Answer:
Explanation:
The formula of the reaction:
KClO₂ → KCl + O₂
To assign oxidation numbers, we have to obey some rules:
- Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
- The charge on simple ions signifies their oxidation number.
- The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.
The oxidation number of K in KClO₂:
K + (-1) + 2(-2) = 0
K-5 = 0
K = +5
The oxidation number of K in KCl:
K + (-1) = 0
K = +1
The oxidation number Cl in KClO₂ is -1
For Cl in KCl, the oxidation number is -1
For O in KClO₂, the oxidation number is (2 x -2) = -4
For O in O₂, the oxidation number is 0
K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.
O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.
I don’t care about it tbh like I don’t even care about that stuff but the answer is 29272728
Aye you have the same class as me bruh I need help on some chemistry qustions
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
