النتائج 1 إلى 4 من 4

الموضوع: Gibbs' Free Energy and Cell potential

  1. #1
    عضو ذهبي الصورة الرمزية ماجده
    تاريخ التسجيل
    Aug 2011
    الدولة
    Jordan - Zerqa
    المشاركات
    902

    افتراضي Gibbs' Free Energy and Cell potential

    Let's now relate the electrochemistry ideas we've explored with the thermodynamic parameter called Gibbs' Free Energy.
    We saw,
    w = –QE (work = – charge times potential),
    where Q is the charge and can be defined as Q = nF where F is the Faraday constant (96485 C/mol), named after Michael Faraday.
    In our case, we're interested in the maximum work since this can be related to the thermodynamic parameter DG.
    Thus, for the case where the work is done infinitely slowly (chemical system is always at equilibrium with the electrodes, electrical resistance is zero since current is essentially zero, etc…) we have
    wmax = – QEmax where Emax for standard conditions is simply E° as calculated from the tabulated half-cell potentials.

    Take for example, a cell with a maximum cell potential of 2.50 V. If 1.33 mol of e passes through the cell at an average potential E = 2.10 V. What is the efficiency?
    w = – Q E = – nFE. = 1.33 mol × 96485 C/mol × 2.10 V (V = J/C)
    w = – 2.69 × 105 J = – 269 kJ
    wmax = – nFEmax = 1.33 mol × 96485 C/mol × 2.50 V
    = – 321 kJ
    efficiency = w/wmax × 100% = –269/–321 × 100% = 83.8 %
    Of course, since wmax is only achievable if the work is done reversibly (infinitely slowly), we can never reach 100% efficiency in any system in the real world.

    We already have seen that DG is a measure of the maximum work obtainable from a system. Thus,
    DG = wmax
    DG = – Q E
    DG = – nFE. In this case, the potential is the cell potential Ecell.
    DG° = – RT lnK = – nFE°cell.
    Thus, we now have a link between free-energy, equilibrium and electrochemical thermodynamic parameters.

    Example:
    Is Fe2+ spontaneously oxidized by the oxygen of the air in acidic solution? Calculate DG and K.
    Two half-reactions can be determined by looking in the table of standard reduction potentials.
    4[Fe2+ ® Fe3+ + e ] E°ox = –0.77 V
    O2 + 4H+ + 4e ® 2 H2O E°red = 1.23 V
    4Fe2+ + O2 + 4H+ ® 4Fe3+ + 2H2O
    n = (4 mol electrons / mol equation)
    E°cell = 0.46 V
    (Positive means spontaneous®)
    DG = –nFE°cell = – 4 mol × 96485 C/mol × 0.46 V = – 1.8×105 J = –180 kJ (per mole of reaction)
    K = e72.7 = 3.7×1031.

    SUMMARY
    DG < 0
    Q < K
    Ecell > 0
    ®
    DG = 0
    Q = K
    Ecell = 0
    Equilibrium
    DG > 0
    Q > K
    Ecell < 0
    ¬

    عضو في نادي ماركا الأكاديمي


  2. #2
    عضو ذهبي الصورة الرمزية م.عبد الرحمن
    تاريخ التسجيل
    Aug 2011
    المشاركات
    261

    افتراضي رد: Gibbs' Free Energy and Cell potential

    عضو في نادي ماركا الأكاديمي


  3. #3
    عضو مميز الصورة الرمزية قمر بلحاج
    تاريخ التسجيل
    Jul 2012
    المشاركات
    5,302

    افتراضي رد: Gibbs' Free Energy and Cell potential

    جزاكم الله خيرا

    عضو في نادي ماركا الأكاديمي


  4. #4
    مراقب عام الصورة الرمزية Eiman
    تاريخ التسجيل
    Jul 2011
    الدولة
    الاردن - ماركا
    المشاركات
    14,750

    افتراضي رد: Gibbs' Free Energy and Cell potential

    Thank you
    من جد وجد ....... ومن سار على الدرب وصل

معلومات الموضوع

الأعضاء الذين يشاهدون هذا الموضوع

الذين يشاهدون الموضوع الآن: 1 (0 من الأعضاء و 1 زائر)

الكلمات الدلالية لهذا الموضوع

المفضلات

ضوابط المشاركة

  • لا تستطيع إضافة مواضيع جديدة
  • لا تستطيع الرد على المواضيع
  • لا تستطيع إرفاق ملفات
  • لا تستطيع تعديل مشاركاتك
  •