H\(_2\)S\(_{(g)}\)  +  Cl\(_2\)\(_{(g)}\) → 2HCl\(_{(g)}\) + S\(_{(s)}\) What is the change in oxidation state of sulphur...

H\(_2\)S\(_{(g)}\) + Cl\(_2\)\(_{(g)}\) → 2HCl\(_{(g)}\) + S\(_{(s)}\)

What is the change in oxidation state of sulphur from reactant to product?

Oxidation state of S in H\(_2\)S = 0

H = +1, (+1 X 2) + S = 0

+ 2 + S = 0

S = - 2 ( from the reactant)

Already, from the equation, S is in solid form,as an element, therefore the oxidation state of elements in their free state is 0

The change in oxidation state of sulphur is from - 2 to 0

There is an explanation video available below.

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