If you intended KClO2 (and not KClO3), here is he scoop.
K is +1. It's ALWAYS +1 because it's in group I of the periodic table. Cl is +3 on the left (remember all compounds are 0 so K + Cl + 2O = 0; therefore, 1 + ? + 2*-2 = 0. That makes Cl +3) and O is -2 for EACH oxygen and as you have it as -4 for the total. USUALLY, you want oxidation states for EACH atom. On the right K is +1, Cl is -1 and O2 is 0.
What is reduced? Reduction is the gain of electrons.
So Cl goes from +3 on the left to -1 on the right or gain of 4 e for each Cl.
O goes from -4 total on the left to 0 on the right or loss of 4e.
So Cl is reduced and O is oxidized.
It is important that you know the rules. Here they are.
http://www.chemteam.info/Redox/Redox-Rules.html
For the following reaction
KClO2->KCl+O2
assign oxidation states to each element on each side of the equation.
Reactants
K=?
Cl=?
O=?
Which element is reduced?
I got K=5, Cl=-1, and O=-4 . The element that reduced is K, but its says its wrong. Help, Please!
2 answers
K in KClO2: +1
Cl in KClO2: +3
O in KClO2: -2
K in KCl: +1
Cl in KCl: -1
O in O2: 0
O is oxidized, Cl is reduced
Cl in KClO2: +3
O in KClO2: -2
K in KCl: +1
Cl in KCl: -1
O in O2: 0
O is oxidized, Cl is reduced
1 answer