dr.wael elhelece electrochemistry 331chem
DESCRIPTION
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.TRANSCRIPT
Electrochemistry
الكهربية الكيمياء
331chem.
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Contents المحتويات
Potentiometric Measurements. . الكهربية الجهود وتقدير قياسات
Electrochemical Reaction and Nernest Equation . نيرنست ومعادلة الكهروكيميائية التفاعالت
Reference Electrodes and standard Potential . القياسى والجهد القياسية األقطاب
Thermodynamics and Electrochemical Reactions . الكهروكيميائية والتفاعالت الحرارية الديناميكا
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Contents المحتويات
Diffusion and Electrochemical Reactions. الكهروكيميائية والتفاعالت االنتشار
Voltammetery and Mechanism of Electrode Reaction
عند التفاعالت وميكانيكية الجهود قياس عملياتاالقطاب.
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
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Electrochemical Reactionsالكهروكيميائية التفاعالت
Electrochemical reactions
الكهروكيميائية التفاعالت
Is a reaction in which electrons are transferred from one species to
another.
. آلخر عنصر من االلكترونات انتقال فيه يتم تفاعل هو
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Electrochemical Cellsالكهروكيميائية الخاليا
two electrodes (electronic conductors).) االلكترونات ) انتقال األقطاب من زوج
electrolyte (ionic conductor).) األيونات ) انتقال الكتروليتى محلول
Electrode and its electrolyte are places in an electrode compartment
حاوي إناء فى يوضعا االلكتروليتى والمحلول القطبلهما.
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Electrochemical Cellsالكهروكيميائية الخاليا
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Salt bridge الملحية القنطرة tube containing a concentrated electrolyte solution .
( حرف شكل على أنبوبة عن على( Uعبارة تحتوىمركز ملحى محلول
alternative way of joining two different electrode compartments.
المختلفة األقطاب من زوج لربط تقليدية .طريقة
Electrochemical Cellsالكهروكيميائية الخاليا
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Electrochemical Cellsالكهروكيميائية الخاليا
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Reference electrodes األقطاب القياسىة
Are used to give a value of potential to which other
potentials can be referred.
المختلفة لألقطاب الجهود قيمة وقياس لتعيين .تستخدم
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Standard Hydrogen Electrode
ال ق ال هطب ياسيقيدروجين2H+(aq) + 2e– H2(g)
a
H2
H+
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The “standard” aspect to this cell is that the activity of H2(g) and that of
H+(aq) are both 1.
ايونات وكذا الهيدروجين غاز فاعلية ان يعنى الخلية لهذه قياسي مصطلح
. واحد الهيدروجين
This means that the pressure of H2 is 1 atm and the concentration of H+ is
1M.
الهيدروجين ايونات وتركيز جو ضغط واحد الهيدروجين غاز ضغط يعنى هذا
. موالر واحد
These are our standard reference states.
. المرجعية القياسية حاالتنا هى هذه
Standard Hydrogen Electrode
ال ق ال هطب ياسيقيدروجين
Voltaic (Galvanic) Cell الجلفانية الخاليا
Electrolytic Cell الكهربي التحليل خاليا
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Two types of electrochemical cells
الكهروكيميائية الخاليا من نوعان
Voltaic (Galvanic) Cells
Reactions are spontaneous . تلقائية الخاليا من النوع هذا فى التفاعالت
Redox reactions produce electrical energy طاقة تنتج واالختزال االكسدة تفاعالت
كهربية. Lets look at an example:
المثال سبيل وعلىCu+2 (aq) +Zn (s) -------> Cu (s) + Zn+2 (aq)
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Daniell Cell دانيال خلية
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CuSO4 (aq)ZnSO4 (aq)
Cu metalZn metalsalt bridge
Zn(s) Zn2+(aq) + 2e– Cu2+(aq) + 2e– Cu(s)
a
Cathode (reduction)+ive
Anode (oxidation)–ive االنود
( أكسدة(اختز( الكاثود
ال)
Electrolytic Cell الخليةالتحليلية
Reactions are non spontaneous.
. تلقائية غير التفاعالت
Redox reactions require electrical energy to occur.
. لتحدث كهربية طاقة الى تحتاج واالختزال االكسدة تفاعالت
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Cell Types الخاليا أنواع
Galvanic cell: produces electricity as a result of spontaneous
reactions
الجلفانية .الخاليا تلقائى: لتفاعل نتيجة كهربا تنتج
Electrolytic cell: a non-sponteneous reaction is driven by an
external source
الكهربى التحليل تيار: خاليا دفع تحت يحدث التلقائى تفاعل
كهربى.
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Reduction and oxidation processes occurring in a cell are separated in
space (anode and cathode compartments)
. مفصولة وتكون خلية فى تحدث التى واالختزال االكسدة عمليات
Anode (oxidation); is the electrode at which oxidation is occurring
االنود :. اكسدة عنده يحدث الذى القطب
Cathode (reduction); is the electrode at which reduction is occurring
الكاثود :. اختزال عنده يحدث الذى القطب
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Reactions at electrodes تحدث التى التفاعالتاألقطاب عند
Oxidation and Reduction واالختزال األكسدة
What is reduced is the oxidizing agent. مؤكسد عامل هو يختزل الذي
H+ oxidizes Zn by taking electrons from it.
What is oxidized is the reducing agent. مختز عامل هو يتأكسد لالذي
Zn reduces H+ by giving it electrons.
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Electro Motive Force (EMF) الدافعة القوةالكهربية
Water only spontaneously flows one way in a waterfall.
فى يسري تلقائيا فقط الماء. الهبوط اتجاه
Likewise, electrons only spontaneously flow one way in a redox reaction from higher to lower potential energy.
تلقائيا االلكترونات بالمثلاألعلى من االتجاه في تسري
األقل إلى
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Electro Motive Force (EMF) الدافعة القوةالكهربية
The potential difference between the anode and cathode in a cell is called:
يسمى خلية في والكاثود االنود بين الجهد فرق
Electro Motive Force (EMF) الكهربية الدافعة القوة
It is also called the cell potential, and is designated Ecell.
أيضا الخلية يسمى بالرمز (جهد له ).Ecellويرمز
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Potentiometric Measurements القياساتالجهدية Potentiometer الجهد قياس جهاز
A device for measuring the potential of an electrochemical cell without drawing a current or altering the cell’s composition.
تركيب في الدخول أو بالتيار إمداد دون الخلية جهد لقياس جهازالخلية.
Potentiometry الجهد قياس عمليةUse of Electrodes to Measure Voltages that Provide Chemical
Information.
. كيميائى مصدر من المأخوذة الجهود لقياس األقطاب استخدام
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Potentiometer الجهد قياس جهاز23
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Potentiometric measurements قياس عمليات الجهود
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.
فى الفرق لتعين البوتنشيمتر بواسطة تتم الجهد قياس عملية.( ) ( مرجعى ( ثابت وقطب كاشف عامل قطب بين الجهد
- Cathode is the working/indicator electrode. (right half-cell)
-.( األيمن ( الخلية نصف العامل القطب هو الكاثود- Anode is the counter/reference electrode. (left half-
cell)-( االيسر ( الخلية نصف المرجعى القطب هو .اآلنود
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Ecell = Ec ─ Ea
Where :
Ec is the reduction potential at the cathode.
Ec . الكاثود عند االختزال جهد
Ea is the reduction potential at the anode
Ea . االنود عند االختزال جهد
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Electrochemical reactions التفاعالتالكهروكيميائية
Electrochemical reactions are made up of two “half-reactions”.
. تفاعل نصفى من تتكون
-At one electrode electrons are lost (oxidation).
) لاللكترونات فقد األقطاب أحد عند ).أكسدةيحدث
-At the other electrode, electrons are gained (reduction).
) لاللكترونات اكتساب يحدث االخر القطب ).اختزالعند
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Electrochemical Half Reactions لتفاعل التفاعل نصفكهروكيميائى
Electrochemical half reactions: لتفاعل التفاعل نصف
كهروكيميائى
Oxidation اكسدة
Zn: Zn (s) Zn⇒ 2+(aq) + 2e-
Reduction اختزال
Cu2+: Cu2+(aq) + 2e- Cu (s)⇒
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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 االلكترونات تحذف ان يجب التالى الحظ
Electrochemical Half Reactions لتفاعل التفاعل نصفكهروكيميائى
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Oxidation number التأكسد عددThe charge the atom would have in a molecule )or an ionic compound( if
electrons were completely transferred to the more electronegative atom.
االلكترونات وانتقلت حدث اذا االيونى المركب فى الذرة تحملها التى الشحنة
. كهربية سالبية االكثر للذرة كامل بشكل
1.Oxidation number equals ionic charge for monoatomic ions in ionic
compound.
. ايونى مركب فى الذرة احادى اليون االيونية 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 التاكسد( Aايون .1عدد موجب ويكون
Li+, Li = +1; Mg+2, Mg = +2Dr. Elhelece W. A.
3. The oxidation number of a transition metal ion is positive, but can vary
in magnitude.
. القيمة فى يختلف 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.
ان اما التاكسد حاالت من كبير عدد لها يكون ان يمكن االنتقالية غير العناصر
. سالب او موجوب يكون
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
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Oxidation number التأكسد عدد
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6. The oxidation number of fluorine is always –1.
دائما - ايون الفلورين تاكسد .1عدد
(unless fluorine is in elemental form, F2).
يكون الفلور جزيء فى الفلور عنصر كان ان .0اال
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.
على الشحنة مساويا يكون االيون او الجزيء فى االكسدة أعداد مجموع
. االيون او الجزيء هذا
IF; F= -1; I = +1
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Oxidation number التأكسد عدد
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Oxidation number التأكسد عدد
8. The oxidation number of hydrogen is +1 except when it is bonded to
metals in binary compounds.
الهيدروجين + تاكسد .1عدد ثنائي ملح فى بفلز مرتبط يكون ان عدا
In these cases, its oxidation number is –1 or when it’s in elemental form
(H2; oxidation # =0).
تاكسده - عدد فان الحاالت هذه يكون 1فى العنصرية حالته فى .0او
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 O22-
it is –1, in elemental form (O2 or O3) it is 0. االكسجين - تأكسد يكون -2عدد الهيدروجين اكسيد فوق مركب وفى
يكون 1 العنصرية حالته .0وفى
Oxidation numbers of all the atoms in HCO3- ?
HCO3-
O = -2 H = +13x(-2) + 1 + C = -1
C = +4
4.4
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Oxidation number التأكسد عدد
NaIO3
Na = +1 O = -23x(-2) + 1 + ? = 0
I = +5
IF7
F =-17x(-1) + ? = 0
I = +7
K2Cr2O7
O = -2 K = +17x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
Oxidation numbers of all the elements in the following ?المركبات؟ فى للعناصر التأكسد أعداد
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Oxidation number التأكسد عدد
Determination of Oxidizing and Reducing Agentsالمختزلة والعوامل المؤكسدة العوامل تعين
I. Determine oxidation # for all atoms in both the reactants and products.
والنواتج المتفاعالت كل فى الذرات لكل األكسدة أعداد تعين
I. Look at same atom in reactants and products and see if oxidation # increased or decreased.
ونحسب والنواتج المتفاعالت فى المماثلة الذرات عن نبحث. االكسدة عدد فى الفرق
If oxidation # decreased; substance reduced
. اختزلت المادة فان التأكسد عد قل اذا If oxidation # increased; substance oxidized
. تأكسدت المادة فان التأكسد عد ازداد اذا
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Oxidizing Agent: Substance that oxidizes the other substance by accepting electrons. It is reduced in reaction.
: بكسب االخرين تأكسد التى المادة المؤكسد العامل.( التفاعل ( فى اختزال لها يحدث الكترونات
Reducing Agent: Substance that reduces the other substance by donating electrons. It is oxidized in reaction.
: االخرين تختزل التى المادة المختزل العامل.( التفاعل ( فى تتأكسد الكترونات باعطاهم
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Determination of Oxidizing and Reducing Agentsالمختزلة والعوامل المؤكسدة العوامل تعين
Balancing Redox Equations األكسدة معادالت وزنواإلختزال
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
فى كرومات الثانى ايون بواسطة الثالثي الحديد الى الثنائي الحديد أكسدةحامضي؟ وسط
1. Write the unbalanced equation for the reaction ion ionic form.
. اآليونى الشكل فى للتفاعل الموزونة غير المعادلة اكتب
Fe2+ + Cr2O72- → Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
. تفاعل نصفي الى المعادلة افصل+2 +3
Oxidation: Fe2+ → Fe3+ اكسدة
+6 +3Reduction: Cr2O7
2- Cr3+ اختزال
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Balancing Redox Equations األكسدة معادالت وزنواإلختزال
3. Balance the atoms other than O and H in each half-reaction.
. نصف كل فى والهيدروجين االكسجين غير االخرى الذرات زن
Cr2O72- → 2Cr3+
4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
. والهيدروجين االكسجين لوزن ماء نضيف الحامض فى للتفاعالت بالنسبة
Cr2O72- → 2Cr3+ + 7H2O
14H+ + Cr2O72- → 2Cr3+ + 7H2O
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Balancing Redox Equations األكسدة معادالت وزنواإلختزال
5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
. التفاعل نصفي فى الشحنات لتعادل المعادلة فى لجهة الكترونات اضف
Fe2+ → Fe3+ + 1e-
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
6. Equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
. منهما كل معامالت فى بالضرب التفاعل نصفي فى االلكترونات عدد عادل
6Fe2+ → 6Fe3+ + 6e-
6e- + 14H+ + Cr2O72- → 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+ + Cr2O72- → 2Cr3+ + 7H2O
6Fe2+ → 6Fe3+ + 6e-Oxidation:
Reduction:
14H+ + Cr2O72- + 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.
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Balancing Redox Equations األكسدة معادالت وزنواإلختزال
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 .
عملية: مثال تقدم لكى موزونة أيونية معادلة اكتب
محلول في البرمنجات ايون بواسطة اليود أكسدة
. الرباعي والمنجنيز الجزيئي اليود لينتج قلوي
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Balancing Redox Equations األكسدة معادالت وزنواإلختزال
Step 1 : Write the unbalanced equation is:
:1الخطوة كالتالي الموزونة غير المعادلة اكتب
MnO4- + I- → MnO2 + I2
Step 2: The two half-reactions are:
:2الخطوة هما التفاعل نصفى -1 0 Oxidation; I- → I2 األكسدة
+7 +4 Reduction: MnO4
- → MnO2 االختزال
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Step 3: We balance each half-reaction for number and type of atoms and charges.
.3الخطوة والشحنة: نوع لكل الذرات بأعداد تفاعل نصف كل نوزن Oxidation half reaction: We first balance the I atoms;
: الذرات عدد أوال األكسدة التفاعل 2I- → I2 نصف
To balance charges, we add two electrons to the right-hand side of the equation: ان يجب الشحنات نزن لكىالتفاعل من ايمنى للجهة الكترون زوج نضيف
2 I- → I2 + 2e-
Reduction half-reaction : To balance the O atoms, we add two H2O molecules on the right:
: نضيف االوكسجين ذرات نزن كى االختزال التفاعل :2نصف اليمنى للجهة ماء جزيء
MnO4- → MnO2 + 2H2O
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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:
4خطوة : الكلى التفاعل على للحصول التفاعل نصفى نجمع: التفاعل نصفى فى االلكترونات عدد مكافئة بعد وذلك
(1) 3( 2I- → I2 + 2e- )
(2) 2(MnO4- + 4H+ + 3e- → MnO2 + 2H2O)
__________________________________
6I- + 2MnO4- + 8H+ + 6e → 3I2 + 2MnO2 + 4H2O + 6e
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Step 5: A final check shows that the equation is balanced in terms of both
atoms and charges.
5خطوة. والشحنات: الذرات 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 االيون بطريقة حامضي وسط فى للتفاعل التالية العادلة زن عملى تمرين
Fe2+ + MnO4- → Fe3+ + Mn2+
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Balancing Redox Equations األكسدة معادالت وزنواإلختزال
Applications تطبيقات
Moving electrons is electric current.. كهربى تيار االلكترونات 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
عند ولكن مفيد شغل اى دون االلكترونات حركة تتم المحلول نفس فى فصلهم
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Connected this way the reaction starts يبدأ التفاعل بالشكل كما التوصيل
Stops immediately because charge builds up.. االلكترونات تزاحم بسبب فورا يتوقف
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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.
الملحية القنطرة. التيار بمرور تسمح
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Fe+2 H+
MnO4-
e-
Electricity travels in a complete circuit.. مغلقة دارة في تنتقل الكهرباء
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Galvanic Cell الجلفانية الخلية
50
Dr. Elhelece W. A.
H+
MnO4-
Fe+2
Porous Disk
اسطوانة منفذة
Instead of a salt bridge القنطرة من بدالالملحية
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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.. االلكترونات يجذب المؤكسد العامل
Reducing agent pushes the electron. . االلكترونات يدفع المختزل العامل
The push or pull (“driving force”) is called the cell potential Ecell.
( الخلية ( جهد تسمى المحركة القوة االلكترونات وجذب .Ecell دفع
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 من كولوم لكل الشغل من جول
الشحنات 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 الهيدروجين قطبالقياسي
This is the reference all other oxidations are compared to.. األكسدة عمليات كل له نسبة تقاس الذى المرجع
Eº = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions. القياسية الظروف º تعنى
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 االختزال لجهود جداول يوجد
One of the reactions must be reversed, so change its sign.. اشارته تعكس ولهذا يعكس ان يجب التفاعالت 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 احسبCu(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 االختزال جهد
More negative Eº بالسالب االعلى 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 االعلى القيمة more easily electron is lost سهولة اكثر االلكترونات 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 ׀solidAnode on the left ‖ Cathode on the right
اليمين على والكاثود المزدوج الخط يسار على االنود Single line different phases.
.( العنصر ( نفس مختلفة حاالت يعنى مفرد خط Double line porous disk or salt bridge.
. قرص او ملحية قنطرة تعنى خطان If all the substances on one side are aqueous, a
platinum electrode is indicated. قطب ان يعنى سائلة جهة فى المواد كل كانت اذا
. مستخدم البالتين
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?
يحدث التالية الكلية التفاعالت من اى القياسية الظروف تحتالخلية؟ فى
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 فى والتغير والشغل الجهداالنثالبى
EMF= potential (V) = work (J) / Charge(C)E = work done by system / charge
الشحنات على مقسوما النظام من المبذول الشغل. الخلية جهد يساوى
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 االنثالبي فى التغير احسبCu+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 فى والتغير والشغل الجهداالنثالبى
Dr. Elhelece W. A.
64
Standard electrode potentials E/V
F2)g( + 2 e- 2 F-)aq( + 2.87
MnO42-)aq( + 4 H+)aq( + 2 e- MnO2)s( + 2 H2O)l( + 1.55
MnO4-)aq( + 8 H+)aq( + 5 e- Mn2+)aq( + 4 H2O)l( + 1.51
Cl2)g( + 2 e- 2 Cl-)aq( + 1.36
Cr2O72-)aq( + 14 H+)aq( + 6 e- 2 Cr3+)aq( + 7 H2O)l( + 1.33
Br2)g( + 2 e- 2 Br-)aq( + 1.09
Ag+)aq( + e- Ag)s( + 0.80
Fe3+)aq( + e- Fe2+)aq( + 0.77
MnO4-)aq( + e- MnO4
2-)aq( + 0.56
I2)g( + 2 e- 2 I-)aq( + 0.54
Cu2+)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
Pb2+)aq( + 2 e- Pb)s( - 0.13
Sn2+)aq( + 2 e- Sn)s( - 0.14
V3+)aq( + e- V2+)aq( - 0.26
Ni2+)aq( + 2 e- Ni)s( - 0.25
Fe2+)aq( + 2 e- Fe)s( - 0.44
Zn2+)aq( + 2 e- Zn)s( - 0.76
Al3+)aq( + 3 e- Al)s( - 1.66
Mg2+)aq( + 2 e- Mg)s( - 2.36
Na+)aq( + e- Na)s( - 2.71
Ca2+)aq( + 2 e- Ca)s( - 2.87
K+)aq( + e- K)s( - 2.93
Increasingreducing
power
Increasingoxidising
power
molesxelectronsofmolesofNo
eofmolesgainOofmolesSince
4.041.0.
2
O mol 1.0mol g /32O g .23
221
2-1
2
C600,38)e C/mol005(96 mol) 4.0(Fzq - Dr. Elhelece W. A.
65
Example مثالIf 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 2221
e2HSHS
OHSHOHS 2221
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?
فى جهد فرق خالل الشحنة من مقار لنقل المبذول الشغل مقدارالخلية؟
Example مثال
Solution الحل
C 2103.2)C/mol96500()mol 3104.2(F.zq
J 101.6C)103.2)(V70.0(q.E WorkElectrical 22
• The greater the difference in potential (V), the stronger the e will be pushed around the circuit.
. , قويا الخلية فى االلكترونات دفع كان كبيرا الجهد فى الفرق كان كلما
• 12V battery “pushes” electrons through a circuit eight times harder than a 1.5V battery.
جهدها الدارة 12خلية فى االلكترونات تدفع جهدها 8فولت خلية أضعاف1.5. .Dr. Elhelece W. Aفولت
66
Example (E from free energy change) الحرة مثال 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
C96500()
Cdmol
emol2(
Cdmol
J310150
Fn
GE
EFnG
A spontaneous chemical reaction of negative G produces a positive voltage.
. موجب الخلية جهد تكون سالب الحرارية الطاقة فى التغير قيمة التلقائي .Dr. Elhelece W. Aالتفاعل
67
Reaction Quotient التفاعل حاصل
the reaction quotient (Q) is equal to the equilibrium constant (k).
. االتزان ظروف غير فى االتزان لثابت مساويا يكون التفاعل حاصل
We can write the reaction quotient Q for any half reaction in terms of the
activities of the species:
درجة بداللة تفاعل نصف الي التفاعل حاصل كتابة يمكننا
التفكك.
Note: electrons do not appear in the reaction quotient.
. التفاعل: حاصل فى االلكترونات تظهر ال الحظ
(a=1 for pure solids and liquids so they do not appear).
.1الكفاءة = 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.
. االتزان 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.
. االتزان 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 قيمة K اذا من , Qأكبر المتفاعالت تحويل الطردى االتجاه فى يسير التفاعل. نواتج الى
If K < Q, the reaction will proceed in the reverse direction, converting products into reactants.
قيمة كانت قيمة Qاذا من , Kأكبر الى النواتج تحويل العكسي االتجاه فى يسير التفاعلمتفاعالت.
If Q = K then the system is already at equilibrium.
قيمة كانت قيمة Qاذا . Kتساوى االتزان 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 الموالرية المختلفة االجزاء كميات
all measured at the same moment in time.. الزمن من اللحظة نفس عند كلها مقاسة
To calculate Q::Qلحساب
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 الموالري التركيز اوجد
3. Subsitute values into the expression and solve.. وحل الصيغة فى القيم ادخل
Reaction Quotient التفاعل حاصل
Dr. Elhelece W. A.
71
Example: 0.035 moles of SO2, 0.500 moles of SO2Cl2, and 0.080 moles of Cl2 are
combined in an evacuated 5.00 L flask and heated to 100oC. What is Q before the reaction begins? Which direction will the reaction proceed in order to establish equilibrium?
SO2Cl2(g) SO2(g) + Cl2(g) Kc = 0.078 at 100oC•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 SO2Cl20.035 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 بداللة جبس دالة عن التعبير
The relation between cell potential E and free energy gives الخلية جهد بين العالقةالحرة والطاقة
73
Dr. Elhelece W. A.
Rearrange and obtain the Nernst Equation معادلة على نحصل الترتيب اعادةنيرنست
G G RT lnQ
n F E n F E RT lnQ
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 .25المعادلة فارادى وثابت العام الثابت قيم وحساب درجة
Note the difference between using natural logarithms and base10 logarithms.بداللة واللوغارتم الطبيعى اللوغارتم بين الفرق .10الحظ
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 للحسابات nكن نتيجة العملية فى المنتقلة االلكترونات موالت عدد. المختارة الكمية
E E 0.0257
nlnQ
E E 0.0592n
logQ
74
Dr. Elhelece W. A.
Nernst Equation نيرنست معادلة
Standard Reduction Potentials القياسية االختزال 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.
: المحاليل كل تكون عندما قطب عند لالختزال المصاحب الجهد فرق هو القياسي االختزال جهدالضغوط 1بتركيز وكل .1موالر جو ضغط
E0 = 0 V
Standard hydrogen electrode (SHE)القياسي الهيدروجين قطب
2e- + 2H+ (1 M) H2 (1 atm)
Reduction Reaction االختزال 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?
الكادميوم من مكونة كهروكيميائية لخلية القياسية الدافعة القوة مقدار هو ماوالكروم.
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 - Eanodecell0 0
E0 = -0.40 – (-0.74) cellE0 = 0.34 V cell
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 عالقة نيرنست معادلة
For the reaction (written as reduction) ( اختزال ( هيئة على مكتوب للتفاعل بالنسبة
]a
Aa
bBa
[lnnF
RTEE
BbenAa -
a
b
Ba
nEE
Aa
logV 05916.0
n is the no. of electrons in either the electrode or cell reaction. او القطب فى االلكترونات عددالخلية .التفاعل
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 الساس 25عادةا اللوغارتم واستخدام .10درجة 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 االكسجين الختزال نيرنست معادلة اكتب
Note that: asolid =1, agas = pressure aH2O =1, aion = molarity
Example (Nernst Equation) ( نيرنست ( معادلة مثال
2]H[log2
05916.023.1E
2]H[
1log
2
05916.023.1E
V1.23E OH2e 2H O 2-
221
2O
2
][H P
]OH[log
2
05916.023.1E
21
2
]H[log05916.023.1E
V1.23E OH24e 4H O 2-
2
4O
22
][H P
]OH[log
4
05916.023.1E
2
][Hlog05916.023.1E
Note that multiplying the reaction by any factor does not affect either Eº or the calculated E. : قيمة على يؤثر ال عامل باى المعادلة ضرب ان او Eºالحظ المحسوبة. Eالقياسية
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).
V 799.0E )s(AgeAg
V 402.0E )s(Cde22Cd
V 781.00.50
1log
1
05916.0799.0E
V461.00.010
1log
2
05916.0402.0E
V 242.1)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.
مباشرة نيرنست معادلة تطبيق خالل من الخلية جهد بحساب ان نالحظ. القيمة نفس يعطى
V242.1
]5.0[
]01.0[log
2
05916.0201.1E
]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
05916.0EE
2cell
2cell
2
2
Cd/CdAg/Agcell
2
22
cellcell
2
Since Ecell is found to be +ve quantity, the reaction as written is spontaneous.
. , تلقائي التفاعل وبالتالى موجبة الخلية جهد قيمةDr. Elhelece W. A.
80
Example (Ecell) مثال( الخلية ( 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 (ECu/Cu2+ = 0.339 V). The Cu wire was attached to
the positive terminal and the calomel to the negative terminal of the potentiometer.
, الموجبة بالنهاية وصل النحاس القياسي الزئبق وقطب نحاس سلك بغمس كونت خلية. السالبة بالنهاية والزئبق للجهاز
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 . 4.864x10-4ما موالر
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.
. االنود هو واالخر الكاثود هو للبتنشيومتر الموجبة بالنهاية الموصول القطب
1. Cu2+(aq) + 2e- Cu(s)
]Cu[log2
05916.0EE 20
V309.0)10.0(log2
05916.0339.0E
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 EoCell الحرة الطاقة بين العالقة
الخلية وجهد االتزان وثابت
Now we can relate Eocell to the equilibrium constant (K) of a redox
reaction.
اكسدة لتفاعل االتزان وثابت الخلية جهد بين العلقة ربط يمكن االن
واختزال.
We saw that the standard free-energy change G° for a reaction is
related to its equilibrium constant as follows:
: كالتالي االتزان 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 Eocell
Go= - RTln K
we obtain
- nFEocell = - RTln K
Solving for Eocell
Eocell = (RT/nF) lnK
When T = 298 K
Dr. Elhelece W. A.85
The equation Eocell = (RT/nF) lnK
by substituting for R, T and F values
Eocell={(8.314J/K.mol)(298K)}/
{n (96,500J/V.mol) ln K}
or Eocell = (0.0257 V/ n) ln K
Alternatively, This equation can be written using the base-
10 logarithm of K:
Eocell = (0.0592 V/n) log K
Spontaneity of Redox Reactions االكسدة تفاعالت تلقائيةواالختزال
G = -nFEcell
G0 = -nFEcell0
n = number of moles of electrons in reaction
F = 96,500J
V • mol = 96,500 C/mol
G0 = -RT ln K = -nFEcell0
Ecell0 =
RT
nFln K
(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)ln K=
=0.0257 V
nln KEcell
0
=0.0592 V
nlog 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 250C?
عن التالي للتفاعل االتزان ثابت درجة؟25ماقيمةFe2+ (aq) + 2Ag (s) → Fe (s) + 2Ag+ (aq)
=0.0257 V
nln 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 /Ag0 0
2+ +
Dr. Elhelece W. A.87
Will the following reaction occur spontaneously at 250C 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 /Cd0 0
2+ 2 +
-0.0257 V
nln QE 0E =
-0.0257 V
2ln -0.04 VE =
0.010
0.60
E = 0.013
E > 0 Spontaneous
19.5Dr. Elhelece W. A.88
Dr. Elhelece W. A.89
Example مثال
Calculate the equilibrium constant for the following reaction at
25°C: التالي للتفاعل االتزان ثابت احسب
Sn(s) + 2Cu2+(aq) ↔ Sn2+
(aq) + 2Cu+(aq)
EoCu2+/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 :
:من الكهربية الدافعة والقوة االتزان ثابت بين العالقة االستراجية
المعادلة.
Eocell = (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 االنود
( االكسدة(
Cathode (reduction): ( االختزال ( الكاثود
2Cu2+(aq) + 2e → 2Cu+(aq)
Eocell = Eo
cathode - Eoanode
Eocell = Eo
Cu2+/Cu+ - EoSn2+/Sn
Eocell = 0.15 V - (-0.14V)
Dr. Elhelece W. A.91
Eocell = 0.29 V
Equation Eocell = (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: للتفاعل القياسية الحرة الطاقة احسب
التالي
2Au(s) +3Ca2+(1.0M) →2Au3+(l.0M) + 3Ca(s)
EoCa2+/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 االنود
( اكسدة(
Cathode(reduction): 3Ca2+(1M) + 6e →3Ca(s) ( اختزال ( الكاثود
Eocell = Eo
cathode - Eoanode
Eocell = Eo
Ca2+/Ca - EoAu3+/Au
Eocell = - 2.87 V - 1.5V
Eocell = - 4.37 V
Now we use the Equation : Go = - nFE°
Dr. Elhelece W. A.94
The overall reaction shows that n = 6, so توضح المعادلة
االلكترونات = عدد 6ان
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). التلقائي
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 ReferenceالقطبelectrodeThe role of the R.E. is to provide a fixed potential which does not vary during the experiment.
خالل اليتغير محدد بجهد يمد ان هو القياسي القطب دورالتجربة.
A good R.E. should be able to maintain a constant potential even if a few microamps are passed through its surface.
ثابت جهد على االبقاء على القادر هو الجيد القياسي القطب. جدا ضعيف التيار لو حتى
Dr. Elhelece W. A.96
The electrolyte solution المحلولااللكتروليتي 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. لكى قياسي وقطب عامل قطب بينالمحسوبة غير المحلول ومقاومة الجهدى الخطأ .نقلل
Dr. Elhelece W. A.97
Faraday’s lawIf W grams of the substance is deposited by Q coulombs of electricity, then
W Q
But Q = it, Hence W i t
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 LawE
By definition Z96500
I . t . EW
96500
E=Equivalent mass of the substance
1 Faraday=96500 coulomb
Na e Na E 23g1F 23g
3Al 3e Al E 9g27g3F
2Cu 2e Cu E 31.75g2F 63.5g
Dr. Elhelece W. A.99
ExampleHow 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
325 10 6096500
325 10 6096500 2
32325 10 60
6.023 1096500 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.
VoltammetryElectrochemistry techniques based on current
(i) measurement as function of voltage (Eappl)Working electrode
(microelectrode )place where redox occurssurface 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. Oxidation – removal of electrons from a species .
Reduction – addition of electrons to a speciesb- Reference electrode . Ans. Reference electrodes, as their name
suggests, are used to give a value of potential to which other potentials can be referred in terms of
a potential difference potentials can only be registered as differences with respect to a chosen
reference value.
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
Mg2+
/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) → Mg2+(1.0M)
+2e EoMg2+/Mg= - 2.37 V
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 25oC ?
Ans. Eocell = Eo
cathode - Eoanode
Eocell = Eo
Ag+/Ag - EoMg2+/Mg
Eocell = 0.80 V - (-2.37V)
Eocell = 3.17 V
Check : The positive value of Eo shows that the forward reaction is favored
105
Dr. Elhelece W. A.
Electrochemical Cells
106 Dr. Elhelece W. A.
Electrochemical cells are Batteries
107 Dr. Elhelece W. A.
Alkaline Batteries
KOH108 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 torque110 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 cellsYes write the following down in your
notes...Electrochemical CellsThere are two types of Electrochemical
cells1)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 cellWhen 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 cellsThe 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 cellAt 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 and build up
) of electrons give the electronsتراكم)energy
For every electron that leaves the negative terminal of the circuit, another
electron must move out of the circuit and onto the positive terminal
Electrons release their energy (voltage) to the load
123
Dr. Elhelece W. A.
124
Dr. Elhelece W. A.
Dry Cells
The electrolyte in dry cells is not a liquid as in wet cells but rather a paste so dry
cells are not exactly dry
125
Dr. Elhelece W. A.
Secondary Cells Secondary cells are also known as
rechargeable batteries and have a second recharging chemical reaction in
addition to the discharging reaction
126
Dr. Elhelece W. A.
127
Dr. Elhelece W. A.
Notes on Electrolytic Cells
An electrolytic cell is a system of two inert (nonreactive) electrodes (C or Pt) and an electrolyte connected to a power supply. It has the following characteristics
1. Nonspontaneous redox reaction 2. Produces chemicals from electricity 3. Forces electrolysis to occur
128 Dr. Elhelece W. A.
When analyzing an electrolytic cell, your first and most important step is to determine the oxidation and reduction reactions.
Electrolytic Cell Main Rule The electrode that is connected to the -ve terminal of the power supply will gain electrons and therefore be the site of reduction.
129 Dr. Elhelece W. A.
Other Rules: For Electrochemical and Electrolytic Cells Oxidation always occurs at the anode and reduction at the cathodeElectrons flow through the wire and go from anode to cathodeAnions (- ions) migrate to the anode and cations (+ions) migrate towards the cathode.
130 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
131 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Draw a beaker, two inert electrodes wired to a power supply.
132 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Draw a beaker, two inert electrodes wired to a power supply.
Power Supply
DC - +
133 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Supply
DC - +
Label the electrode with Pt or C.
134 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Supply
DC - +
Pt Pt
Label the electrode with Pt or C.
135 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Supply
DC - +
PtPt
Add the electrolyte
136 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Supply
DC - +
PtPt
Add the electrolyte Molten or liquid means no water!
Na+
Br-
137 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Supply
DC - +
PtPt
Label the negative and positive electrodes
Na+
Br-
138 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Supply
DC - +
PtPt
Label the negative and positive electrodes
Na+
Br-
_ +
139 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The negative is reduction and the positive is oxidation.
Na+
Br-
_ +
140 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The negative is reduction and the positive is oxidation.
Na+
Br-
_reduction
+oxidation
141 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The anode is oxidation and the cathode is reduction.
Na+
Br-
_reductioncathode
+oxidationanode
142 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The anion migrates to the anode and the cation to the cathode.
Na+
Br-
_reductioncathode
+oxidationanode
143 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The anode reaction is the oxidation of the anion.
Na+
Br-
_reductioncathode
+oxidationanode
144 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The anode reaction is the oxidation of the anion.
Na+
Br-
_reductioncathode
+oxidationanode2Br- → Br2(g)+ 2e-
145 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The Cathode reaction is the reduction of the cation.
Na+
Br-
_reductioncathode
+oxidationanode2Br- → Br2(g)+ 2e-
146 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The Cathode reaction is the reduction of the cation.
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
+oxidationanode2Br- → Br2(g)+ 2e-
147 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
Gas Br2 is produced at the anode.
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
+oxidationanode2Br- → Br2(g)+ 2e-
148 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
Liquid Na is produced at the cathode.
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
+oxidationanode2Br- → Br2(g)+ 2e-
149 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The potential for each half reaction is calculated and the oxidation sign is reversed
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
+oxidationanode2Br- → Br2(g)+ 2e-
150 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The potential for each half reaction is listed and the oxidation sign is reversed
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
-2.71 v
+oxidationanode2Br- → Br2(g)+ 2e-
-1.09 v
151 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The overall redox reaction is written.
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
-2.71 v
+oxidationanode2Br- → Br2(g)+ 2e-
-1.09 v
152 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The overall redox reaction is written.
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
-2.71 v
+oxidationanode2Br- → Br2(g)+ 2e-
-1.09 v
2Na+ + 2Br- → Br2(g)+ 2Na(l) E0 = -3.80 v153 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
-2.71 v
+oxidationanode2Br- → Br2(g)+ 2e-
-1.09 v
2Na+ + 2Br- → Br2(g)+ 2Na(s) E0 = -3.80 v
The minimum theoretical voltage MTV required to force this nonspontaneous reaction to occur is the negative of the cell potential.
154 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
The minimum theoretical voltage MTV required to force this nonspontaneous reaction to occur is the negative of the cell potential.
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
-2.71 v
+oxidationanode2Br- → Br2(g)+ 2e-
-1.09 v
2Na+ + 2Br- → Br2(g)+ 2Na(s) E0 = -3.80 v MTV = +3.80 v
155 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
-2.71 v
+oxidationanode2Br- → Br2(g)+ 2e-
-1.09 v
2Na+ + 2Br- → Br2(g)+ 2Na(s) E0 = -3.80 v MTV = +3.80 v
Electrons flow through the wire from anode to cathode.
156 Dr. Elhelece W. A.
1. Draw and completely analyze a molten NaBr electrolytic cell.
Power Source - +
PtPt
Electrons flow through the wire from anode to cathode.
Na+
Br-
_reductioncathode2Na+ + 2e- → 2Na(l)
-2.71 v
+oxidationanode2Br- → Br2(g)+ 2e-
-1.09 v
2Na+ + 2Br- → Br2(g)+ 2Na(s) E0 = -3.80 v MTV = +3.80 v
e-
e-
157 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
158 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
159 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
Add the ions. (aq) or M
or solution means water.
160 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
Label the -, +, anode, cathode, oxidation, and reduction.
161 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
Label the -, +, anode, cathode, oxidation, and reduction.
-Cathodereduction
+Anodeoxidation
162 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
The cation and water migrate to the cathode
-Cathodereduction
+Anodeoxidation
163 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
The cation and water migrate to the cathode
-Cathodereduction
+Anodeoxidation
164 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
The cation or water reduces. The higher one on the chart is most spontaneous and occurs.
-Cathodereduction
+Anodeoxidation
165 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v
1/2O2 + 2H+(10-7M) + 2e- → H20 0.82 v
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
K+ + 1e- → K(s) -2.93 v166 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v
1/2O2 + 2H+(10-7M) → H20 0.32 v
Reduction of water
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
K+ + 1e- → K(s) -2.93 v167 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v
1/2O2 + 2H+(10-7M) → H20 0.32 vOxidation of water
Reduction of water
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
K+ + 1e- → K(s) -2.93 v168 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v
1/2O2 + 2H+(10-7M) → H20 0.32 vOxidation of water
Reduction of water
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
Reduction of K+
K+ + 1e- → K(s) -2.93 v169 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v
1/2O2 + 2H+(10-7M) → H20 0.32 vOxidation of water
strongest oxidizing agent or highestReduction of water select most spontaneous reaction
2H2O + 2e- → 2H2(g) + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
Reduction of KK+ + 1e- → K(s) -2.93 v
Overpotential Effect- treat water as if it were just below Zn
170 Dr. Elhelece W. A.
The overpotential effect is a higher than normal voltage required for the half reaction. This is often due to extra voltage required to produce a gas bubble in solution.
171 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
The cation or water reduces. The higher one on the chart is most spontaneous and occurs.
-CathodeReduction
+Anodeoxidation
172 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
The cation or water reduces. The higher one on the chart is most spontaneous and occurs.
-CathodeReduction2H2O+2e- → 2H2+ 2OH--0.41 v
+Anodeoxidation
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2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
The anion + water goes to the anode.
+Anodeoxidation
-CathodeReduction2H2O+2e- → 2H2+ 2OH--0.41 v
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2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
For oxidation the most spontaneous reaction is found on the redox chart and is lowest.
+Anodeoxidation
-CathodeReduction2H2O+2e- → 2H2+ 2OH--0.41 v
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Cl2 + 2e- → 2Cl- 1.36 v
1/2O2 + 2H+(10-7M) + 2e- → H20 0.82 vOxidation of water
I2(s) + 2e- → 2I- 0.54 v
Reduction of water
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
K+ + 1e- → K(s) -2.93 v176 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v
1/2O2(g) + 2H+(10-7M) → H20 0.82 vOxidation of water
I2(s) + 2e- → 2I- 0.54 v Oxidation of I-
Reduction of water
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
K+ + 1e- → K(s) -2.93 v177 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v overpotential effect means water is here
1/2O2 + 2H+(10-7M) → H20 0.82 vOxidation of water
I2(s) + 2e- → 2I- 0.54 v Oxidation of I-
Reduction of water
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
K+ + 1e- → K(s) -2.93 v178 Dr. Elhelece W. A.
Cl2 + 2e- → 2Cl- 1.36 v overpotential effect means water is here
1/2O2 + 2H+(10-7M) → H20 0.82 vOxidation of water
I2(s) + 2e- → 2I- 0.54 v Oxidation of I-
pick strongest reducing agent- lowerReduction of water
2H2O + 2e- → 2H2 + 2OH- -0.42 v
Zn2+ + 2e- → Zn(s) -0.76 v
K+ + 1e- → K(s) -2.93 v179 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
For oxidation the most spontaneous reaction is found on the redox chart and is lowest.
+AnodeOxidation
-CathodeReduction
2H2O+2e- → 2H2+ 2OH--0.41 v
180 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
For oxidation the most spontaneous reaction is found on the redox chart and is lowest.
+AnodeOxidation2I- → I2(s) + 2e-
-0.54 v
-CathodeReduction
2H2O+2e- → 2H2+ 2OH--0.41 v
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2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
Write the overall reaction with the cell potential.
+AnodeOxidation2I- → I2(s) + 2e-
-0.54 v
-CathodeReduction
2H2O +2e- → 2H2+ 2OH-
-0.41 v
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2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
Write the overall reaction with the cell potential.
+AnodeOxidation2I- → I2(s) + 2e-
-0.54 v
-CathodeReduction
2H2O+2e- → 2H2+ 2OH--0.41 v
2H2O+ 2I- → 2H2+ I2(s) + 2OH- E0 = -0.95 v
183 Dr. Elhelece W. A.
2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source - +
Pt Pt
K+
I-
H2O
Write the overall reaction with the cell potential.
+AnodeOxidation2I- → I2(s) + 2e-
-0.54 v
-CathodeReduction
2H2O+2e- → H2+ 2OH--0.41 v
2H2O+ 2I- → H2+ I2(s) + 2OH- E0 = -0.95 v MTV = +0.95v
e-
e-
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Electrolysis
Running a galvanic cell backwards.Put a voltage bigger than the potential
and reverse the direction of the redox reaction.
Used for electroplating.
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1.0 M
Zn+2
e- e-
Anode Cathode
1.10
Zn Cu1.0 M
Cu+2
186 Dr. Elhelece W. A.
1.0 M
Zn+2
e- e-
AnodeCathode
A battery >1.10V
Zn Cu1.0 M
Cu+2
187 Dr. Elhelece W. A.
Calculating plating
Have to count charge.Measure current I (in amperes)1 amp = 1 coulomb of charge per secondq = I x tq/nF = moles of metalMass of plated metalHow long must 5.00 amp current be
applied to produce 15.5 g of Ag from Ag+
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Dr. Elhelece W. A.
Calculating plating
.1Current x time = charge
.2Charge ∕Faraday = mole of e-
.3Mol of e- to mole of element or compound
.4Mole to grams of compoundOr the reverse if you want time to plate
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How long would it take to plate 5.00 g Fe from an aqueous solution of Fe(NO3)3 at
a current of 2.00 A ?
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Dr. Elhelece W. A.
Dr. Elhelece W. A.191
Corrosion
Not all spontaneous redox reaction are beneficial.Not all spontaneous redox reaction are beneficial.
Natural redox process that oxidizes metal to their oxides Natural redox process that oxidizes metal to their oxides and sulfides runs billions of dollars annually. Rust for and sulfides runs billions of dollars annually. Rust for example is not the direct product from reaction between example is not the direct product from reaction between iron and oxygen but arises through a complex iron and oxygen but arises through a complex electrochemical process.electrochemical process.
Rust: FeRust: Fe22OO3 3 • X H• X H22OO
Anode: Fe(s) Fe+2 + 2e- E° = 0.44 V
Cathode: O2 (g) + 4H+ + 4e- 2H2O (l) E° = 1.23 V
Net:Net: FeFe+2+2 will further oxidized to Fe will further oxidized to Fe22OO3 3 • X H• X H22OO
Corrosion
Rusting - spontaneous oxidation.Most structural metals have reduction
potentials that are less positive than O2.
Fe Fe+2 +2e- Eº= 0.44 V
O2 + 2H2O + 4e- 4OH-Eº= 0.40 V
Fe+2 + O2 + H2O Fe2O3 + H+
Reactions happens in two places.
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We have all seen corrosion and know that the process produces a new and
less desirable material from the original metal and can result in a loss of function
of the component or system. The corrosion product we see most
commonly is the rust which forms on the surface of steel and somehow
Steel → Rust
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TWO REACTIONS
For this to happen the major component of steel, iron (Fe) at the surface of a component undergoes a number of
simple changes. Firstly,
Fe → Fen+ + n electronsThe iron atom can lose some electrons
and become a positively charged ion. This allows it to bond to other groups of
atoms that are negatively charged.
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Dr. Elhelece W. A.
We know that wet steel rusts to give a variant of iron oxide so the other half of
the reaction must involve water (H2O) and oxygen (O2) something like this
O2 + 2H2O + 4e- → 4OH -
This makes sense as we have a negatively charged material that can combine with the iron and electrons,
2Fe + O2 + 2H2O → 2Fe(OH)2
Iron + Water with oxygen → Iron Hydroxide
dissolved in it
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THE NEXT STEP
Oxygen dissolves quite readily in water and because there is usually an excess
of it, reacts with the iron hydroxide.
4Fe(OH)2 + O2 → 2H2O + 2Fe2O3.H2O
Iron hydroxide + oxygen → water + Hydrated iron
oxide
( brown rust)
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Dr. Elhelece W. A.
THE PROCESS (Five facts)
This series of steps tells us a lot about the corrosion process.
( 1 )Ions are involved and need a medium to move in (usually water)
( 2 )Oxygen is involved and needs to be supplied
( 3 )The metal has to be willing ) استعداد) to علىgive up electrons to start the process
( 4 )A new material is formed and this may react again or could be protective of the original
metal
( 5 )A series of simple steps are involved and a driving force is needed to achieve them
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The most important fact is that interfering with the steps allows the corrosion reaction to be stopped or
slowed to a manageable ) .rate معقول)
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Water
Rust
Iron Dissolves-
Fe Fe+2
e-
Salt speeds up process by increasing conductivity
O2 + 2H2O +4e- 4OH-
Fe2+ + O2 + 2H2O Fe2O3 + 8 H+
Fe2+
199 Dr. Elhelece W. A.
Types of metallic corrosion
Uniform corrosion, occurs over the majority of the surface of a metal at a
steady and often predictable rate.Uniform corrosion can be slowed or
stopped by using the five basic facts;
(1 )Slow down or stop the movement of electrons
(a )Coat the surface with a non-conducting medium such as paint,
lacquer or oil
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)b (Reduce the conductivity of the solution in contact with the metal an extreme case being to keep it dry.
Wash away conductive pollutants regularly.)c (Apply a current to the material (cathodic
protection) .)2 (Slow down or stop oxygen from reaching the
surface. Difficult to do completely but coatings can help.
)3 (Prevent the metal from giving up electrons by using a more corrosion resistant metal higher in the
electrochemical series. Use a sacrificial coating which gives up its electrons more easily than the metal being
protected. Apply cathodic protection. Use inhibitors.
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Dr. Elhelece W. A.
(4 )Select a metal that forms an oxide that is protective and stops the reaction.
Localised corrosion 1 -GALVANIC CORROSION
2 -PITTING CORROSION, Pitting corrosion occurs in materials that have a protective film such as a
corrosion product or when a coating breaks down.
3-SELECTIVE ATTACK 4 -MICROBIAL CORROSION, This general
class covers the degradation of materials by bacteria or fungi or their by-products.
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Corrosion inhibitors
Corrosion inhibitors are organic or inorganic species added to the solution
in low concentration and that reduce the rate of corrosion. Inhibition can
function in three different ways: 1 .A reagent that promotes the appearance
of a precipitate on the metal surface, possibly catalysing the formation of a
passive layer, for example hydroxyl ion, phosphate,
carbonate, and silicate.
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