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    (c) The cell represented below was set up under standard conditions.
    Pt|Fe2+(aq), Fe3+(aq)||Tl3+(aq),Tl+(aq)|Pt Cell e.m.f. = + 0.48 V

    (i) Deduce the standard electrode potential for the following half-reaction.
    Tl3+(aq) + 2e– → Tl+(aq)
    ................................ ................................ ................................ ...........................
    ................................ ................................ ................................ ...........................
    (ii) Write an equation for the spontaneous cell reaction.
    ................................ ................................ ................................ ...........................



    Hi everyone i did the Q did it well - apart from spontaneous reaction i got 2fe3+ + Tl3+ ====) 2Fe2+ + Tl+

    This was by using the anti-clockwise rule = however - the real answer for the Q was

    Tl3+ + 2 Fe2+ ===) 2Fe3+ + T1+

    Need an explanation why thanks!
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    (Original post by ahsan_ijaz)
    (c) The cell represented below was set up under standard conditions.
    Pt|Fe2+(aq), Fe3+(aq)||Tl3+(aq),Tl+(aq)|Pt Cell e.m.f. = + 0.48 V

    (i) Deduce the standard electrode potential for the following half-reaction.
    Tl3+(aq) + 2e– → Tl+(aq)
    ................................ ................................ ................................ ...........................
    ................................ ................................ ................................ ...........................
    (ii) Write an equation for the spontaneous cell reaction.
    ................................ ................................ ................................ ...........................



    Hi everyone i did the Q did it well - apart from spontaneous reaction i got 2fe3+ + Tl3+ ====) 2Fe2+ + Tl+

    This was by using the anti-clockwise rule = however - the real answer for the Q was

    Tl3+ + 2 Fe2+ ===) 2Fe3+ + T1+

    Need an explanation why thanks!
    Forget the anticlockwise "rule". Examiners can throw in a spanner by writing the list in a different order.

    Spontaneous reaction requires that E(reduced half-equation) - E(oxidised half-equation) = positive value

    Mnemonic E = E(red-ox)
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    Yeah I realised the cycle works - you just have to right them both as reduction equations


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    (Original post by charco)
    Forget the anticlockwise "rule". Examiners can throw in a spanner by writing the list in a different order.

    Spontaneous reaction requires that E(reduced half-equation) - E(oxidised half-equation) = positive value

    Mnemonic E = E(red-ox)
    i think there some information is missing there
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    (Original post by tinashe dongo)
    i think there some information is missing there
    No, there isn't ...
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    (Original post by tinashe dongo)
    i think there some information is missing there
    The equation you get using anti clockwise rule is correct - but you need to remember to right the more negative e cell on top and less negative on bottom and both as reduction reactions so species at non pointy end of arrow will react to give the species at pointy end



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