N' in [tex]NO[/tex] is, (+2)
That means, the oxidation state of 'S' in reactant is (-2) and product side is (0) and the oxidation state of 'N' in reactant side is (+5) and product side is (+2).
In this reaction, sulfur shows oxidation because its oxidation state changes from (-2) to (0) and nitrogen shows reduction because its oxidation state changes from (+5) to (+2).
The S element experiences an oxidation reaction: from oxidation number -2 to 0
The N element experiences a reduction reaction: from oxidation number + 5 to + 2
Further explanation
The formula for determining Oxidation Numbers in general:
1. Single element = 0. Examples of Ar, Mg, Cu, Fe, N₂, O₂, etc. = 0
Group IA (Li, Na, K, Rb, Cs, and Fr): +1
Group IIA (Be, Mg, Ca, Sr and Ba): +2
H in compound = +1, except metal hydride compounds (Hydrogen which binds to IA or IIA groups) oxidation number H = -1, for example, LiH, MgH₂, etc.
2. Oxidation number O in compound = -2, except OF2 = + 2 and in peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.
3 The oxidation number in an uncharged compound = 0,
Total oxidation number in ion = ion charge, Example NO₃⁻ = -1
Let see the oxidation number of all the elements (from left to right)
Reaction :
H₂s + 2H⁺ + 2NO₃⁻ ⇒ 3S + 2NO + 4H₂O
H₂S: H⁺ = +1, S = -2
H₂S = (2 x H) + (1 x S) =(2x1) + (1x-2) = 2 -2 =0
2H⁺ : H⁺ = +1
2NO₃⁻ : N = + 5, O = -2,
NO₃⁻ = (1 x N) + (3 x O) = (1 x 5) +(3x-2)= 5-6 = -1
3S : S = 0
2NO : N = +2, O = -2
NO = (1 x N) + (1 x O) = (1 x 2) + (1 x -2) = 2 -2 = 0
4H₂O : H⁺ = +1, O = -2
H₂O = (2 x H) + (1 x O) = (2 x 1) + (1 x -2) = 0
Learn more
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