Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    4
    ReputationRep:
    use oxidation states to balance following equation:
    Cl2 + NaOH (goes to) NaClO3 + NaCl + H2O

    Any help appreciated
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (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.
    • Thread Starter
    Offline

    4
    ReputationRep:
    (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
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (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-

    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-
    • Thread Starter
    Offline

    4
    ReputationRep:
    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-

    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-
    Thanks a lot Thats great cheers
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.