1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why there are infinite ways to balance this Redox equation?

Tags:
  1. Nov 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Balance :-
    $${Cl_2O_7} + {H_2O_2} {\longrightarrow} {ClO_2^-} +{O_2} + {H^+}$$
    Medium :- basic


    2. Relevant equations


    3. The attempt at a solution
    $${\stackrel{+7}{Cl_2}\stackrel{-2}{O_7}} + {\stackrel{-1}{H_2}\stackrel{-1}{O_2}} {\longrightarrow} {\stackrel{+3}{Cl}\stackrel{-2}{O_2^-}} +{\stackrel{0}{O_2}} + {\stackrel{+1}{H^+}}$$

    Since both Oxygen and Chlorine reduces, but Oxygen also oxidizes in equation,

    $${(1/2)\stackrel{+7}{Cl_2}\stackrel{-2}{O_7}} + {\stackrel{-1}{H_2}\stackrel{-1}{O_2}} + 3{\stackrel{-1}{H_2}\stackrel{-1}{O_2}} {\longrightarrow} {\stackrel{+3}{Cl}\stackrel{0}{O_2^-}} +{3\stackrel{-2}{O_2}} + {\stackrel{+1}{H^+}}$$

    After that i balanced Oxygen and Hydrogen to get
    $${Cl_2O_7} + {8H_2O_2} + 2OH^- {\longrightarrow} {2ClO_2^-} +{6O_2} + {9H^2O}$$

    $$
    \begin{array}{c|c|c|}

    \text{Element} & \text{RHS} & \text{LHS}\\
    Cl & 2 & 2\\
    \hline
    H & 18 & 18\\
    \hline
    O & 25 & 25\\

    \end{array}$$

    But given answer is something else,
    $${Cl_2O_7} + {4H_2O_2} + 2OH^- {\longrightarrow} {2ClO_2^-} +{4O_2} + {5H^2O}$$

    $$
    \begin{array}{c|c|c|}

    \text{Element} & \text{RHS} & \text{LHS}\\
    Cl & 2 & 2\\
    \hline
    H & 10 & 10\\
    \hline
    O & 17 & 17\\
    \end{array}$$

    Both equations seems balanced but i don't understand why ?
    Also http://www.webqc.org/balance.php says there are infinite ways to balance this equation.

    So how can I get the second equation instead of mine with Oxidation Number method ?
     
  2. jcsd
  3. Nov 12, 2016 #2

    Borek

    User Avatar

    Staff: Mentor

    Problem is (as it often happens with the hydrogen peroxide reactions), you have two reactions taking place in parallel. One is the reaction between Cl2O7 and H2O2, the other is the reaction of the H2O2 decomposition. Any linear combination of both is still a correctly balanced reaction.

    See http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure for more examples.
     
  4. Nov 12, 2016 #3
    What should I change in my approach to get the equation in the book's solution ?
    I don't know what should I change. If i change the order of steps I get an unbalanced equation. :(
     
  5. Nov 12, 2016 #4

    Borek

    User Avatar

    Staff: Mentor

    What you wrote is difficult to follow, as some of the oxidation numbers are wrong (oxygen in O2 is not -2, hydrogen in H2O2 is not -1) - thus it is hard to check what is going on and why you get this particular solution.

    These are often difficult problems and the stoichiometry is best checked through the experiment, as it depends on the details of the reaction mechanism (impossible to predict just by looking at the reaction equation). Note that you can wrote two separate reaction equations:

    Cl2O7 + 2OH- → 2ClO2- + 2O2 + H2O

    and

    2H2O2 → 2H2O + O2

    In general, every linear combination (that is, multiply the first equation by any number and add the second equation multiplied by any number) of these will produce a correctly balanced equation.

    Check that both equations you listed can be produced this way.
     
  6. Nov 12, 2016 #5
    Sorry I was not familiar with these Latex commands and, I also copy-pasted some lines which caused the problem.

    I did this :-

    ##{\stackrel{+7}{Cl_2}\stackrel{-2}{O_7}} + {\stackrel{+1}{H_2}\stackrel{-1}{O_2}} {\longrightarrow} {\stackrel{+3}{Cl}\stackrel{-2}{O_2^-}} +{\stackrel{0}{O_2}} + {\stackrel{+1}{H^+}}##

    Then I balanced the increase and decrease in oxidation numbers.
    ##{(1/2)\stackrel{+7}{Cl_2}\stackrel{-2}{O_7}} + {\stackrel{+1}{H_2}\stackrel{-1}{O_2}} + 3{\stackrel{+1}{H_2}\stackrel{-1}{O_2}} {\longrightarrow} {\stackrel{+3}{Cl}\stackrel{-2}{O_2^-}} +{3\stackrel{0}{O_2}} + {\stackrel{+1}{H^+}}##

    I multiplied ##\stackrel{+7}{Cl_2}\stackrel{-2}{O_7}## by 1/2 to balance the number of Cl on both sides.
    I took two ##{\stackrel{+1}{H_2}\stackrel{-1}{O_2}}## as Oxygen is both reduced and oxidised.

    After that I balanced oxygen and hydrogen by adding ##H_2O## and ##OH^-##.
    I got ##{Cl_2O_7} + {8H_2O_2} + 2OH^- {\longrightarrow} {2ClO_2^-} +{6O_2} + {9H^2O}##

    Is this correct ?
     
  7. Nov 13, 2016 #6

    Borek

    User Avatar

    Staff: Mentor

    Why do you think oxygen gets both oxidized and reduced?

    What would you get if you assume it gets only oxidized?

    You are all the time concentrating on finding "how to balance the reaction to get the answer given in the book" instead of understanding that "the picture here is more complicated and the answer given in the book doesn't have to be correct". Won't get you far in general.

    This example should be never used in the intro level book, as it is misleading.
     
  8. Nov 13, 2016 #7
    Oxygen get reduced because its oxidation number decrease from -1 ##H_2O_2## to -2 in ##ClO_2^-## . It get oxidized because it increases from -1 ##H_2O_2## and -2 in ##Cl_2O_7## to 0 in ##O_2##.
     
  9. Nov 13, 2016 #8

    Borek

    User Avatar

    Staff: Mentor

    Why do you assume oxygen "moved" from H2O2 to the chlorine containing molecule? Cl2O7 already contains enough "own" oxygen atoms to produce 2ClO2- without a need for additional oxygen:

    Cl2O7 → 2ClO2-

    (this is not balanced, but if you do the counting you will find there are 3 excess oxygen atoms on the left, no need to add more of them).

    And that's why you got a different result - you added unnecessary molecules of water peroxide. Then you balanced the equation with water and oxygen - trick is you can add ANY number of additional H2O2 molecules on the left and still balance the reaction with water and oxygen, that's because you are just adding more hydrogen peroxide decomposition "reactions".

    But: just because you will not add excess hydrogen peroxide doesn't mean you will get a CORRECT answer. As you have seen the "final" equation depends on whether you put more or less hydrogen peroxide molecules on the LHS. You can as well don't use them at all and get the reaction I already posted earlier:

    Cl2O7 + 2OH- → 2ClO2- + 2O2 + H2O

    And I will repeat myself: the only sure way to tell which of these reactions is a correct one is to check the stoichiometry by experiment.
     
  10. Nov 13, 2016 #9
    Yes you are right. Thanks for clarifying.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Why there are infinite ways to balance this Redox equation?
  1. Balancing Redox Equation (Replies: 13)

Loading...