1.
You solved it as if it were in acidic conditions, but as you can see from the fact that it's reacting with sodium hydroxide it must be in base. In this case you use water to balance the half equations:
Cl2 + 6H2O --> 2ClO3- + 12H+ + 10e
Cl2 + 2e --> 2Cl-
equalise the electrons and add:
Cl2 + 6H2O --> 2ClO3- + 12H+ + 10e
5Cl2 + 10e --> 10Cl-
-------------------------------- add
Cl2 + 6H2O + 5Cl2 --> 2ClO3- + 12H+ + 10Cl-
collect terms
6Cl2 + 6H2O ---> 2ClO3- + 12H+ + 10Cl-
divide through by 2
3Cl2 + 3H2O ---> ClO3- + 6H+ + 5Cl-
Now add base ions (hydroxide) to remove the hydrogen ions to both sides and cancel down water.
3Cl2 + 3H2O + 6OH- ---> ClO3- + 6H+ + 5Cl- + 6OH-
3Cl2 + 3H2O + 6OH- ---> ClO3- + 6H2O + 5Cl-
3Cl2 + 6OH- ---> ClO3- + 3H2O + 5Cl-
Last reply 1 week ago
Im confused about this chemistry question, why does it form these productsLast reply 1 week ago
Im confused about this chemistry question, why does it form these products