What is the oxidizing agent here?
The oxidation state of K remains +1, while the oxidation state of some of the Cl atoms changes from -1 to 0. What is the oxidizing agent here?
The oxidation state of K remains +1, while the oxidation state of some of the Cl atoms changes from -1 to 0. What is the oxidizing agent here?
2 Answers
The oxidizing agent is chlorate anion,
Explanation:
Chlorates (and more so perchlorates) are potent oxidizing agents. In chlorate anion the sum of the oxidation numbers MUST equal the negative charge of the ion: i.e.
What is being oxidized in the reaction above, and what is the oxidation product?
What is the oxidation number of
The oxidation state of chlorine in
#6"HCl"(aq) + "KClO"_3(aq) -> "KCl"(aq) + 3"H"_2"O"(l) + 3"Cl"_2(g)#
The oxygen actually takes priority when assigning oxidation states, and it gets a
Therefore, chlorine gets reduced and oxidized, but do know that we are talking about two separate chlorine atoms.
SHORTCUT TO SEEING OXIDATION/REDUCTION
It's easier to tell if you look at the number of oxygen atoms on
From the relationship of
DETERMINING HALF-REACTIONS FROM THE ENTIRE REACTION
We can dissect the equation and extract a half reaction to balance the overall reaction from the beginning, after comparing
- Add water on the right side to balance the oxygens.
- Add protons on the left side to balance the hydrogens, since the overall reaction has
#"HCl"# , hydrochloric acid, and balancing things in base would neutralize some of it.
#color(green)(6e^(-) + 6"H"^(+)(aq) + "KClO"_3(aq) -> "KCl"(aq)+ 3"H"_2"O"(l))#
Here you should notice that
The other reaction naturally should involve
#color(green)(2"HCl"(aq) -> "Cl"_2(g) + 2"H"^(+)(aq) + 2e^(-))#
And you can see the imbalance of electrons, so you should scale it up by three when you sum these up to get the full reaction.
#3(2"HCl"(aq) -> "Cl"_2(g) + cancel(2"H"^(+)(aq)) + cancel(2e^(-)))#
#cancel(6e^(-)) + cancel(6"H"^(+)(aq)) + "KClO"_3(aq) -> "KCl"(aq)+ 3"H"_2"O"(l)#
#"---------------------------------------------------------------------"#
#color(blue)(6"HCl"(aq) + "KClO"_3(aq) -> "KCl"(aq) + 3"H"_2"O"(l) + 3"Cl"_2(g))#