Identify the oxidation numbers for all the elements in the reactants and products for 3h2s(aq)+2h+(aq)+2no−3(aq)→3s(s)+2n o(g)+4h2o(l) identify the oxidation numbers by dragging the appropriate labels to their respective targets.

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

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

the reduction process

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an element's oxidation

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an element's oxidation number increase

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answer: your mm

explanation: