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    can someone calculate step by step the emf of this?

    I2 + 2Br- ---> Br2 + 2I-

    I2 + 2e- ---> 2I- =+0.54V
    Br2 + 2e- ---> 2Br- = +1.07V

    Its question 2b page 207 on NT book btw

    thanks.
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    (Original post by JSN)
    can someone calculate step by step the emf of this?

    I2 + 2Br- ---> Br2 + 2I-

    I2 + 2e- ---> 2I- =+0.54V
    Br2 + 2e- ---> 2Br- = +1.07V

    Its question 2b page 207 on NT book btw

    thanks.
    Put the half equations the right way to show reduction and oxidation. A redox reaction cannot have both reactions as oxidising, one must reduce and one must oxidise!
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    (Original post by Booyah)
    Put the half equations the right way to show reduction and oxidation. A redox reaction cannot have both reactions as oxidising, one must reduce and one must oxidise!
    would it be..

    2Br- ---> Br2 + 2e- = +1.07V
    I2 + 2e- ---> 2I- = -0.54V

    ?
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    (Original post by JSN)
    would it be..

    2Br- ---> Br2 + 2e- = +1.07V
    I2 + 2e- ---> 2I- = -0.54V

    ?
    Exactly, and as easy as that was all you have to do is add the values, if they are above o then the reaction is feasible
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    (Original post by Booyah)
    Exactly, and as easy as that was all you have to do is add the values, if they are above o then the reaction is feasible
    The answer im getting is +0.53V, the textbook says the answer is -0.53V where am I going wrong do you think?

    Thanks for the help..
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    (Original post by JSN)
    The answer im getting is +0.53V, the textbook says the answer is -0.53V where am I going wrong do you think?

    Thanks for the help..
    I'm an idiot, forgive me for telling you you were correct,
    The full equation is:

    I2 + 2Br- ---> Br2 + 2I-

    The half equations are:

    I2+ 2e- ----> 2I- = +0.54
    and

    2Br- ----> Br2 + 2e- = -1.07

    Which gives you -0.53

    Sorry for the confusion, I misread the half equations earlier.
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    (Original post by Booyah)
    I'm an idiot, forgive me for telling you you were correct,
    The full equation is:

    I2 + 2Br- ---> Br2 + 2I-

    The half equations are:

    I2+ 2e- ----> 2I- = +0.54
    and

    2Br- ----> Br2 + 2e- = -1.07

    Which gives you -0.53

    Sorry for the confusion, I misread the half equations earlier.
    thank you so much, cleared up alot there!
    appreciate it REPPED!
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    (Original post by JSN)
    thank you so much, cleared up alot there!
    appreciate it REPPED!
    No problem
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    (Original post by Booyah)
    No problem
    last question why would the answer for this be +0.26v not +0.13?!

    2H+ + Pb ---> Pb2+ + H2

    2H+ + 2e- ---> H2 = 0.00V
    Pb2+ + 2e- ---> Pb =-0.13V

    therefore

    Pb ---> Pb2+ + 2e- = +0.13V
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    (Original post by JSN)
    last question why would the answer for this be +0.26v not +0.13?!

    2H+ + Pb ---> Pb2+ + H2

    2H+ + 2e- ---> H2 = 0.00V
    Pb2+ + 2e- ---> Pb =-0.13V

    therefore

    Pb ---> Pb2+ + 2e- = +0.13V
    I'm gonna be honest and say I do not know... I am sure +0.13 would be correct, but check it out with your teacher, he or she can probably tell you why, and post it on here I would like to know why it is right!
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    (Original post by Booyah)
    I'm gonna be honest and say I do not know... I am sure +0.13 would be correct, but check it out with your teacher, he or she can probably tell you why, and post it on here I would like to know why it is right!
    Haha no worries and I will do tomorrow and I will post it on here

    and for the first question what would the overall sum be to get the answer just do know what to minus together?
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    (Original post by JSN)
    Haha no worries and I will do tomorrow and I will post it on here

    and for the first question what would the overall sum be to get the answer just do know what to minus together?
    Well I have never used an equation, I tend to just add the values as one will be reducing and the other oxidising, I don't know any equation as such...
 
 
 
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