H\(_2\)S\(_{(g)}\) + Cl\(_2\)\(_{(g)}\) → 2HCl\(_{(g)}\) + S\(_{(s)}\)
What is the change in oxidation state of sulphur from reactant to product?
- -3 to -1
- -3 to 1
-
-2 to 0
- -1 to 0
The correct answer is: C
Explanation
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
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