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    How does it work?
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    (Original post by mfc1993DP)
    How does it work?
    cell potential cannot be measured to an absolute value, so they need to be quoted against a standard reference electrode.

    It so happens that H+/H2 is commonly used as a standard reference electrode.
    H+(aq) + e- ----> 1/2H2(g)
    and the potential here is taken as 0 V.

    So every other standard reduction potential is measure relative to this point.

    However, it must not be forgotten that in certain circumstances where this might not be possible(as Pt electrode required as part of the setup for H2 part of the cell), then other alternatives like
    1) Ag/AgCl
    2) HgCl2/Hg Calomel electrode

    could also work
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    (Original post by shengoc)
    cell potential cannot be measured to an absolute value, so they need to be quoted against a standard reference electrode.

    It so happens that H+/H2 is commonly used as a standard reference electrode.
    H+(aq) + e- ----> 1/2H2(g)
    and the potential here is taken as 0 V.

    So every other standard reduction potential is measure relative to this point.

    However, it must not be forgotten that in certain circumstances where this might not be possible(as Pt electrode required as part of the setup for H2 part of the cell), then other alternatives like
    1) Ag/AgCl
    2) HgCl2/Hg Calomel electrode

    could also work
    ok, but where does the electron in the equation (H+ + e- -------> 1/2H2) come from?
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    (Original post by mfc1993DP)
    ok, but where does the electron in the equation (H+ + e- -------> 1/2H2) come from?
    ah hah, this is just a half cell. redox reactions mean there is a transfer of electrons, so reduction and oxidation results from combining two half cells together.

    ie 1/2F2 + e- ---> F-
    combine that with H+ + e - ----> H+

    you can write them as 1/2F2 + 1/2H2 ---> HF
    F has been reduced, H has been oxidised.
    The standard electrode reduction potential for the CELL REACTION would be greater than 0(actual value = +2.87V), implying that forward reaction is thermodynamically spontaneous, but it doesn't tell you how fast.
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    (Original post by shengoc)
    ah hah, this is just a half cell. redox reactions mean there is a transfer of electrons, so reduction and oxidation results from combining two half cells together.

    ie 1/2F2 + e- ---> F-
    combine that with H+ + e - ----> H+

    you can write them as 1/2F2 + 1/2H2 ---> HF
    F has been reduced, H has been oxidised.
    The standard electrode reduction potential for the CELL REACTION would be greater than 0(actual value = +2.87V), implying that forward reaction is thermodynamically spontaneous, but it doesn't tell you how fast.
    oh, I get it now , your explaination is much less complicated than the one my teacher gave me. Thankyou for your help
 
 
 
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