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# AQA AS Level Redox question Watch

1. use oxidation states to balance following equation:
Cl2 + NaOH (goes to) NaClO3 + NaCl + H2O

Any help appreciated
2. (Original post by MEPS1996)
use oxidation states to balance following equation:
Cl2 + NaOH (goes to) NaClO3 + NaCl + H2O

Any help appreciated
You need to construct two half equations (you can use water to add oxygen atoms as the conditions are alkaline):

1. showing the chorine being oxidised to chlorate ions
2. showing chlorine being reduced to chloride ions

Then you multiply equations 1 &2 by appropriate numbers so that the electrons are balanced.

Then you add them together and cancel out any common terms.
3. (Original post by charco)
You need to construct two half equations (you can use water to add oxygen atoms as the conditions are alkaline):

1. showing the chorine being oxidised to chlorate ions
2. showing chlorine being reduced to chloride ions

Then you multiply equations 1 &2 by appropriate numbers so that the electrons are balanced.

Then you add them together and cancel out any common terms.
using the method of adding h+ ions to balance hydrogen and water to balance oxygen:
5/2Cl2 + 5e-(goes to) 5Cl-
which is correct according to my textbook
Then for the other half equation i get:
1/2Cl2 + 3H2O (goes to) ClO3- + 6H+ +5e-
However the book has the answer
1/2Cl2 + 6OH- (goes to)ClO3- +5e- + 3H20 +5e-
It seems the answer in the book has been obtained by taking the equation i got, splitting the H2O into its H+ and OH- ions, to get
1/2Cl2 + 3H+ + 3OH- (goes to) CLO3- + 6H+ + 5e-
cancelling the common 3H+ ions
1/2Cl2 + 3OH- (goes to) ClO3- + 3H+ +5e-
then added 3OH- to each side to get water on the right
1/2Cl2 +6OH- (goes to) ClO3- + 3H2O + 5e-
Is this necessary, it seems they have left H+ ions in in the answer to the next question?
Thanks a lot
4. (Original post by MEPS1996)
using the method of adding h+ ions to balance hydrogen and water to balance oxygen:
5/2Cl2 + 5e-(goes to) 5Cl-
which is correct according to my textbook
Then for the other half equation i get:
1/2Cl2 + 3H2O (goes to) ClO3- + 6H+ +5e-
However the book has the answer
1/2Cl2 + 6OH- (goes to)ClO3- +5e- + 3H20 +5e-
It seems the answer in the book has been obtained by taking the equation i got, splitting the H2O into its H+ and OH- ions, to get
1/2Cl2 + 3H+ + 3OH- (goes to) CLO3- + 6H+ + 5e-
cancelling the common 3H+ ions
1/2Cl2 + 3OH- (goes to) ClO3- + 3H+ +5e-
then added 3OH- to each side to get water on the right
1/2Cl2 +6OH- (goes to) ClO3- + 3H2O + 5e-
Is this necessary, it seems they have left H+ ions in in the answer to the next question?
Thanks a lot
The issue is that the half equations fare different for acidic and basic conditions.

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-

Cl2 + 6H2O --> 2ClO3- + 12H+ + 10e
5Cl2 + 10e --> 10Cl-
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-
1. (Original post by charco)
The issue is that the half equations fare different for acidic and basic conditions.
(Original post by charco)

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-

Cl2 + 6H2O --> 2ClO3- + 12H+ + 10e
5Cl2 + 10e --> 10Cl-
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-
Thanks a lot Thats great cheers

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