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    if a half equation is: Zn2+ + 2e- ----> Zn and it's E is -0.76 V
    would: Zn------> 2e- +Zn2+ also be -0.76V? or would it be 0.76V?
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    (Original post by frozo123)
    if a half equation is: Zn2+ + 2e- ----> Zn and it's E is -0.76 V
    would: Zn------> 2e- +Zn2+ also be -0.76V? or would it be 0.76V?
    If you are talking about standard electrode potentials then they are equilibria expressed (by convention) as reductions.

    Zn2+(aq) + 2e <==> Zn(s) ........... E = -0.76 V

    The negative sign shows that there is a tendency for the equilibrium to move to the side of releasing electrons compared with the standard hydrogen electrode.

    It is not a value for an equation.
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    (Original post by charco)
    If you are talking about standard electrode potentials then they are equilibria expressed (by convention) as reductions.

    Zn2+(aq) + 2e <==> Zn(s) ........... E = -0.76 V

    The negative sign shows that there is a tendency for the equilibrium to move to the side of releasing electrons compared with the standard hydrogen electrode.

    It is not a value for an equation.
    If I balanced this half equation with another equation and E standard came out as positive, could I reverse the equation and e standard would be negative?


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    (Original post by frozo123)
    If I balanced this half equation with another equation and E standard came out as positive, could I reverse the equation and e standard would be negative?


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    In this case you are dealing with a difference between two Eº values and yes the positive from E(red) - E(ox) will give the direction of spontaneous change.
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    (Original post by charco)
    In this case you are dealing with a difference between two Eº values and yes the positive from E(red) - E(ox) will give the direction of spontaneous change.
    Just to clarify if there was a reaction between potassium dichromate ( +1.33 V) and Chlorine ( +1.36)

    How would you explain why the answer would be -0.03V? I know how to do it but don't know the reasoning behind it
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    (Original post by frozo123)
    Just to clarify if there was a reaction between potassium dichromate ( +1.33 V) and Chlorine ( +1.36)

    How would you explain why the answer would be -0.03V? I know how to do it but don't know the reasoning behind it
    As I said, in terms of standard reduction potentials, the E(cell) is given by E(red) - E(ox), where E(red) is the species being reduced and E(ox) is the species being oxidised.

    Your two reagents will not react as they are both oxidising agents and can only be reduced.
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    (Original post by charco)
    As I said, in terms of standard reduction potentials, the E(cell) is given by E(red) - E(ox), where E(red) is the species being reduced and E(ox) is the species being oxidised.

    Your two reagents will not react as they are both oxidising agents and can only be reduced.
    According to the exam question, they do react if solid potassium dichromate is used and hot concentrated sulphuric acid is added?
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    (Original post by frozo123)
    According to the exam question, they do react if solid potassium dichromate is used and hot concentrated sulphuric acid is added?
    Not with chlorine it won't...
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    (Original post by charco)
    Not with chlorine it won't...
    here's the link to the question:


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    (Original post by frozo123)
    here's the link to the question:


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    where?
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    (Original post by charco)
    where?
    http://papers.xtremepapers.com/Edexc...6%20Unit-6.pdf

    Q20)
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    You must learn to read questions carefully:

    An acidic solution of potassium manganate(VII) will liberate chlorine from dilute sodium
    chloride solution but an acidic solution of potassium dichromate(VI) will not. Solid potassium dichromate(VI) will liberate chlorine gas from concentrated hydrochloric acid


    This does not say that dichromate ions are reacting with chlorine,it says that dichromate ions react with chloride ions (from hydrochloric acid,not sulfuric acid!!).

    In this case the dichromate ions are being reduced and the chloride ions oxidised.

    You get the value for E(cell) and find that it is negative, :woo:

    However, E values only apply to standard conditions and the conditions given are anything but ...

    Hence under these conditions there is reaction.
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    (Original post by charco)
    You must learn to read questions carefully:

    An acidic solution of potassium manganate(VII) will liberate chlorine from dilute sodium
    chloride solution but an acidic solution of potassium dichromate(VI) will not. Solid potassium dichromate(VI) will liberate chlorine gas from concentrated hydrochloric acid


    This does not say that dichromate ions are reacting with chlorine,it says that dichromate ions react with chloride ions (from hydrochloric acid,not sulfuric acid!!).

    In this case the dichromate ions are being reduced and the chloride ions oxidised.

    You get the value for E(cell) and find that it is negative, :woo:

    However, E values only apply to standard conditions and the conditions given are anything but ...

    Hence under these conditions there is reaction.
    sorry about the confusion was tired

    wait may you tell me the equation? because I thought you do the most postive minus the negative one to get the E value? so you would get +0.03? or

    my equation was 2cr3+ + 3Cl2 + 7H20 ------> cr207- + 6cl- + 7H+ something like that

    or has you say Er- Eox= so the reducing agent is on the left of the E cell diagram?
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    (Original post by frozo123)
    sorry about the confusion was tired

    wait may you tell me the equation? because I thought you do the most postive minus the negative one to get the E value? so you would get +0.03? or

    my equation was 2cr3+ + 3Cl2 + 7H20 ------> cr207- + 6cl- + 7H+ something like that

    or has you say Er- Eox= so the reducing agent is on the left of the E cell diagram?
    Hmm no.
    If your equation is 2Cr^{3+} + 3Cl_2 + 7H_20 ------&gt; Cr_2O_7^- + 6Cl^- + 7H^+


    You need to consider the half reactions. From the equation you can see that
    Cr^{3+} ---&gt; Cr_2O_7^- which is an oxidation
    and Cl_2 ---&gt; 2Cl^- which is a reduction

    So therefore you do \Delta E^{\circ} = E_{chlorine} - E_{chromium}

    To answer your question, you can't reverse the sign on a standard reduction potential, it's fundamentally wrong
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    (Original post by langlitz)
    Hmm no.
    If your equation is 2Cr^{3+} + 3Cl_2 + 7H_20 ------&gt; Cr_2O_7^- + 6Cl^- + 7H^+


    You need to consider the half reactions. From the equation you can see that
    Cr^{3+} ---&gt; Cr_2O_7^- which is an oxidation
    and Cl_2 ---&gt; 2Cl^- which is a reduction

    So therefore you do \Delta E^{\circ} = E_{chlorine} - E_{chromium}

    To answer your question, you can't reverse the sign on a standard reduction potential, it's fundamentally wrong
    My chemistry teacher did chromium- Chlorine? so i'm confused


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    (Original post by frozo123)
    My chemistry teacher did chromium- Chlorine? so i'm confused


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    What's the full question? Are you sure you have the right reaction equation?


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