Question

Find the oxidation number of S in H2SO4​

Answer 1

Answer:

Explanation:

Explanation:In H2SO4 , hydrogen exists in its usual +1 state, and oxygen exists in its −2 state.

Explanation:In H2SO4 , 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:

Explanation:In H2SO4 , 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.

Explanation:In H2SO4 , 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.

Explanation:In H2SO4 , 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.

Explanation:In H2SO4 , 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.

Explanation:In H2SO4 , 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.

Explanation:

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Answer 2

Explanation:

Let X be the oxidation number of S in H

2

SO

4

.

The oxidation number of hydrogen is +1 and the oxidation number of oxygen is -2.

2(+1)+X+4(−2)=0

2+X−8=0

X−6=0

X=+6

Hence, the oxidation number of S in H

2

SO

4

is +6.