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To find the correct oxidations state of S in H2SO4 (Sulfuric acid), and each element in the molecule, we use a few rules and some simple math.First, since th...
In sulfuric acid (H2SO4), the oxidation number of sulfur is +6. We find this by knowing that hydrogen has an oxidation number of +1 (totaling +2 for the two hydrogen atoms), and oxygen has an oxidation number of -2 (totaling -8 for the four oxygen atoms).
The sulfurous acid, thus formed reacts with oxygen to generate sulfur trioxide, (9.8.13), which reacts with water to form sulfuric acid, (9.8.14). \[ \rm O_3 + h\nu \rightarrow O* + O_2\] \[ \rm O* + H_2O \rightarrow 2 HO\cdot\] \[ \rm HSO_3 + O_2 \rightarrow HO_2 + SO_3\] \[ \rm SO_3 + H_2O \rightarrow H_2SO_4\]
Enter the formula of a chemical compound to find the oxidation number of each element. A net ionic charge can be specified at the end of the compound between { and }. For example: ZnCl4 {2-} or NH2NH3 {+}. Enter just an element symbol to show the common and uncommon oxidation states of the element.
The oxidation number of the oxygen atoms in hydrogen peroxide is -1. H 2 O 2 can therefore act as an oxidizing agent and capture two more electrons to form a pair of hydroxide ions, in which the oxygen has an oxidation number of -2.
23 Απρ 2018 · Explanation: In H 2SO4, hydrogen exists in its usual +1 state, and oxygen exists in its −2 state. Let x be the oxidation number of sulfur in sulfuric acid. Then we got: 1 ⋅ 2 + x −2 ⋅ 4 = 0. 2 + x − 8 = 0. x − 6 = 0. x = 6. So, sulfur exists in the +6 state.
29 Αυγ 2023 · The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.
