Respuesta :
Answer:
The oxygen element in H2O2 is the specie that is reduced in H2O and oxidized into O2.
Explanation:
5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)
Oxidation is defined as the outright loss of electrons. Oxidation leads to an increase in an element's oxidation number.
If this reaction is broken down into reduction and oxidation halves
It is observed that, of the reactants above,
H202 becomes H2O and O2
MnO4- + H+ becomes Mn2+ and H2O
Oxidation number of Mn changes from +7 In MnO4- to +2 In Mn2+ (evidently reduction)
The Oxygen in MnO4- doesn't change oxidation numbers as its oxidation number stays at -2
Oxidation number of Oxygen changes from -1 in H2O2 to -2 In H2O and 0 in O2
The Hyrogen in H2O2 doesn't change oxidation numbers, itsoxidation number stays at +1 in H2O2 and H2O.
This indicates that H2O2 undergoes oxidation and reduction; more specifically, the oxygen element in H2O2 is the specie that is reduced in H2O and oxidized into O2.
Hope this Helps
Answer:
In the reaction [tex]\rm H_2O_2[/tex] oxidizes and the oxidation number of oxygen changes.
Explanation:
The ionic equation for the reaction of [tex]\text H_2O_2[/tex] with [tex]\text MnO_4^-[/tex] is given below:
[tex]\rm 5\;H_{2} O_{2}\;+\;\text 2\;MnO_{4}\; +\; 6 H^+\rightarrow\; 2\; Mn^{+2}\; +\; 8 H_{2}O\; + \; 5O_{2}[/tex]
The oxidation is defined as the loss of electrons and gain in oxidation state. In the above reaction, the change in oxidation state of the elements are:
1. In [tex]\rm H_2O_2[/tex], substrate is broken down into [tex]\rm H_2O[/tex] and [tex]\rm O_2[/tex].
The oxidation state of Hydrogen remains same i.e. +1 in [tex]\rm H_2O[/tex] and [tex]\text H_2O_2[/tex]. The oxidation number of Oxygen in [tex]\rm H_2O_2[/tex] is -1 which decreases to -2 in [tex]\rm H_2O[/tex] and increases to the value of 0 in [tex]\rm O_2[/tex].
2. In [tex]\rm MnO_4^-[/tex], substrate gains the [tex]\rm H^+[/tex] and forms [tex]\rm Mn^2^+[/tex] and [tex]\rm H_2O[/tex]. The oxidation number of [tex]\rm Mn[/tex] in [tex]\rm MnO_4^-[/tex] is +7. The reduction of product.
From the equation it is evident that [tex]\rm H_2O_2[/tex] oxidizes in the reaction and the oxidation number of oxygen changes from -1 state to -2 and 0 oxidation number.
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https://brainly.com/question/15686033?referrer=searchResults
