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.
    Answer and Explanation:
    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.

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