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?

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2 Answers
Feb 7, 2016

Answer:

The oxidizing agent is chlorate anion, #ClO_3^-#. The formal oxidation state of #Cl# in this species is #+V#.

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. #Cl_(ON) + 3xxO_(ON) = -1#. Since oxygen generally has an oxidation number of #-II# (and it does here), the oxidation number of #Cl# is #+V#.

What is being oxidized in the reaction above, and what is the oxidation product?

What is the oxidation number of #Cl# in perchlorate anion, #ClO_4^-#?

Feb 7, 2016

The oxidation state of chlorine in #"KClO"_3# is NOT #-1#.

#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 #-2# oxidation state; the charge of #"ClO"_3^(-)#, with three oxygens (#3xx-2=-6#), adds up such that chlorine has an oxidation state of #\mathbf(+5)#.

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 #"KCl"# and compare it to #"KClO"_3#. The more oxygens there are, the more oxidized a compound is.

From the relationship of #"KCl"# to #"KClO"_3# (2 alike atoms), it naturally means you should inspect #"KClO"_3# more closely.

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 #"KCl"# to #"KClO"_3# and recognizing the difference in the number of oxygens.

  1. Add water on the right side to balance the oxygens.
  2. 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 #"KClO"_3# as a whole was reduced because it lost oxygen atoms (or, if you prefer, because six electrons were added). Therefore, it must be the oxidizing agent.

The other reaction naturally should involve #"HCl"# and #"Cl"_2#, as they're the only reactants remaining:

#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))#