Dr.wael elhelece electrochemistry 331chem

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Information about Dr.wael elhelece electrochemistry 331chem
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Published on March 10, 2014

Author: drwaelelhelece

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Potentiometric measurements, electrochemical reactions and Nernest equation, reference electrodes, standard potentials, thermodynamics of electrochemical reactions, diffusion and electrochemical reactions, voltammetry, mechanism of electrode reactions, physical and chemical meaning of corrosion, study of the effect of media on the corrosion.

Electrochemistry ‫الكيمياء‬‫الكهربية‬ 331chem. 1 Dr. Elhelece W. A.

Contents ‫المحتويات‬  Potentiometric Measurements. .‫الكهربية‬ ‫الجهود‬ ‫وتقدير‬ ‫قياسات‬  Electrochemical Reaction and Nernest Equation .‫نيرنست‬ ‫ومعادلة‬ ‫الكهروكيميائية‬ ‫التفاعلت‬  Reference Electrodes and standard Potential .‫القياسى‬ ‫والجهد‬ ‫القياسية‬ ‫القطاب‬  Thermodynamics and Electrochemical Reactions .‫الكهروكيميائية‬ ‫والتفاعلت‬ ‫الحرارية‬ ‫الديناميكا‬ 2 Dr. Elhelece W. A.

Contents ‫المحتويات‬  Diffusion and Electrochemical Reactions .‫الكهروكيميائية‬ ‫والتفاعلت‬ ‫النتشار‬  Voltammetery and Mechanism of Electrode Reaction .‫القطاب‬ ‫عند‬ ‫التفاعلت‬ ‫وميكانيكية‬ ‫الجهود‬ ‫قياس‬ ‫عمليات‬  Physical and Chemical Meaning of Corrosion .‫للتآكل‬ ‫والكيميائى‬ ‫الفيزيائى‬ ‫المعنى‬  Study of the Effect of Media on the Corrosion. .‫التآكل‬ ‫عمليات‬ ‫على‬ ‫الوسط‬ ‫تأثير‬ ‫دراسة‬ 3 Dr. Elhelece W. A.

Electrochemical Reactions ‫الكهروكيميائية‬ ‫التفاعل ت‬ Electrochemical reactions ‫الكهروكيميائية‬ ‫التفاعل ت‬ Is a reaction in which electrons are transferred from one species to another. .‫لرخر‬ ‫عنصر‬ ‫من‬ ‫اللكترونات‬ ‫انتقال‬ ‫فيه‬ ‫يتم‬ ‫تفاعل‬ ‫هو‬ 4 Dr. Elhelece W. A.

Electrochemical Cells ‫الكهروكيميائية‬ ‫الخليا‬  two electrodes (electronic conductors) .(‫اللكترونات‬ ‫)انتقال‬ ‫القطاب‬ ‫من‬ ‫زوج‬  electrolyte (ionic conductor) .(‫اليونات‬ ‫)انتقال‬ ‫الكتروليتى‬ ‫محلول‬  Electrode and its electrolyte are places in an electrode compartment .‫لهما‬ ‫حاوي‬ ‫إناء‬ ‫فى‬ ‫يوضعا‬ ‫اللكتروليتى‬ ‫والمحلول‬ ‫القطب‬ 5 Dr. Elhelece W. A.

Electrochemical Cells ‫الكهروكيميائية‬ ‫الخليا‬ Dr. Elhelece W. A. 6 Salt bridge ‫الملحية‬ ‫القنطرة‬  tube containing a concentrated electrolyte solution . ) ‫حرف‬ ‫شكل‬ ‫على‬ ‫أنبوبة‬ ‫عن‬ ‫عبارة‬U‫ملحى‬ ‫محلول‬ ‫على‬ ‫تحتوى‬ ( ‫مركز‬  alternative way of joining two different electrode compartments. ‫المختلفة‬ ‫القطاب‬ ‫من‬ ‫زوج‬ ‫لربط‬ ‫تقليدية‬ ‫طريقة‬.

Electrochemical Cells ‫الكهروكيميائية‬ ‫الخلايا‬ Dr. Elhelece W. A. 7

Electrochemical Cells ‫الكهروكيميائية‬ ‫الخليا‬ Dr. Elhelece W. A. 8

Reference electrodes ‫القياسىة‬ ‫الطقطاب‬ Are used to give a value of potential to which other potentials can be referred. ‫المختلفة‬ ‫للقطاب‬ ‫الجهود‬ ‫قيمة‬ ‫وقياس‬ ‫لتعيين‬ ‫تستخدم‬. 9 Dr. Elhelece W. A.

Standard Hydrogen Electrode ‫طق‬‫ال‬ ‫طب‬‫ه‬‫ال‬ ‫يدروجين‬‫ق‬‫ياسي‬ 2H+ (aq) + 2e– H2(g) H2 H+ Pt Dr. Elhelece W. A. 10

Dr. Elhelece W. A. 11 The “standard” aspect to this cell is that the activity of H2(g) and that of H+ (aq) are both 1. .‫واحد‬ ‫الهيدروجين‬ ‫اايونا ت‬ ‫وكذا‬ ‫الهيدروجين‬ ‫غاز‬ ‫فاعلية‬ ‫ان‬ ‫ايعنى‬ ‫الخلية‬ ‫لهذه‬ ‫قياسي‬ ‫مصطلح‬ This means that the pressure of H2 is 1 atm and the concentration of H+ is 1M. .‫مولر‬ ‫واحد‬ ‫الهيدروجين‬ ‫اايونا ت‬ ‫وتركيز‬ ‫جو‬ ‫ضغط‬ ‫واحد‬ ‫الهيدروجين‬ ‫غاز‬ ‫ضغط‬ ‫ايعنى‬ ‫هذا‬ These are our standard reference states. .‫المرجعية‬ ‫القياسية‬ ‫حالتنا‬ ‫هى‬ ‫هذه‬ Standard Hydrogen Electrode ‫طق‬‫ال‬ ‫طب‬‫ه‬‫ال‬ ‫يدروجين‬‫ق‬‫ياسي‬

 Voltaic (Galvanic) Cell ‫الجلفانية‬ ‫الخلايا‬  Electrolytic Cell ‫الكهربي‬ ‫التحليل‬ ‫خلايا‬ 12 Dr. Elhelece W. A. Two types of electrochemical cells ‫الكهروكيميائية‬ ‫الخلايا‬ ‫من‬ ‫نوعان‬

Voltaic (Galvanic) Cells  Reactions are spontaneous .‫تلقائية‬ ‫الخليا‬ ‫من‬ ‫النوع‬ ‫هذا‬ ‫فى‬ ‫التفاعلت‬  Redox reactions produce electrical energy .‫كهربية‬ ‫طاقة‬ ‫تنتج‬ ‫والتختزال‬ ‫الكسدة‬ ‫تفاعلت‬  Lets look at an example: ‫المثال‬ ‫سبيل‬ ‫وعلى‬ Cu+2 (aq) +Zn (s) -------> Cu (s) + Zn+2 (aq) 13 Dr. Elhelece W. A.

Daniell Cell ‫دانيال‬ ‫خلية‬ Dr. Elhelece W. A. 14 CuSO4 (aq)ZnSO4 (aq) Cu metalZn metal salt bridge Zn(s) → Zn2+ (aq) + 2e– Cu2+ (aq) + 2e– → Cu(s) Cathode (reduction) +ive Anode (oxidation) –ive‫)أكسدة‬ ‫)النود‬ ‫)الكاثود)اختزال‬

Electrolytic Cell ‫التحليلية‬ ‫الخلية‬ Reactions are non spontaneous. .‫تلقائية‬ ‫غير‬ ‫التفاعلت‬ Redox reactions require electrical energy to occur. .‫لتحدث‬ ‫كهربية‬ ‫طاقة‬ ‫الى‬ ‫تحتاج‬ ‫والرختزال‬ ‫الكسدة‬ ‫تفاعلت‬ 15 Dr. Elhelece W. A.

Cell Types ‫الخلايا‬ ‫أنواع‬ Galvanic cell: produces electricity as a result of spontaneous reactions ‫الجلفانية‬ ‫الخلايا‬.‫تلقائى‬ ‫لتفاعل‬ ‫نتيجة‬ ‫كهربا‬ ‫تنتج‬ : Electrolytic cell: a non-sponteneous reaction is driven by an external source ‫الكهربى‬ ‫التحليل‬ ‫خلايا‬.‫كهربى‬ ‫تيار‬ ‫دفع‬ ‫تحت‬ ‫ايحدث‬ ‫لتلقائى‬ ‫تفاعل‬ : 16 Dr. Elhelece W. A.

Reduction and oxidation processes occurring in a cell are separated in space (anode and cathode compartments) .‫مفصولة‬ ‫وتكون‬ ‫خلية‬ ‫فى‬ ‫تحدث‬ ‫التى‬ ‫والختزال‬ ‫الكسدة‬ ‫عمليا ت‬  Anode (oxidation); is the electrode at which oxidation is occurring ‫النود‬:.‫اكسدة‬ ‫عنده‬ ‫ايحدث‬ ‫الذى‬ ‫القطب‬  Cathode (reduction); is the electrode at which reduction is occurring ‫الكاثود‬:.‫اختزال‬ ‫عنده‬ ‫ايحدث‬ ‫الذى‬ ‫القطب‬ 17 Dr. Elhelece W. A. Reactions at electrodes ‫الطقطاب‬ ‫عند‬ ‫تحدث‬ ‫التى‬ ‫التفاعل ت‬

Oxidation and Reduction ‫والختزال‬ ‫الكسدة‬  What is reduced is the oxidizing agent. ‫مؤكسد‬ ‫عامل‬ ‫هو‬ ‫ايختزل‬ ‫الذي‬  H+ oxidizes Zn by taking electrons from it.  What is oxidized is the reducing agent. ‫مختز‬ ‫عامل‬ ‫هو‬ ‫ايتأكسد‬ ‫الذي‬‫ل‬  Zn reduces H+ by giving it electrons. 18 Dr. Elhelece W. A.

19 Dr. Elhelece W. A.

Electro Motive Force (EMF) ‫الكهربية‬ ‫الدافعة‬ ‫القوة‬  Water only spontaneously flows one way in a waterfall. ‫اتجاه‬ ‫فى‬ ‫سري‬‫س‬‫ايس‬ ‫سا‬‫س‬‫تلقائي‬ ‫سط‬‫س‬‫فق‬ ‫الماء‬ .‫الهبوط‬  Likewise, electrons only spontaneously flow one way in a redox reaction from higher to lower potential energy. ‫في‬ ‫تسسري‬ ‫تلقائيسا‬ ‫اللكترونا ت‬ ‫بالمثسل‬ ‫القل‬ ‫إلى‬ ‫العلى‬ ‫من‬ ‫التجاه‬ 20 Dr. Elhelece W. A.

Electro Motive Force (EMF) ‫الكهربية‬ ‫الدافعة‬ ‫القوة‬  The potential difference between the anode and cathode in a cell is called: ‫ايسمى‬ ‫خلية‬ ‫في‬ ‫والكاثود‬ ‫النود‬ ‫بين‬ ‫الجهد‬ ‫فرق‬ Electro Motive Force (EMF) ‫الكهربية‬ ‫الدافعة‬ ‫القوة‬  It is also called the cell potential, and is designated Ecell. ‫أايضا‬ ‫ايسمى‬‫الخلية‬ ‫جهد‬) ‫بالرمز‬ ‫له‬ ‫وايرمز‬Ecell.( 21 Dr. Elhelece W. A.

Potentiometric Measurements ‫الجهداية‬ ‫القياسا ت‬  Potentiometer ‫الجهد‬ ‫طقياس‬ ‫جهاز‬ A device for measuring the potential of an electrochemical cell without drawing a current or altering the cell’s composition. .‫الخلية‬ ‫تركيب‬ ‫في‬ ‫الدخول‬ ‫أو‬ ‫بالتيار‬ ‫إمداد‬ ‫دون‬ ‫الخلية‬ ‫جهد‬ ‫لقياس‬ ‫جهاز‬  Potentiometry ‫الجهد‬ ‫طقياس‬ ‫عملية‬ Use of Electrodes to Measure Voltages that Provide Chemical Information. .‫كيميائى‬ ‫مصدر‬ ‫من‬ ‫المأخوذة‬ ‫الجهود‬ ‫لقياس‬ ‫القطاب‬ ‫استخدام‬ 22 Dr. Elhelece W. A.

Potentiometer ‫الجهد‬ ‫قياس‬ ‫جهاز‬23 Dr. Elhelece W. A.

 Potentiometric measurements ‫الجهود‬ ‫طقياس‬ ‫عمليا ت‬ Potentiometric measurements are made using a potentiometer to determine the difference in potential between a working (an indicator) electrode and a counter (a reference) electrode. ‫عامل‬ ‫قطب‬ ‫بين‬ ‫الجهد‬ ‫فى‬ ‫الفرق‬ ‫لتعين‬ ‫البوتنشيمتر‬ ‫بواسطة‬ ‫تتم‬ ‫الجهد‬ ‫قياس‬ ‫عملية‬ .(‫)مرجعى‬ ‫ثابت‬ ‫وقطب‬ (‫)كاشف‬ - Cathode is the working/indicator electrode. (right half-cell) -.(‫الايمن‬ ‫الخلية‬ ‫)نصف‬ ‫العامل‬ ‫القطب‬ ‫هو‬ ‫الكاثود‬ - Anode is the counter/reference electrode. (left half- cell) -(‫الايسر‬ ‫الخلية‬ ‫)نصف‬ ‫المرجعى‬ ‫القطب‬ ‫هو‬ ‫النود‬. 24 Dr. Elhelece W. A.

Ecell = Ec ─ Ea Where : Ec is the reduction potential at the cathode. Ec‫الكاثود‬ ‫عند‬ ‫الختزال‬ ‫.جهد‬ Ea is the reduction potential at the anode Ea‫النود‬ ‫عند‬ ‫الختزال‬ ‫.جهد‬ 25 Dr. Elhelece W. A.

Electrochemical reactions ‫الكهروكيميائية‬ ‫التفاعل ت‬ Electrochemical reactions are made up of two “half-reactions”. .‫تفاعل‬ ‫نصفى‬ ‫من‬ ‫تتكون‬ -At one electrode electrons are lost (oxidation). ) ‫لللكترونا ت‬ ‫فقد‬ ‫القطاب‬ ‫أحد‬ ‫عند‬ ‫ايحدث‬‫أكسدة‬.( -At the other electrode, electrons are gained (reduction). ) ‫لللكترونا ت‬ ‫اكتساب‬ ‫ايحدث‬ ‫الخر‬ ‫القطب‬ ‫عند‬‫اختزال‬.( 26 Dr. Elhelece W. A.

Electrochemical Half Reactions ‫كهروكيميائى‬ ‫لتفاعل‬ ‫التفاعل‬ ‫نصف‬  Electrochemical half reactions: ‫كهروكيميائى‬ ‫لتفاعل‬ ‫التفاعل‬ ‫نصف‬ Oxidation ‫اكسدة‬ Zn: Zn (s) Zn⇒ 2+ (aq)+ 2e- Reduction ‫اختزال‬ Cu2+ : Cu2+ (aq) + 2e- Cu (s)⇒ 27 Dr. Elhelece W. A.

28 Dr. Elhelece W. A. The overall reaction is the difference between the two half- reactions: ‫التفاعل‬ ‫نصفى‬ ‫بين‬ ‫الفرق‬ ‫هو‬ ‫الكلى‬ ‫التفاعل‬ Zn(s) + Cu2+ (aq) Zn⇒ 2+ (aq) + Cu(s) (note that the electrons have to cancel in the overall reaction). .‫الكلى‬ ‫التفاعل‬ ‫من‬ ‫اللكترونا ت‬ ‫تحذف‬ ‫ان‬ ‫ايجب‬ ‫التالى‬ ‫لحظ‬ Electrochemical Half Reactions ‫كهروكيميائى‬ ‫لتفاعل‬ ‫التفاعل‬ ‫نصف‬

29 Oxidation number ‫التأكسد‬ ‫عدد‬ The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred to the more electronegative atom. ‫الكثر‬ ‫للذرة‬ ‫كامل‬ ‫بشكل‬ ‫اللكترونات‬ ‫وانتقلت‬ ‫حدث‬ ‫اذا‬ ‫اليونى‬ ‫المركب‬ ‫فى‬ ‫الذرة‬ ‫تحملها‬ ‫التى‬ ‫الشحنة‬ .‫كهربية‬ ‫سالبية‬ 1.Oxidation number equals ionic charge for monoatomic ions in ionic compound. .‫ايونى‬ ‫مركب‬ ‫فى‬ ‫الذرة‬ ‫احادى‬ ‫ليون‬ ‫اليونية‬ ‫الشحنة‬ ‫يكافئ‬ ‫التأكسد‬ ‫عد‬ CaBr2; Ca = +2, Br = -1 2. Metal ions in Family A have one, positive oxidation number; Group IA metals are +1, IIA metals are +2. ) ‫المجموعة‬ ‫فى‬ ‫المعدن‬ ‫ايون‬A‫التاكسد‬ ‫عدد‬ (1.‫موجب‬ ‫ويكون‬ Li+ , Li = +1; Mg+2 , Mg = +2 Dr. Elhelece W. A.

3. The oxidation number of a transition metal ion is positive, but can vary in magnitude. .‫القيمة‬ ‫فى‬ ‫ايختلف‬ ‫ولكن‬ ‫موجب‬ ‫دائما‬ ‫انتقالى‬ ‫لعنصر‬ ‫التاكسد‬ ‫عدد‬ 4. Nonmetals can have a variety of oxidation numbers, both positive and negative numbers which can vary in magnitude. ‫او‬ ‫موجوب‬ ‫ايكون‬ ‫ان‬ ‫اما‬ ‫التاكسد‬ ‫حال ت‬ ‫من‬ ‫كبير‬ ‫عدد‬ ‫لها‬ ‫ايكون‬ ‫ان‬ ‫ايمكن‬ ‫النتقالية‬ ‫غير‬ ‫العناصر‬ .‫سالب‬ 5. Free elements (uncombined state) have an oxidation number of zero. Each atom in O2, F2, H2, Cl2, K, Be has the same oxidation number; zero. .‫صفر‬ ‫التاكسد‬ ‫عدد‬ ‫يكون‬ (‫الترابط‬ ‫عدم‬ ‫حالة‬ ‫)فى‬ ‫الحرة‬ ‫العناصر‬ Dr. Elhelece W. A. Oxidation number ‫التأكسد‬ ‫عدد‬ 30

6. The oxidation number of fluorine is always –1. - ‫دائما‬ ‫اايون‬ ‫الفلوراين‬ ‫تاكسد‬ ‫عدد‬1. (unless fluorine is in elemental form, F2). ‫ايكون‬ ‫الفلور‬ ‫جزيء‬ ‫فى‬ ‫الفلور‬ ‫عنصر‬ ‫كان‬ ‫ان‬ ‫ال‬0. 7. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. ‫او‬ ‫الجزيء‬ ‫هذا‬ ‫على‬ ‫الشحنة‬ ‫مسساوايا‬ ‫ايكون‬ ‫الايون‬ ‫او‬ ‫الجزيسء‬ ‫فسى‬ ‫الكسدة‬ ‫أعداد‬ ‫مجموع‬ .‫الايون‬ IF; F= -1; I = +1 31 Dr. Elhelece W. A. Oxidation number ‫التأكسد‬ ‫عدد‬

Dr. Elhelece W. A. 32 Oxidation number ‫التأكسد‬ ‫عدد‬ 8. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. + ‫الهيدروجين‬ ‫تاكسد‬ ‫عدد‬1.‫ثنائي‬ ‫ملح‬ ‫فى‬ ‫بفلز‬ ‫مرتبط‬ ‫ايكون‬ ‫ان‬ ‫عدا‬ In these cases, its oxidation number is –1 or when it’s in elemental form (H2; oxidation # =0). - ‫تاكسده‬ ‫عدد‬ ‫فان‬ ‫الحال ت‬ ‫هذه‬ ‫فى‬1‫ايكون‬ ‫العنصراية‬ ‫حالته‬ ‫فى‬ ‫او‬0. HF; F= -1, H= +1 NaH; Na= +1, H = -1

H2O; H=+1, O= -2 SO3; O = -2; S = +6 9. The oxidation number of oxygen is usually –2. In H2O2 and O2 2- it is –1, in elemental form (O2 or O3) it is 0. - ‫الكسجين‬ ‫تأكسد‬ ‫عدد‬2- ‫يكون‬ ‫الهيدروجين‬ ‫اكسيد‬ ‫فوق‬ ‫مركب‬ ‫وفى‬1‫يكون‬ ‫العنصرية‬ ‫حالته‬ ‫وفى‬0. Oxidation numbers of all the atoms in HCO3 - ? HCO3 - O = -2 H = +1 3x(-2) + 1 + C = -1 C = +4 4.4 33 Dr. Elhelece W. A. Oxidation number ‫التأكسد‬ ‫عدد‬

NaIO3 Na = +1 O = -2 3x(-2) + 1 + ? = 0 I = +5 IF7 F =-1 7x(-1) + ? = 0 I = +7 K2Cr2O7 O = -2 K = +1 7x(-2) + 2x(+1) + 2x(?) = 0 Cr = +6 Oxidation numbers of all the elements in the following ? ‫المركبات؟‬ ‫فى‬ ‫للعناصر‬ ‫التأكسد‬ ‫أعداد‬ 34 Dr. Elhelece W. A. Oxidation number ‫التأكسد‬ ‫عدد‬

Determination of Oxidizing and Reducing Agents ‫المختزلة‬ ‫والعوامل‬ ‫المؤكسدة‬ ‫العوامل‬ ‫تعين‬ I. Determine oxidation # for all atoms in both the reactants and products. ‫والنواتج‬ ‫المتفاعل ت‬ ‫كل‬ ‫فى‬ ‫الذرا ت‬ ‫لكل‬ ‫الكسدة‬ ‫أعداد‬ ‫تعين‬ I. Look at same atom in reactants and products and see if oxidation # increased or decreased. ‫عدد‬ ‫فى‬ ‫الفرق‬ ‫ونحسب‬ ‫والنواتج‬ ‫المتفاعل ت‬ ‫فى‬ ‫المماثلة‬ ‫الذرا ت‬ ‫عن‬ ‫نبحث‬ .‫الكسدة‬  If oxidation # decreased; substance reduced .‫اختزلت‬ ‫المادة‬ ‫فان‬ ‫التأكسد‬ ‫عد‬ ‫قل‬ ‫اذا‬  If oxidation # increased; substance oxidized .‫تأكسد ت‬ ‫المادة‬ ‫فان‬ ‫التأكسد‬ ‫عد‬ ‫ازداد‬ ‫اذا‬ 35 Dr. Elhelece W. A.

 Oxidizing Agent: Substance that oxidizes the other substance by accepting electrons. It is reduced in reaction. ‫الكترونا ت‬ ‫بكسب‬ ‫الخراين‬ ‫تأكسد‬ ‫التى‬ ‫المادة‬ :‫المؤكسد‬ ‫العامل‬ .(‫التفاعل‬ ‫فى‬ ‫اختزال‬ ‫لها‬ ‫)ايحدث‬  Reducing Agent: Substance that reduces the other substance by donating electrons. It is oxidized in reaction. ‫الكترونا ت‬ ‫باعطاهم‬ ‫الخراين‬ ‫تختزل‬ ‫التى‬ ‫المادة‬ :‫المختزل‬ ‫العامل‬ .(‫التفاعل‬ ‫فى‬ ‫)تتأكسد‬ 36 Dr. Elhelece W. A. Determination of Oxidizing and Reducing Agents ‫المختزلة‬ ‫والعوامل‬ ‫المؤكسدة‬ ‫العوامل‬ ‫تعين‬

Balancing Redox Equations ‫والختزال‬ ‫الكسدة‬ ‫معادل ت‬ ‫وزن‬ The oxidation of Fe2+ to Fe3+ by Cr2O7 2- in acid solution? ‫حامضي؟‬ ‫وسط‬ ‫فى‬ ‫كرومات‬ ‫الثانى‬ ‫ايون‬ ‫بواسطة‬ ‫الثليثي‬ ‫الحديد‬ ‫الى‬ ‫الثنائي‬ ‫الحديد‬ ‫أكسدة‬ 1. Write the unbalanced equation for the reaction ion ionic form. .‫اليونى‬ ‫الشكل‬ ‫فى‬ ‫للتفاعل‬ ‫الموزونة‬ ‫غير‬ ‫المعادلة‬ ‫اكتب‬ Fe2+ + Cr2O7 2- → Fe3+ + Cr3+ 2. Separate the equation into two half-reactions. .‫تفاعل‬ ‫نصفي‬ ‫الى‬ ‫المعادلة‬ ‫افصل‬ +2 +3 Oxidation: Fe2+ → Fe3+ ‫اكسدة‬ +6 +3 Reduction: Cr2O7 2- Cr3+ ‫اتختزال‬ 37 Dr. Elhelece W. A.

Dr. Elhelece W. A. 38 Balancing Redox Equations ‫والختزال‬ ‫الكسدة‬ ‫معادل ت‬ ‫وزن‬ 3. Balance the atoms other than O and H in each half-reaction. .‫نصف‬ ‫كل‬ ‫فى‬ ‫والهيدروجين‬ ‫الكسجين‬ ‫غير‬ ‫الخرى‬ ‫الذرا ت‬ ‫زن‬ Cr2O7 2- → 2Cr3+ 4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. .‫والهيدروجين‬ ‫الكسجين‬ ‫لوزن‬ ‫ماء‬ ‫نضيف‬ ‫الحامض‬ ‫فى‬ ‫للتفاعل ت‬ ‫بالنسبة‬ Cr2O7 2- → 2Cr3+ + 7H2O 14H+ + Cr2O7 2- → 2Cr3+ + 7H2O

39 Dr. Elhelece W. A. Balancing Redox Equations ‫والختزال‬ ‫الكسدة‬ ‫معادل ت‬ ‫وزن‬ 5. Add electrons to one side of each half-reaction to balance the charges on the half- reaction. .‫التفاعل‬ ‫نصفي‬ ‫فى‬ ‫الشحنا ت‬ ‫لتعادل‬ ‫المعادلة‬ ‫فى‬ ‫لجهة‬ ‫الكترونا ت‬ ‫اضف‬ Fe2+ → Fe3+ + 1e- 6e- + 14H+ + Cr2O7 2- → 2Cr3+ + 7H2O 6. Equalize the number of electrons in the two half-reactions by multiplying the half- reactions by appropriate coefficients. .‫منهما‬ ‫كل‬ ‫معامل ت‬ ‫فى‬ ‫بالضرب‬ ‫التفاعل‬ ‫نصفي‬ ‫فى‬ ‫اللكترونا ت‬ ‫عدد‬ ‫عادل‬ 6Fe2+ → 6Fe3+ + 6e- 6e- + 14H+ + Cr2O7 2- → 2Cr3+ + 7H2O

7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. 6e- + 14H+ + Cr2O7 2- → 2Cr3+ + 7H2O 6Fe2+ → 6Fe3+ + 6e-Oxidation: Reduction: 14H+ + Cr2O7 2- + 6Fe2+ → 6Fe3+ + 2Cr3+ + 7H2O 8. Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. 40 Dr. Elhelece W. A. Balancing Redox Equations ‫والختزال‬ ‫الكسدة‬ ‫معادل ت‬ ‫وزن‬

 Example : Write a balanced ionic equation to represent the oxidation of iodide ion (I- ) by permanganate ion (MnO4 - ) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide . ‫مثال‬‫بواسطة‬ ‫اليود‬ ‫أكسسدة‬ ‫عمليسة‬ ‫تقدم‬ ‫لكى‬ ‫موزونسة‬ ‫أايونيسة‬ ‫معادلسة‬ ‫اكتسب‬ : .‫الرباعي‬ ‫والمنجنيز‬ ‫الجزايئي‬ ‫اليود‬ ‫لينتج‬ ‫طقلوي‬ ‫محلول‬ ‫في‬ ‫البرمنجا ت‬ ‫اايون‬ 41 Dr. Elhelece W. A. Balancing Redox Equations ‫والختزال‬ ‫الكسدة‬ ‫معادل ت‬ ‫وزن‬

 Step 1 : Write the unbalanced equation is: ‫الخطوة‬1:‫كالتالي‬ ‫الموزونة‬ ‫غير‬ ‫المعادلة‬ ‫اكتب‬ MnO4 - + I- → MnO2 + I2  Step 2: The two half-reactions are: ‫الخطوة‬2:‫هما‬ ‫التفاعل‬ ‫نصفى‬ -1 0  Oxidation; I- → I2‫الكسدة‬ +7 +4  Reduction: MnO4 - → MnO2 ‫الختزال‬ 42 Dr. Elhelece W. A.

 Step 3: We balance each half-reaction for number and type of atoms and charges. ‫الخطوة‬3.‫والشحنة‬ ‫نوع‬ ‫لكل‬ ‫الذرا ت‬ ‫بأعداد‬ ‫تفاعل‬ ‫نصف‬ ‫كل‬ ‫نوزن‬ :  Oxidation half reaction: We first balance the I atoms; ‫الذرا ت‬ ‫عدد‬ ‫أول‬ :‫الكسدة‬ ‫التفاعل‬ ‫نصف‬2I- → I2  To balance charges, we add two electrons to the right-hand side of the equation: ‫التفاعل‬ ‫من‬ ‫اايمنى‬ ‫للجهة‬ ‫الكترون‬ ‫زوج‬ ‫نضيف‬ ‫ان‬ ‫ايجب‬ ‫الشحنا ت‬ ‫نزن‬ ‫لكى‬ 2 I- → I2 + 2e-  Reduction half-reaction : To balance the O atoms, we add two H2O molecules on the right: ‫نضيف‬ ‫الوكسجين‬ ‫ذرا ت‬ ‫نزن‬ ‫كى‬ :‫الختزال‬ ‫التفاعل‬ ‫نصف‬2:‫اليمنى‬ ‫للجهة‬ ‫ماء‬ ‫جزيء‬ MnO4 - → MnO2 + 2H2O 43 Dr. Elhelece W. A.

 Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 as follows: ‫خطوة‬4‫مكافئة‬ ‫بعد‬ ‫وذلك‬ ‫الكلى‬ ‫التفاعل‬ ‫على‬ ‫للحصول‬ ‫التفاعل‬ ‫نصفى‬ ‫نجمع‬ : :‫التفاعل‬ ‫نصفى‬ ‫فى‬ ‫اللكترونا ت‬ ‫عدد‬ (1) 3( 2I- → I2 + 2e- ) (2) 2(MnO4 - + 4H+ + 3e- → MnO2 + 2H2O) __________________________________ 6I- + 2MnO4- + 8H+ + 6e → 3I2 + 2MnO2 + 4H2O + 6e 44 Dr. Elhelece W. A.

 Step 5: A final check shows that the equation is balanced in terms of both atoms and charges. ‫خطوة‬5.‫والشحنا ت‬ ‫الذرا ت‬ ‫حيث‬ ‫من‬ ‫موزونة‬ ‫المعادلة‬ ‫ان‬ ‫تتأكد‬ ‫نهائية‬ ‫مراجعة‬ : Practice Exercise: Balance the following equation for the reaction in an acidic medium by the ion-electron method : .‫الكترون‬ ‫الايون‬ ‫بطرايقة‬ ‫حامضي‬ ‫وسط‬ ‫فى‬ ‫للتفاعل‬ ‫التالية‬ ‫العادلة‬ ‫زن‬ :‫عملى‬ ‫تمراين‬ Fe2+ + MnO4 - → Fe3+ + Mn2+ 45 Dr. Elhelece W. A. Balancing Redox Equations ‫والختزال‬ ‫الكسدة‬ ‫معادل ت‬ ‫وزن‬

Applications ‫تطبيقا ت‬  Moving electrons is electric current. .‫كهربى‬ ‫تيار‬ ‫اللكترونا ت‬ ‫حركة‬  8H+ +MnO4 - + 5Fe+2 +5e- Mn+2 + 5Fe+3 +4H2O  Helps to break the reactions into half reactions. .‫نصفين‬ ‫الى‬ ‫التفاعل‬ ‫قسمة‬ ‫على‬ ‫تساعد‬  8H+ +MnO4 - +5e- Mn+2 +4H2O  5(Fe+2 Fe+3 + e- )  In the same mixture it happens without doing useful work, but if separate ‫فصلهم‬ ‫عند‬ ‫ولكن‬ ‫مفيد‬ ‫شغل‬ ‫اى‬ ‫دون‬ ‫اللكترونا ت‬ ‫حركة‬ ‫تتم‬ ‫المحلول‬ ‫نفس‬ ‫فى‬ 46 Dr. Elhelece W. A.

 Connected this way the reaction starts ‫ايبدأ‬ ‫التفاعل‬ ‫بالشكل‬ ‫كما‬ ‫التوصيل‬  Stops immediately because charge builds up. .‫اللكترونا ت‬ ‫تزاحم‬ ‫بسبب‬ ‫فورا‬ ‫ايتوقف‬ 47 Dr. Elhelece W. A. H+ MnO4 - Fe+2 e- e- e- e- e- Galvanic Cell ‫الجلفانية‬ ‫الخلية‬

H+ MnO4 - Fe+2 Galvanic Cell ‫الجلفانية‬ ‫الخلية‬ Salt Bridge allows current to flow. ‫تسمح‬ ‫الملحية‬ ‫القنطرة‬ .‫التيار‬ ‫بمرور‬ 48 Dr. Elhelece W. A.

Fe+2 H+ MnO4 - e- Electricity travels in a complete circuit. .‫مغلقة‬ ‫دارة‬ ‫في‬ ‫تنتقل‬ ‫الكهرباء‬ 49 Dr. Elhelece W. A. Galvanic Cell ‫الجلفانية‬ ‫الخلية‬

50 Dr. Elhelece W. A.

H+ MnO4 - Fe+2 Porous Disk ‫منفذة‬ ‫اسطوانة‬ Instead of a salt bridge ‫الملحية‬ ‫القنطرة‬ ‫من‬ ‫بدل‬ 51 Dr. Elhelece W. A. Galvanic Cell ‫الجلفانية‬ ‫الخلية‬

Reducing Agent Oxidizing Agent e- e- e- e- e- e- Anode Cathode 52 Dr. Elhelece W. A. Galvanic Cell ‫الجلفانية‬ ‫الخلية‬

Cell Potential ‫الخلية‬ ‫جهد‬  Oxidizing agent pulls the electron. .‫اللكترونات‬ ‫يجذب‬ ‫المؤكسد‬ ‫العامل‬  Reducing agent pushes the electron. .‫اللكترونات‬ ‫يدفع‬ ‫المختزل‬ ‫العامل‬  The push or pull (“driving force”) is called the cell potential Ecell. ‫الخلية‬ ‫جهد‬ ‫تسمى‬ (‫المحركة‬ ‫)القوة‬ ‫اللكترونات‬ ‫وجذب‬ ‫دفع‬Ecell.  Also called the electromotive force (emf). .‫الكهربية‬ ‫الدافعة‬ ‫القوة‬ ‫تسمى‬ ‫وايضا‬  Unit is the volt(V) ‫الفولت‬ ‫هى‬ ‫الوحدة‬  = 1 joule of work/coulomb of charge ‫الشحنات‬ ‫من‬ ‫كولوم‬ ‫لكل‬ ‫الشغل‬ ‫من‬ ‫جول‬  Measured with a voltmeter ‫)الفولتميتر‬ ‫الفولت‬ ‫قياس‬ ‫بجهاز‬ ‫)تقاس‬ 53 Dr. Elhelece W. A.

Zn+2 SO4 - 2 1 M HCl Anode 0.76 1 M ZnSO4 H+ Cl- H2 in Cathode 54 Dr. Elhelece W. A. Cell Potential ‫الخلية‬ ‫جهد‬

1 M HCl H+ Cl- H2 in Standard Hydrogen Electrode ‫القياسي‬ ‫الهيدروجين‬ ‫قطب‬  This is the reference all other oxidations are compared to. .‫الكسدة‬ ‫عمليات‬ ‫كل‬ ‫له‬ ‫نسبة‬ ‫تقاس‬ ‫الذى‬ ‫المرجع‬  Eº = 0  º indicates standard states of 25ºC, 1 atm, 1 M solutions.  ‫القياسية‬ ‫الظروف‬ ‫تعنى‬ º 55 Dr. Elhelece W. A.

Cell Potential ‫الخلية‬ ‫جهد‬ Zn(s) + Cu+2 (aq) Zn+2 (aq) + Cu(s)  The total cell potential is the sum of the potential at each electrode. .‫قطب‬ ‫كل‬ ‫عند‬ ‫الجهود‬ ‫مجموع‬ ‫هو‬ ‫الكلي‬ ‫الخلية‬ ‫جهد‬ Eºcell = EºZn→ Zn +2 + EºCu +2 →Cu  We can look up reduction potentials in a table. .‫للعناصر‬ ‫التختزال‬ ‫لجهود‬ ‫جداول‬ ‫يوجد‬  One of the reactions must be reversed, so change its sign. .‫اشارته‬ ‫تعكس‬ ‫ولهذا‬ ‫يعكس‬ ‫ان‬ ‫يجب‬ ‫التفاعلت‬ ‫احد‬ 56 Dr. Elhelece W. A.

 Determine the cell potential for a galvanic cell based on the redox reaction: :‫التفاعل‬ ‫على‬ ‫تعتمد‬ ‫جلفانية‬ ‫لخلية‬ ‫الجهد‬ ‫احسب‬ Cu(s) + Fe+3 (aq)→Cu+2 (aq) + Fe+2 (aq)  Fe+3 (aq)+ e- → Fe+2 (aq) Eº = 0.77 V  Cu+2 (aq)+2e- →Cu(s) Eº = 0.34 V ‫الحل‬  Cu(s) →Cu+2 (aq)+2e- Eº = -0.34 V  2Fe+3 (aq)+ 2e- →2Fe+2 (aq) Eº = 0.77 V 57 Dr. Elhelece W. A. Cell Potential ‫الخلية‬ ‫جهد‬

Reduction potential ‫التختزال‬ ‫جهد‬  More negative Eº ‫بالسالب‬ ‫العلى‬ ‫القيمة‬  more easily electron is added ‫سهولة‬ ‫اكثر‬ ‫الكترونات‬ ‫اضافة‬  More easily reduced ‫يختزل‬ ‫سهولة‬ ‫اكثر‬  Better oxidizing agent ‫قوى‬ ‫مؤكسد‬ ‫عامل‬  More positive Eº ‫بالموجب‬ ‫العلى‬ ‫القيمة‬  more easily electron is lost ‫سهولة‬ ‫اكثر‬ ‫اللكترونات‬ ‫فقد‬  More easily oxidized ‫يتأكسد‬ ‫سهولة‬ ‫اكثر‬  Better reducing agent ‫قوى‬ ‫مختزل‬ ‫عامل‬ 58 Dr. Elhelece W. A.

Line Notation ‫الخطى‬ ‫التعبير‬ solid ‫׀‬ Aqueous ‖ Aqueous ‫׀‬solid Anode on the left ‖ Cathode on the right ‫اليمين‬ ‫على‬ ‫والكاثود‬ ‫المزدوج‬ ‫الخط‬ ‫يسار‬ ‫على‬ ‫النود‬  Single line different phases. .(‫العنصر‬ ‫)نفس‬ ‫مختلفة‬ ‫حالت‬ ‫يعنى‬ ‫مفرد‬ ‫تخط‬  Double line porous disk or salt bridge. . ‫قرص‬ ‫او‬ ‫ملحية‬ ‫قنطرة‬ ‫تعنى‬ ‫تخطان‬  If all the substances on one side are aqueous, a platinum electrode is indicated. .‫مستخدم‬ ‫البلتين‬ ‫قطب‬ ‫ان‬ ‫يعنى‬ ‫سائلة‬ ‫جهة‬ ‫فى‬ ‫المواد‬ ‫كل‬ ‫كانت‬ ‫اذا‬ 59 Dr. Elhelece W. A.

Cu2+ Fe+2 For the last reaction ‫السابق‬ ‫للتفاعل‬ Cu(s)‫׀‬Cu+2 (aq) ‖ Fe+2 (aq),Fe+3 (aq)‫׀‬Pt(s) 60 Dr. Elhelece W. A. Cu+2 Line Notation ‫الخطى‬ ‫التعبير‬

Under standard conditions, which of the following is the net reaction that occurs in the cell? ‫الخلية؟‬ ‫فى‬ ‫يحدث‬ ‫التالية‬ ‫الكلية‬ ‫التفاعلت‬ ‫من‬ ‫اى‬ ‫القياسية‬ ‫الظروف‬ ‫تحت‬ Cd|Cd2+ || Cu2+ |Cu a. Cu2+ + Cd → Cu + Cd2+ b. Cu + Cd → Cu2+ + Cd2+ c. Cu2+ + Cd2+ → Cu + Cd d. Cu + Cd 2+ → Cd + Cu2+ 61 Dr. Elhelece W. A. Line Notation ‫الخطى‬ ‫التعبير‬

Potential, Work and ΔG ‫النثالبى‬ ‫فى‬ ‫والتغير‬ ‫والشغل‬ ‫الجهد‬  EMF= potential (V) = work (J) / Charge(C)  E = work done by system / charge ‫جهد‬ ‫يساوى‬ ‫الشحنات‬ ‫على‬ ‫مقسوما‬ ‫النظام‬ ‫من‬ ‫المبذول‬ ‫الشغل‬ .‫الخلية‬ E = -w / q  Charge is measured in coulombs. ‫بالكولوم‬ ‫تقاس‬ ‫الشحنة‬ -w = q E  Faraday = 96,485 C/mol e-  q = nF = zF= moles of e- x charge/mole e- w = -qE = -nFE = zFE = ΔG 62 Dr. Elhelece W. A.

ΔGº = -nFEº  if Eº > 0, then ΔGº < 0 spontaneous  if Eº< 0, then ΔGº > 0 nonspontaneous  Calculate ΔGº for the following reaction: :‫للتفاعل‬ ‫النثالبي‬ ‫فى‬ ‫التغير‬ ‫احسب‬ Cu+2 (aq)+ Fe(s) →Cu(s)+ Fe+2 (aq)  Fe+2 (aq)+ 2e- → Fe(s) Eº = 0.44 V  Cu+2 (aq)+2e- → Cu(s) Eº = 0.34 V 63 Dr. Elhelece W. A. Potential, Work and ΔG ‫النثالبى‬ ‫فى‬ ‫والتغير‬ ‫والشغل‬ ‫الجهد‬

Dr. Elhelece W. A. 64 Standard electrode potentials E/V F2(g) + 2 e -  2 F - (aq) + 2.87 MnO4 2- (aq) + 4 H + (aq) + 2 e -  MnO2(s) + 2 H2O(l) + 1.55 MnO4 - (aq) + 8 H + (aq) + 5 e -  Mn 2+ (aq) + 4 H2O(l) + 1.51 Cl2(g) + 2 e -  2 Cl - (aq) + 1.36 Cr2O7 2- (aq) + 14 H + (aq) + 6 e -  2 Cr 3+ (aq) + 7 H2O(l) + 1.33 Br2(g) + 2 e -  2 Br - (aq) + 1.09 Ag + (aq) + e -  Ag(s) + 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) + 0.77 MnO4 - (aq) + e -  MnO4 2- (aq) + 0.56 I2(g) + 2 e -  2 I - (aq) + 0.54 Cu 2+ (aq) + 2 e -  Cu(s) + 0.34 Hg2Cl2(aq) + 2 e -  2 Hg(l) + 2 CI - (aq) + 0.27 AgCl(s) + e-  Ag(s) + Cl- (aq) + 0.22 2 H + (aq) + 2 e -  H2(g) 0.00 Pb 2+ (aq) + 2 e -  Pb(s) - 0.13 Sn 2+ (aq) + 2 e -  Sn(s) - 0.14 V 3+ (aq) + e -  V 2+ (aq) - 0.26 Ni 2+ (aq) + 2 e -  Ni(s) - 0.25 Fe 2+ (aq) + 2 e -  Fe(s) - 0.44 Zn 2+ (aq) + 2 e -  Zn(s) - 0.76 Al 3+ (aq) + 3 e -  Al(s) - 1.66 Mg 2+ (aq) + 2 e -  Mg(s) - 2.36 Na + (aq) + e -  Na(s) - 2.71 Ca 2+ (aq) + 2 e -  Ca(s) - 2.87 K + (aq) + e -  K(s) - 2.93 Increasing reducing power Increasing oxidising power

molesxelectronsofmolesofNo eofmolesgainOofmolesSince 4.041.0. 2 Omol1.0molg/32Og.23 22 1 2 -1 2 == = − C600,38)eC/mol005(96mol)4.0(Fzq - === Dr. Elhelece W. A. 65 Example ‫مثال‬ If 3.2 g of O2 were reduced in the overall reaction with HS− : how many coulombs have been transferred from HS− to O2 or how many charges pass in the circuit? ‫الخلية؟‬ ‫عبر‬ ‫تنقل‬ ‫التى‬ ‫الشحنات‬ ‫عدد‬ ‫كم‬ OHe2H2O 222 1 →++ −+ −+++→− e2HSHS OHSHOHS 222 1 +→++ +− Solution ‫الحل‬

How much work can be done if 2.4 mmol (z) of electrons go through a potential difference of 0.70 V (E) in the ocean-floor battery? ‫الخلية؟‬ ‫فى‬ ‫جهد‬ ‫فرق‬ ‫خلل‬ ‫الشحنة‬ ‫من‬ ‫مقار‬ ‫لنقل‬ ‫المبذول‬ ‫الشغل‬ ‫مقدار‬ Example ‫مثال‬ Solution ‫الحل‬ C2103.2)C/mol96500()mol3104.2(F.zq ×=−×== J101.6C)103.2)(V70.0(q.EWorkElectrical 22 ×=×== • The greater the difference in potential (V), the stronger the e− will be pushed around the circuit. .‫قويا‬ ‫الخلية‬ ‫فى‬ ‫اللكترونات‬ ‫دفع‬ ‫كان‬ ,‫كبيرا‬ ‫الجهد‬ ‫فى‬ ‫الفرق‬ ‫كان‬ ‫كلما‬ • 12V battery “pushes” electrons through a circuit eight times harder than a 1.5V battery. ‫جهدها‬ ‫خلية‬12‫الدارة‬ ‫فى‬ ‫اللكترونات‬ ‫تدفع‬ ‫فولت‬8‫جهدها‬ ‫خلية‬ ‫أضعاف‬1.5.‫فولت‬ Dr. Elhelece W. A. 66

Example (E from free energy change) ‫مثال‬‫الحرة‬ ‫الطاقة‬ ‫فى‬ ‫التغير‬ ‫من‬ ‫الجهد‬ ‫حساب‬ Calculate the voltage that will be measured by the potentiometer in the figure, Knowing that the free energy change (ΔG) for the net reaction is −150 kJ/mol of Cd. )(2)(2)()(2)(:Re 2)()(: )(2)(22)(2:Re 2 2 aqClsAgaqCdsAgClsCdactionNet eaqCdsCdOxidation aqClsAgesAgClduction −+ −+ −− ++→+ +→ +→+ Solution )C/J(V777.0 ) emol C 96500() Cdmol emol 2( Cdmol J310150 Fn G E EFnG += − − ×− −= ∆ −= −=∆ A spontaneous chemical reaction of negative ∆G produces a positive voltage. .‫موجب‬ ‫الخلية‬ ‫جهد‬ ‫تكون‬ ‫سالب‬ ‫الحرارية‬ ‫الطاقة‬ ‫فى‬ ‫التغير‬ ‫قيمة‬ ‫التلقائي‬ ‫التفاعل‬Dr. Elhelece W. A. 67

Reaction Quotient ‫التفاعل‬ ‫حاصل‬ the reaction quotient (Q) is equal to the equilibrium constant (k). .‫التزان‬ ‫ظروف‬ ‫غير‬ ‫فى‬ ‫التزان‬ ‫لثابت‬ ‫مساويا‬ ‫يكون‬ ‫التفاعل‬ ‫حاصل‬ We can write the reaction quotient Q for any half reaction in terms of the activities of the species: .‫التفكك‬ ‫درجة‬ ‫بدللة‬ ‫تفاعل‬ ‫نصف‬ ‫ل ي‬ ‫التفاعل‬ ‫حاصل‬ ‫كتابة‬ ‫يمكننا‬ Note: electrons do not appear in the reaction quotient. .‫التفاعل‬ ‫حاصل‬ ‫فى‬ ‫اللكترونات‬ ‫تظهر‬ ‫ل‬ :‫لحظ‬ (a=1 for pure solids and liquids so they do not appear). = ‫الكفاءة‬1.‫النقية‬ ‫والسوائل‬ ‫الصلبة‬ ‫للمواد‬ 68 Dr. Elhelece W. A.

Dr. Elhelece W. A.69 Calculating the Reaction Quotient, Q ‫التفاعل‬ ‫حاصل‬ ‫حساب‬ Q can be calculated for any set of conditions, not just for equilibrium. .‫التزان‬ ‫عند‬ ‫فقط‬ ‫ليس‬ ‫ظروف‬ ‫اى‬ ‫عند‬ ‫التفاعل‬ ‫حاصل‬ ‫حساب‬ ‫يمكن‬ Q can be used to determine which direction a reaction will shift to reach equilibrium. .‫التزان‬ ‫الى‬ ‫التفاعل‬ ‫يزيح‬ ‫ان‬ ‫يمكن‬ ‫اتجاه‬ ‫اى‬ ‫فى‬ ‫يوضح‬ ‫ان‬ ‫يمكن‬ ‫التفاعل‬ ‫حاصل‬ If K > Q, a reaction will proceed forward, converting reactants into products. ‫قيمة‬ ‫كانت‬ ‫اذا‬K‫قيمة‬ ‫من‬ ‫أكبر‬Q.‫نواتج‬ ‫الى‬ ‫المتفاعلت‬ ‫تحويل‬ ,‫الطردى‬ ‫التجاه‬ ‫فى‬ ‫يسير‬ ‫التفاعل‬ If K < Q, the reaction will proceed in the reverse direction, converting products into reactants. ‫قيمة‬ ‫كانت‬ ‫اذا‬Q‫قيمة‬ ‫من‬ ‫أكبر‬K.‫متفاعلت‬ ‫الى‬ ‫النواتج‬ ‫تحويل‬ ,‫العكسي‬ ‫التجاه‬ ‫فى‬ ‫يسير‬ ‫التفاعل‬ If Q = K then the system is already at equilibrium. ‫قيمة‬ ‫كانت‬ ‫اذا‬Q‫قيمة‬ ‫تساوى‬K.‫التزان‬ ‫عند‬ ‫يكون‬ ‫التفاعل‬ Reaction Quotient ‫التفاعل‬ ‫حاصل‬

Dr. Elhelece W. A. 70 In order to determine Q we need to know: :‫نعرف‬ ‫ان‬ ‫يجب‬ ‫التفاعل‬ ‫حاصل‬ ‫قيمة‬ ‫لحساب‬ the equation for the reaction, including the physical states, ,‫الفزيائية‬ ‫الحالة‬ ‫متضمنة‬ ‫التفاعل‬ ‫معادلة‬ the quantities of each species (molarities and/or pressures), .(‫الضغوط‬ ‫او‬ ‫)المولرية‬ ‫المختلفة‬ ‫الجزاء‬ ‫كميات‬ all measured at the same moment in time. .‫الزمن‬ ‫من‬ ‫اللحظة‬ ‫نفس‬ ‫عند‬ ‫كلها‬ ‫مقاسة‬ To calculate Q: ‫لحساب‬Q: 1. Write the expression for the reaction quotient. .‫التفاعل‬ ‫حاصل‬ ‫لحساب‬ (‫)الصيغة‬ ‫القانون‬ ‫اكتب‬ 2. Find the molar concentrations or partial pressures of each species involved. .‫التفاعل‬ ‫فى‬ ‫جزء‬ ‫لكل‬ ‫الجزيئية‬ ‫الضغوط‬ ‫او‬ ‫المولر ي‬ ‫التركيز‬ ‫اوجد‬ 3. Subsitute values into the expression and solve. .‫وحل‬ ‫الصيغة‬ ‫فى‬ ‫القيم‬ ‫ادتخل‬ Reaction Quotient ‫التفاعل‬ ‫حاصل‬

Dr. Elhelece W. A. 71 Example: 0.035 moles of SO2 , 0.500 moles of SO2 Cl2 , and 0.080 moles of Cl2 are combined in an evacuated 5.00 L flask and heated to 100o C. What is Q before the reaction begins? Which direction will the reaction proceed in order to establish equilibrium? SO2 Cl2 (g) SO2 (g) + Cl2 (g) Kc = 0.078 at 100o C •Write the expression to find the reaction quotient, Q. Since Kc is given, the amounts must be expressed as moles per liter (molarity). The amounts are in moles so a conversion is required. 0.500 mole SO2Cl2/5.00 L = 0.100 M SO2Cl2 0.035 mole SO2/5.00 L = 0.070 M SO2 0.080 mole Cl2/5.00 L = 0.016 M Cl2 Reaction Quotient ‫التفاعل‬ ‫حاصل‬

Dr. Elhelece W. A.72 Substitute the values in to the expression and solve for Q. Compare the answer to the value for the equilibrium constant and predict the shift. 0.078 (K) > 0.011 (Q) Since K >Q, the reaction will proceed in the forward direction in order to increase the concentrations of both SO2 and Cl2 and decrease that of SO2Cl2 until Q = K. Reaction Quotient ‫التفاعل‬ ‫حاصل‬

Nernst Equation ‫نيرنست‬ ‫معادلة‬ Take the expression for the Gibbs dependence on activity and turn this around for an expression in terms of the cell potential. .‫الخلية‬ ‫جهد‬ ‫فى‬ ‫التفاعل‬ ‫حاصل‬ ‫بدللة‬ ‫جبس‬ ‫دالة‬ ‫عن‬ ‫التعبير‬ The relation between cell potential E and free energy gives ‫الحرة‬ ‫والطاقة‬ ‫الخلية‬ ‫جهد‬ ‫بين‬ ‫العلقة‬ 73 Dr. Elhelece W. A. Rearrange and obtain the Nernst Equation ‫نيرنست‬ ‫معادلة‬ ‫على‬ ‫نحصل‬ ‫الترتيب‬ ‫اعادة‬ ∆G = ∆Go + RTlnQ −nFE = −nFEo + RTlnQ

The equation is sometimes streamlined by restricting discussion to T = 25 °C and inserting the values for the constants, R and F. ‫الحرارة‬ ‫درجة‬ ‫عندة‬ ‫المناقشة‬ ‫بقصر‬ ‫تشرط‬ ‫احيانا‬ ‫المعادلة‬25.‫فارادى‬ ‫وثابت‬ ‫العام‬ ‫الثابت‬ ‫قيم‬ ‫وحساب‬ ‫درجة‬ Note the difference between using natural logarithms and base10 logarithms. ‫بدللة‬ ‫واللوغارتم‬ ‫الطبيعى‬ ‫اللوغارتم‬ ‫بين‬ ‫الفرق‬ ‫لحظ‬10. Be aware of the significance of “n” – the number of moles of electrons transferred in the process according to the stoichiometry chosen. ‫لمدلول‬ ‫حذر‬ ‫على‬ ‫كن‬n.‫المختارة‬ ‫الكمية‬ ‫للحسابات‬ ‫نتيجة‬ ‫العملية‬ ‫فى‬ ‫المنتقلة‬ ‫اللكترونات‬ ‫مولت‬ ‫عدد‬ E =Eo − 0.0257 n ln Q E =Eo − 0.0592 n log Q 74 Dr. Elhelece W. A. Nernst Equation ‫نيرنست‬ ‫معادلة‬

Standard Reduction Potentials ‫القياسية‬ ‫التختزال‬ ‫جهود‬ 19.3 Standard reduction potential (E0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. ‫بتركيز‬ ‫المحاليل‬ ‫كل‬ ‫تكون‬ ‫عندما‬ ‫قطب‬ ‫عند‬ ‫للختزال‬ ‫المصاحب‬ ‫الجهد‬ ‫فرق‬ ‫هو‬ :‫القياسي‬ ‫الختزال‬ ‫جهد‬1‫وكل‬ ‫مولر‬ ‫الضغوط‬1.‫جو‬ ‫ضغط‬ E0 = 0 V Standard hydrogen electrode (SHE) ‫القياسي‬ ‫الهيدروجين‬ ‫قطب‬ 2e- + 2H+ (1 M) H2 (1 atm) Reduction Reaction ‫الختزال‬ ‫تفاعل‬ 75 Dr. Elhelece W. A.

What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? .‫والكروم‬ ‫الكادميوم‬ ‫من‬ ‫مكونة‬ ‫كهروكيميائية‬ ‫لخلية‬ ‫القياسية‬ ‫الدافعة‬ ‫القوة‬ ‫مقدار‬ ‫هو‬ ‫ما‬ Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V 2e- + Cd2+ (1 M) → Cd (s) Cr (s) → Cr3+ (1 M) + 3e- Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd2+ (1 M) → 3Cd (s) + 2Cr3+ (1 M) E0 = Ecathode - Eanodecell 0 0 E0 = -0.40 – (-0.74)cell E0 = 0.34 Vcell 76 Dr. Elhelece W. A.

Dr. Elhelece W. A. 77  The Nernst Equation relates the potential of the half cell to the activities of the chemical species (their concentrations). .(‫)التراكيز‬ (‫)التفكك‬ ‫النشاط‬ ‫ومعامل‬ ‫اللية‬ ‫نصف‬ ‫جهد‬ ‫بين‬ ‫علقة‬ ‫نيرنست‬ ‫معادلة‬  For the reaction (written as reduction) ‫اتختزال‬ ‫هيئة‬ ‫على‬ ‫)مكتوب‬ ‫للتفاعل‬ ‫)بالنسبة‬ ] a Aa b Ba [ln nF RT EE −°= BbenAa - ↔+ a b B a n EE A a log V05916.0 −= ° n is the no. of electrons in either the electrode or cell reaction. ‫التفاعل‬ ‫او‬ ‫القطب‬ ‫فى‬ ‫اللكترونات‬ ‫عدد‬ ‫.الخلية‬ We usually calculate half-reactions at 25º C, substituting that in with the gas constant and to base 10 log gives: ‫عند‬ ‫التفاعل‬ ‫نصف‬ ‫نحسب‬ ‫عادةا‬25‫لساس‬ ‫اللوغارتم‬ ‫واستخدام‬ ‫درجة‬10.‫الغازات‬ ‫وثابت‬ Where a is the activities of species A and B. Nernst Equation ‫نيرنست‬ ‫معادلة‬

Dr. Elhelece W. A. 78  Write the Nernst equation for the reduction of O2 to water: :‫لماء‬ ‫الكسجين‬ ‫لتختزال‬ ‫نيرنست‬ ‫معادلة‬ ‫اكتب‬  Note that: asolid =1, agas = pressure aH2O =1, aion = molarity Example (Nernst Equation) ‫نيرنست‬ ‫)معادلة‬ ‫)مثال‬ 2 ]H[log 2 05916.0 23.1E + += 2 ]H[ 1 log 2 05916.0 23.1E + −= V1.23EOH2e2HO 2 - 22 1 =°↔++ + 2 O 2 ][HP ]OH[ log 2 05916.0 23.1E 2 1 2 + −= ]H[log05916.023.1E + += V1.23EOH24e4HO 2 - 2 =°↔++ + 4 O 2 2 ][HP ]OH[ log 4 05916.0 23.1E 2 + −= ][Hlog05916.023.1E + += Note that multiplying the reaction by any factor does not affect either Eº or the calculated E. ‫قيمة‬ ‫على‬ ‫يؤثر‬ ‫ل‬ ‫عامل‬ ‫باى‬ ‫المعادلة‬ ‫ضرب‬ :‫ان‬ ‫لحظ‬Eº ‫او‬ ‫القياسية‬E.‫المحسوبة‬

Dr. Elhelece W. A. 79  Find the voltage for the Ag-Cd cell and state if the reaction is spontaneous if the right cell contained 0.50 M AgNO3(aq) and the left contained 0.010M Cd(NO3)2(aq). V799.0E)s(AgeAg =↔+ ° + −+ V402.0E)s(Cde22Cd −=° −↔−++ V781.0 0.50 1 log 1 05916.0 799.0E =−=+ V461.0 0.010 1 log 2 05916.0 402.0E −=−−=− V242.1)461.0(781.0EEEcell +=−−=−= −+ For Ag/Ag+ electrode For Cd/Cd2+ electrode For the Cell

)s(Ag2e2Ag2 ↔+ −+ −+ +↔ e2CdCd 2 )s(Ag2CdAg2)s(Cd 2 +↔+ ++  Note that you will get the same value of cell potential if you apply the Nernst equation directly to the overall cell reaction. .‫القيمة‬ ‫نفس‬ ‫يعطى‬ ‫مباشرة‬ ‫نيرنست‬ ‫معادلة‬ ‫تطبيق‬ ‫تخلل‬ ‫من‬ ‫الخلية‬ ‫جهد‬ ‫بحساب‬ ‫ان‬ ‫نلحظ‬ V242.1 ]5.0[ ]01.0[ log 2 05916.0 201.1E ]5.0[ ]01.0[ log 2 05916.0 ))402.0(799.0(E ]Ag[ ]Cd[ log 2 05916.0 )EE(E ]Ag][Cd[ ]Ag][Cd[ log 2 05916.0 EE 2cell 2cell 2 2 Cd/CdAg/Agcell 2 22 cellcell 2 = −= −−−= −−= −= + + + + ++ oo o Since Ecell is found to be +ve quantity, the reaction as written is spontaneous. .‫تلقائي‬ ‫التفاعل‬ ‫وبالتالى‬ ,‫موجبة‬ ‫الخلية‬ ‫جهد‬ ‫قيمة‬ Dr. Elhelece W. A. 80

Example (Ecell) ‫مثال‬‫الخلية‬ ‫)جهد‬ ) A cell was prepared by dipping a Cu wire and a saturated calomel electrode (ESCE = 0.241 V) into 0.10 M CuSO4 solution (E°Cu/Cu2+ = 0.339 V). The Cu wire was attached to the positive terminal and the calomel to the negative terminal of the potentiometer. ‫بالنهاية‬ ‫والزئبق‬ ‫للجهاز‬ ‫الموجبة‬ ‫بالنهايية‬ ‫وصيل‬ ‫النحاس‬ ,‫القياسيي‬ ‫الزئبيق‬ ‫وقطيب‬ ‫نحاس‬ ‫سيلك‬ ‫بغميس‬ ‫كونيت‬ ‫تخليية‬ .‫السالبة‬ 1- Write the half-cell reaction of Cu electrode. .‫النحاس‬ ‫لقطب‬ ‫التفاعل‬ ‫نصف‬ ‫اكتب‬ 2- Write the Nernst equation for the Cu electrode. .‫النحاس‬ ‫لقطب‬ ‫نيرنست‬ ‫معادلة‬ ‫اكتب‬ 3- Calculate the cell voltage. .‫الخلية‬ ‫جهد‬ ‫احسب‬ 4- What would happen if the [Cu2+ ] were 4.864x10-4 M? ‫النحاس‬ ‫تركيز‬ ‫كان‬ ‫اذا‬ ‫يحدث‬ ‫الذى‬ ‫ما‬4.864x10-4 .‫مولر‬ Dr. Elhelece W. A. 81

Dr. Elhelece W. A. 82 Solution ‫الحل‬ Electrode connected to the positive terminal of the potentiometer is the cathode and the other is the anode. .‫النود‬ ‫هو‬ ‫والتخر‬ ‫الكاثود‬ ‫هو‬ ‫للبتنشيومتر‬ ‫الموجبة‬ ‫بالنهاية‬ ‫الموصول‬ ‫القطب‬ 1. Cu2+ (aq) + 2e- → Cu(s) ]Cu[log 2 05916.0 EE 20 + += V309.0)10.0(log 2 05916.0 339.0E =+= 3. Ecell = ECu/Cu2+ − ESCE = 0.309 V − 0.241 V = 0.068 V 2.

Dr. Elhelece W. A. 83 Relationships Among ∆Go , K and Eo Cell ‫وثابت‬ ‫الحرة‬ ‫الطاقية‬ ‫بين‬ ‫العلقية‬ ‫الخلية‬ ‫وجهد‬ ‫التزان‬  Now we can relate Eo cell to the equilibrium constant (K) of a redox reaction. .‫واتختزال‬ ‫اكسدة‬ ‫لتفاعل‬ ‫التزان‬ ‫وثابت‬ ‫الخلية‬ ‫جهد‬ ‫بين‬ ‫العلقة‬ ‫ربط‬ ‫يمكن‬ ‫الن‬  We saw that the standard free-energy change ∆G° for a reaction is related to its equilibrium constant as follows: :‫كالتالي‬ ‫التزان‬ ‫بثابت‬ ‫مرتبط‬ ‫لتفاعل‬ ‫القياسية‬ ‫الحرة‬ ‫الطاقة‬ ‫ان‬ ‫اتضح‬ ‫كما‬ ∆Go = - RTln K

Dr. Elhelece W. A. 84  Therefore, if we combine Equations:  ∆Go =- n F Eo cell  ∆Go = - RTln K  we obtain  - nFEo cell = - RTln K  Solving for Eo cell Eo cell = (RT/nF) lnK  When T = 298 K

Dr. Elhelece W. A.85  The equation Eo cell = (RT/nF) lnK  by substituting for R, T and F values  Eo cell={(8.314J/K.mol)(298K)}/  {n (96,500J/V.mol) ln K}  or Eo cell = (0.0257 V/ n) ln K  Alternatively, This equation can be written using the base- 10 logarithm of K:  Eo cell = (0.0592 V/n) log K

Spontaneity of Redox Reactions ‫والتختزال‬ ‫الكسدة‬ ‫تفاعلت‬ ‫تلقائية‬ ∆G = -nFEcell ∆G0 = -nFEcell 0 n = number of moles of electrons in reaction F = 96,500 J V • mol = 96,500 C/mol ∆G0 = -RT ln K = -nFEcell 0 Ecell 0 = RT nF ln K (8.314 J/K•mol)(298 K) n (96,500 J/V•mol) ln K= = 0.0257 V n ln KEcell 0 = 0.0592 V n log KEcell 0 Dr. Elhelece W. A.86

2e- + Fe2+ Fe 2Ag 2Ag+ + 2e-Oxidation: Reduction: What is the equilibrium constant for the following reaction at 250 C? ‫عن‬ ‫التالي‬ ‫للتفاعل‬ ‫التزان‬ ‫ثابت‬ ‫ماقيمة‬25‫درجة؟‬ Fe2+ (aq) + 2Ag (s) → Fe (s) + 2Ag+ (aq) = 0.0257 V n ln KEcell 0 19.4 E0 = -0.44 – (0.80) E0 = -1.24 V 0.0257 V x nE0 cellexpK = n = 2 0.0257 V x 2-1.24 V = exp K = 1.23 x 10-42 E0 = EFe /Fe – EAg /Ag 0 0 2+ + Dr. Elhelece W. A.87

Will the following reaction occur spontaneously at 250 C if [Fe2+ ] = 0.60 M and [Cd2+ ] = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq) 2e- + Fe2+ 2Fe Cd Cd2+ + 2e-Oxidation: Reduction: n = 2 E0 = -0.44 – (-0.40) E0 = -0.04 V E0 = EFe /Fe – ECd /Cd 0 0 2+ 2 + - 0.0257 V n ln QE 0E = - 0.0257 V 2 ln-0.04 VE = 0.010 0.60 E = 0.013 E > 0 Spontaneous 19.5 Dr. Elhelece W. A.88

Dr. Elhelece W. A.89  Example ‫مثال‬  Calculate the equilibrium constant for the following reaction at 25°C: ‫التالي‬ ‫للتفاعل‬ ‫التزان‬ ‫ثابت‬ ‫احسب‬  Sn(s) + 2Cu2+ (aq) ↔ Sn2+ (aq) + 2Cu+ (aq)  Eo Cu2+/Cu+ = 0.15 V, Eo Sn2+/Sn = - 0.14V  Strategy : The relationship between the equilibrium constant K and the standard emf is given by equation : .‫المعادلة‬ ‫من‬ ‫الكهربية‬ ‫الدافعة‬ ‫والقوة‬ ‫التزان‬ ‫ثابت‬ ‫بين‬ ‫العلقة‬ :‫الستراجية‬ Eo cell = (0.257 V/n) ln K

Dr. Elhelece W. A.90  Solution: ‫الحل‬  The half-cell reactions are ‫هما‬ ‫الخلية‬ ‫نصفى‬  Anode (oxidation): Sn(s) → Sn2+ (aq) + 2e ‫)الكسدة‬ ‫)النود‬  Cathode (reduction): ‫)التختزال‬ ‫)الكاثود‬  2Cu2+ (aq) + 2e → 2Cu+ (aq)  Eo cell = Eo cathode - Eo anode  Eo cell = Eo Cu2+/Cu+ - Eo Sn2+/Sn  Eo cell = 0.15 V - (-0.14V)

Dr. Elhelece W. A.91  Eo cell = 0.29 V  Equation Eo cell = (0.0257 V/ n) ln K  can be written :  In K = nE°/ 0.0257 V  In the overall reaction we find n = 2. Therefore,  ln K = {(2)(0.29V)} / 0.0257 V = 22.6  K = e22.6 = 7x 109

Dr. Elhelece W. A.92  Example ‫مثال‬  Calculate the standard free-energy change for the following reaction at 25°C: ‫التالي‬ ‫للتفاعل‬ ‫القياسية‬ ‫الحرة‬ ‫الطاقة‬ ‫احسب‬  2Au(s) +3Ca2+ (1.0M) →2Au3+ (l.0M) + 3Ca(s)  Eo Ca2+/Ca = - 2.87 V, Eo Au3+/Au = 1.5V

Dr. Elhelece W. A.93  Solution : ‫الحل‬  The half cell reactions are ‫هما‬ ‫التفاعل‬ ‫نصفي‬  Anode (oxidation): 2Au(s) → 2Au3+ (1M) + 6e ‫)اكسدة‬ ‫)النود‬  Cathode(reduction): 3Ca2+ (1M) + 6e →3Ca(s) ‫)اتختزال‬ ‫)الكاثود‬ Eo cell = Eo cathode - Eo anode  Eo cell = Eo Ca2+/Ca - Eo Au3+/Au  Eo cell = - 2.87 V - 1.5V  Eo cell = - 4.37 V  Now we use the Equation :  ∆Go = - nFE°

Dr. Elhelece W. A.94  The overall reaction shows that n = 6, so ‫ان‬ ‫توضح‬ ‫المعادلية‬ = ‫اللكترونات‬ ‫عدد‬6  ∆Go = - (6)(96,500 J/V .mol)( -4.37 V)  ∆Go = 2.53 X l06 J/mol  ∆Go = 2.53 X I03 kJ/mol (nonspontaneous). ‫لتلقائي‬  Check:The large positive value of ∆G° tells us that the reaction favors the reactants at equilibrium. The result is comparable with the fact that E° for the cell is negative.

Dr. Elhelece W. A.95 ‫القياسي‬ ‫القطب‬Reference electrode The role of the R.E. is to provide a fixed potential which does not vary during the experiment. .‫التجربة‬ ‫تخلل‬ ‫ليتغير‬ ‫محدد‬ ‫بجهد‬ ‫يمد‬ ‫ان‬ ‫هو‬ ‫القياسي‬ ‫القطب‬ ‫دور‬ A good R.E. should be able to maintain a constant potential even if a few microamps are passed through its surface. ‫التيار‬ ‫لو‬ ‫حتى‬ ‫ثابت‬ ‫جهد‬ ‫على‬ ‫البقاء‬ ‫على‬ ‫القادر‬ ‫هو‬ ‫الجيد‬ ‫القياسي‬ ‫القطب‬ .‫جدا‬ ‫ضعيف‬

Dr. Elhelece W. A.96 The electrolyte solution ‫اللكتروليتي‬ ‫المحلول‬  it consists of solvent and a high concentration of an ionized salt and electroactive species .‫كهربيا‬ ‫نشطة‬ ‫وأجزاء‬ ‫عالى‬ ‫بتركيز‬ ‫ماين‬ ‫وملح‬ ‫مذيب‬ ‫من‬ ‫يتكون‬  to increase the conductivity of the solution, to reduce the resistance between: :‫بين‬ ‫المقاومة‬ ‫نقلل‬ ,‫المحلول‬ ‫توصيلية‬ ‫نزيد‬ ‫لكى‬  W.E. and C.E. (to help maintain a uniform current and potential distribution). ‫جهد‬ ‫وفرق‬ ‫موحد‬ ‫تيار‬ ‫على‬ ‫)للحصول‬ ‫حالي‬ ‫وقطب‬ ‫عامل‬ ‫)قطب‬  and between W.E. and R.E. to minimize the potential error due to the uncompensated solution resistance iRu. ‫قياسي‬ ‫وقطب‬ ‫عامل‬ ‫قطب‬ ‫بين‬ ‫المحسوبة‬ ‫غير‬ ‫المحلول‬ ‫ومقاومة‬ ‫الجهدى‬ ‫الخطأ‬ ‫نقلل‬ ‫.لكى‬

Dr. Elhelece W. A.97 Faraday’s law If W grams of the substance is deposited by Q coulombs of electricity, then W Qα But Q = it, Hence W i tα or W = Z it I = current in amperes t = time in seconds. Z = constant of proportionality (electrochemical equivalent.)

Dr. Elhelece W. A.98 Faraday’s Law E By definition Z 96500 = I . t . E W 96500 = E=Equivalent mass of the substance 1 Faraday=96500 coulomb Na e Na E 23g 1F 23g + + → = 3Al 3e Al E 9g 27g3F + + → = 2Cu 2e Cu E 31.75g 2F 63.5g + + → =

Dr. Elhelece W. A.99 Example How many atoms of calcium will be deposited from a solution of CaCl2 by a current of 25 milliamperes flowing for 60 s? Number of Faraday of electricity passed 3 25 10 60 96500 − × × = 3 25 10 60 96500 2 − × × = × 3 2325 10 60 6.023 10 96500 2 − × × = × × × = 4.68 × 1018 atoms of calcium. Solution: moles of Ca atoms atoms of Ca

2-Oxidants such as nitrite and chromate which function by shifting the surface potential of the metal in the positive direction until the passive zone. 3.A reagent that is adsorbed on the metal surface, diminishing either metal dissolution or the reduction of H2O/O2/H+ . In either of these cases, corrosion is reduced. Substances that inhibit metallic dissolution are organic and include aromatic and aliphatic amines, sulphur compounds, and those containing carbonyl groups; the release of hydrogen is inhibited by compounds containing phosphorus, arsenic, and antimony. 100 Dr. Elhelece W. A.

Preventing Corrosion Coating to keep out air and water. Galvanizing - Putting on a zinc coat Has a lower reduction potential, so it is more easily oxidized. Alloying with metals that form oxide coats. Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead. 101 Dr. Elhelece W. A.

Voltammetry Electrochemistry techniques based on current (i) measurement as function of voltage (Eappl) Working electrode (microelectrode) place where redox occurs surface area few mm2 to limit current flow Reference electrode constant potential reference (SCE) Counter electrode inert material (Hg, Pt) plays no part in redox but completes circuit Supporting electrolyte alkali metal salt does not react with electrodes but has conductivity 102 Dr. Elhelece W. A.

-Define the Following: a- Oxidation and reduction in terms of electron transfer. Ans. O xidatio n – re m o valo f e le ctro ns fro m a spe cie s. Re ductio n – additio n o f e le ctro ns to a spe cie s b- Reference electrode. Ans. Re fe re nce e le ctro de s, as the ir nam e sug g e sts, are use d to g ive a value o f po te ntialto which o the r po te ntials can be re fe rre d in te rm s o f a po te ntialdiffe re nce po te ntials can o nly be re g iste re d as diffe re nce s with re spe ct to a cho se n 103 Dr. Elhelece W. A.

1-A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 solution and a Ag electrode in a 1.0 M AgNO3 solution. E°Ag+/Ag = 0.80V , Eo Mg 2+ /Mg= - 2.37 V Write the Redox and Overall reactions The standard reduction potentials are: Ag+ (1.0M) + e → Ag(s) E° = 0.80V Mg2+ (1.0M) + 2e → Mg(s) Eo = - 2.37 V Applying the diagonal rule, we see that Ag+ will oxidize Mg: Anode(oxidation) :Mg(s) Mg→ 2+ (1.0M) +2e 104 Dr. Elhelece W. A.

Cathode (reduction): 2Ag+ (l.0M)+2e → 2Ag(s) 0.8V E°Ag+/Ag = 0.80V Overall reaction Mg(s)+2Ag+ (1.0M) → Mg2+ (1.0M) + 2Ag(s) b- Calculate the standard emf of this cell at 25o C? Ans. Eo cell = Eo cathode - Eo anode Eo cell = Eo Ag+/Ag - Eo Mg2+/Mg Eo cell = 0.80 V - (-2.37V) Eo cell = 3.17 V Che ck : The po sitive value o f Eo sho ws that the fo rward re actio n is favo re d 105 Dr. Elhelece W. A.

Electrochemical Cells 106 Dr. Elhelece W. A.

Electrochemical cells are Batteries 107 Dr. Elhelece W. A.

Alkaline Batteries KOH 108 Dr. Elhelece W. A.

Car Batteries Pb-Acid H2SO4 109 Dr. Elhelece W. A.

Mitsubishi iMiEV - Pure Electric Car Powered by a 330 v Li-Ion Rechargeable battery Plugs into your house and takes 14 hours to charge -100 km for $ 0.60 $ 50,000 Can $ 36,000 US Top Speed 130 km/h 63 hp and 133 lb.-ft. of torque 110 Dr. Elhelece W. A.

Cell Phone batteries Lithium Ion Rechargeable battery 111 Dr. Elhelece W. A.

Lithium Coin Cell 112 Dr. Elhelece W. A.

Space Ship Batteries Powered by Radioisotopes 113 Dr. Elhelece W. A.

Ni-Metal Hydride 114 Dr. Elhelece W. A.

Notes on Electrochemical Cells An electrochemical cell – a system of electrodes, electrolytes, and salt bridge that allow oxidation and reduction reactions to occur and electrons to flow through an external circuit. The salt bridge allows ions to migrate from one half- cell to the other without allowing the solutions to mix. 1. Spontaneous redox reaction 2. Produces electricity from chemicals 3. Is commonly called a battery 115 Dr. Elhelece W. A.

Analyzing Electrochemical Cells The reaction that is higher on the reduction chart is the reduction and the lower is oxidation and is written in reverse. 116 Dr. Elhelece W. A.

Electrochemical cells Yes write the following down in your notes... Electrochemical Cells There are two types of Electrochemical cells 1(Primary (disposable( 2(Secondary (rechargeable( In secondary cells two reactions can occur, one discharges the cell and another occurs when the cell is recharged 117 Dr. Elhelece W. A.

Primary cells(write in notes( In a primary cell, chemical reactions use up some of the materials in the cell as electrons flow from the cell When the materials have been used up the cell is said to be discharged and can not be recharged There are two basic types of primary cells The primary wet cell and... The primary dry cell 118 Dr. Elhelece W. A.

Primary wet cell (Do not take notes( The wet cell, also known as a voltaic cell, was invented in 1800 by Volta 119 Dr. Elhelece W. A.

Voltaic cell The voltaic cell is called a wet cell because it is made of two pieces of metal (e.g. magnesium and copper) that are placed in a liquid (e.g. hydrochloric acid( The metal pieces are called electrodes, while the liquid is called an electrolyte The magnesium electrode reacts with the acid, and the energy released separates electrons from the magnesium atoms. These electrons collect on the magnesium electrode (negative terminal( 120 Dr. Elhelece W. A.

Voltaic cell At the same time, positive charges collect on the copper plate (the positive terminal( Current only flows when connected to a circuit Disadvantages: Danger of spilling electrolyte(acid( Continual need to replace zinc plate and acid(consumed( 121 Dr. Elhelece W. A.

The wet cell(voltaic cell( The wet cell(voltaic cell( Consists of two metal electrodes (magnesium and copper) placed in a solution known as an electrolyte (usually an acid, e.g. HCl( The magnesium (Mg) reacts with the acid releasing energy that separates the electrons from the magnesium atoms (the negative terminal( Positive charges build up on the copper (the positive terminal( 122 Dr. Elhelece W. A.

The wet cell(voltaic cell( The chemical reaction an

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