Cl 2 + KOH rightarrow KCI + KCIO 3 + H2O Solution In solution KCl and KClO3 exis
ID: 684868 • Letter: C
Question
Cl 2 + KOH rightarrow KCI + KCIO 3 + H2O
Explanation / Answer
In solution KCl and KClO3 exist as Cl- and ClO3-. KOH existsas OH-. So K+ is just a spectator ion Cl2 + OH- => Cl- + ClO3- +H2O (Stillunbalanced) Balance by half-reaction method Cl2 => Cl- Rxn1 Cl2 => ClO3- Rxn2 To balance Rxn 1, put 2 in front of Cl- on the right Cl2 => 2Cl- Add 2 e- to the left, because 2 Cl atoms become 2Cl- Cl2 + 2e- => 2Cl- halfreaction is balanced To balance reaction 2, put 2 in front of ClO3-, to balance Cl's Cl2 = > 2ClO3- Add 10e- to the right, b/c Cl is +5 in ClO3- and there are 2ClO3- Cl2 => 2ClO3- + 10e- Add 12 OH- on the left and 6H2O on the right to balance H's andO's Cl2 + 12 OH- => 2ClO3- + 10e- + 6H2O Take the two bolded reactions, multiply the first by 5, sothat you can eliminate electrons 5(Cl2 + 2e- => 2Cl-) Cl2 + 12 OH- => 2ClO3- + 10e- + 6H2O ---------------------------------------------- 6Cl2 + 12OH- => 10Cl- + 2ClO3- + 6H2O Reduce by common factors 3Cl2 + 6OH- => 5 Cl- + ClO3- + 3H2O Add back K+, because they are just spectator ions. Add 6 K+ 3Cl2 + 6KOH = > 5KCl + KClO3 + 3H2O It is balanced.