## For the following reaction KClO2 -> KCl + O2 assign oxidation states to each element on each side.

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1. The two Reactants AND Products
Reactants
K
Cl
O
WHICH ELEMENT IS OXIDIZED?
Items K
Cl
O
WHICH ELEMENT IS REDUCED?
Oxidation States and Reduction/Oxidation:
The oxidation condition of a molecule can be controlled by adhering to specific guidelines as demonstrated as follows:
1. Soluble base metals generally have an oxidation condition of +1
2. Oxygen as a rule has an oxidation condition of – 2, except if it is clung to itself
In a response, we state that if a particle lost electrons, it wound up oxidized. On the other hand, in the event that it picked up electrons, it wound up decreased.
Utilizing the principles above, we can decide the oxidation conditions of the reactants and items. To start with, we do the reactants underneath:
KClO2
K = +1
O = – 2
At that point, we discover the oxidation territory of Cl by taking note of that the general particle has a net charge of 0 so the oxidation number of Cl must counteract the oxidation quantities of the remainder of the atom:
Cl = – (+1 + 2*-2) = +3
Presently we can do likewise for the items.
KCl
K = +1
Cl = – 1
O2
O = 0
Since K began with an oxidation number of +1 and finished with an oxidation of +1, it was neither decreased nor oxidized. Notwithstanding, Cl went from +3 to – 1 which means it picked up electrons and was decreased. O, then again, went from – 2 to 0 which means it lost electrons and was oxidized.