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    I don't know how to use oxidation states to balance the following equation:


    CL2 + NaOH ----> NaClO3 + NaCL + H2O


    I know that first you list the compounds that have been reduced/oxidised so:

    Cl2 + e ----> Cl- (reduced)


    BUt then NaOH is more complicated, how do I write a half equation for that, and the rest of the equation?
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    1st work out the oxidation state of Na in that compound and go from there
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    (Original post by xXxiKillxXx)
    I don't know how to use oxidation states to balance the following equation:


    CL2 + NaOH ----> NaClO3 + NaCL + H2O


    I know that first you list the compounds that have been reduced/oxidised so:

    Cl2 + e ----> Cl- (reduced)


    BUt then NaOH is more complicated, how do I write a half equation for that, and the rest of the equation?
    You have to look for species tat have changed their oxidation states.


    On the left hand side (LHS) chlorine has oxidation state zero

    On the RHS there are two chlorines, one with oxidation state -1 and the other with oxidation state +5

    So now you construct half equations:

    1/2Cl2 + 1e --> Cl-
    1/2Cl2 --> Cl5+ + 5e

    You can see that the number of electrons is not equal so now multiply equation 1 by 5 and add them together:

    5/2Cl2 + 5e --> 5Cl-
    1/2Cl2 --> Cl5+ + 5e
    -------------------------------------------------------- add
    5/2Cl2 + 1/2Cl2 --> 5Cl- + Cl5+

    You can now gather terms and put in all of the other components from the reaction and balance spectator ions:

    3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O

    ---------------------------------------------------------------------

    This is only one way to do it. You could also construct the full half equation first and then balance and add.

    1/2Cl2 + 1e --> Cl-
    1/2Cl2 + 6OH- --> ClO3- + 5e + 3H2O

    MTB 5 in equation 1

    5/2Cl2 + 5e --> 5Cl-
    1/2Cl2 + 6OH- --> ClO3- + 5e + 3H2O
    --------------------------------------------------------------- add together

    3Cl2 + 6OH- --> 5Cl- + ClO3- + 3H2O

    and now put in the spectator ions (sodium) to remove the negative charges

    3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O
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    (Original post by charco)
    You have to look for species tat have changed their oxidation states.


    On the left hand side (LHS) chlorine has oxidation state zero

    On the RHS there are two chlorines, one with oxidation state -1 and the other with oxidation state +5

    So now you construct half equations:

    1/2Cl2 + 1e --> Cl-
    1/2Cl2 --> Cl5+ + 5e

    You can see that the number of electrons is not equal so now multiply equation 1 by 5 and add them together:

    5/2Cl2 + 5e --> 5Cl-
    1/2Cl2 --> Cl5+ + 5e
    -------------------------------------------------------- add
    5/2Cl2 + 1/2Cl2 --> 5Cl- + Cl5+

    You can now gather terms and put in all of the other components from the reaction and balance spectator ions:

    3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O

    ---------------------------------------------------------------------

    This is only one way to do it. You could also construct the full half equation first and then balance and add.

    1/2Cl2 + 1e --> Cl-
    1/2Cl2 + 6OH- --> ClO3- + 5e + 3H2O

    MTB 5 in equation 1

    5/2Cl2 + 5e --> 5Cl-
    1/2Cl2 + 6OH- --> ClO3- + 5e + 3H2O
    --------------------------------------------------------------- add together

    3Cl2 + 6OH- --> 5Cl- + ClO3- + 3H2O

    and now put in the spectator ions (sodium) to remove the negative charges

    3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O
    THANKSS!!!!!!!!!!!
 
 
 
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