dr.wael elhelece electrochemistry 331chem

Electrochemistry

الكهربية الكيمياء

331chem.

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Dr. Elhelece W. A.

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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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


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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

Dr. Elhelece W. A.

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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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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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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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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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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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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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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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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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


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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H+

MnO4-

Fe+2

Porous Disk

اسطوانة منفذة

Instead of a salt bridge القنطرة من بدالالملحية

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Reducing Agent

Oxidizing Agent

e-

e-

e- e-

e-

e-

Anode Cathode



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



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.


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.


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


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.


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


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


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



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.


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


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


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


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+ +


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


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


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)


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


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


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°


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.


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


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


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.)


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


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.

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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.

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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

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-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.

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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

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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

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Electrochemical Cells


Electrochemical cells are Batteries


Alkaline Batteries

KOH108 Dr. Elhelece W. A.

Car Batteries

Pb-Acid H2SO4


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


Lithium Coin Cell


Space Ship Batteries

Powered by Radioisotopes


Ni-Metal Hydride


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


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.


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

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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

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Primary wet cell(Do not take notes)

The wet cell, also known as a voltaic cell, was invented in 1800 by Volta

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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)

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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)

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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)

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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

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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

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Secondary Cells Secondary cells are also known as

rechargeable batteries and have a second recharging chemical reaction in

addition to the discharging reaction

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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


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.


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.


1. Draw and completely analyze a molten NaBr electrolytic cell.



Draw a beaker, two inert electrodes wired to a power supply.



Draw a beaker, two inert electrodes wired to a power supply.

Power Supply

DC - +



Power Supply

DC - +

Label the electrode with Pt or C.



Power Supply

DC - +

Pt Pt

Label the electrode with Pt or C.



Power Supply

DC - +

PtPt

Add the electrolyte



Power Supply

DC - +

PtPt

Add the electrolyte Molten or liquid means no water!

Na+

Br-



Power Supply

DC - +

PtPt

Label the negative and positive electrodes

Na+

Br-



Power Supply

DC - +

PtPt

Label the negative and positive electrodes

Na+

Br-

_ +



Power Source - +

PtPt

The negative is reduction and the positive is oxidation.

Na+

Br-

_ +



Power Source - +

PtPt

The negative is reduction and the positive is oxidation.

Na+

Br-

_reduction

+oxidation



Power Source - +

PtPt

The anode is oxidation and the cathode is reduction.

Na+

Br-

_reductioncathode

+oxidationanode



Power Source - +

PtPt

The anion migrates to the anode and the cation to the cathode.

Na+

Br-

_reductioncathode

+oxidationanode



Power Source - +

PtPt

The anode reaction is the oxidation of the anion.

Na+

Br-

_reductioncathode

+oxidationanode



Power Source - +

PtPt

The anode reaction is the oxidation of the anion.

Na+

Br-

_reductioncathode

+oxidationanode2Br- → Br2(g)+ 2e-



Power Source - +

PtPt

The Cathode reaction is the reduction of the cation.

Na+

Br-

_reductioncathode




Power Source - +

PtPt

The Cathode reaction is the reduction of the cation.

Na+

Br-

_reductioncathode2Na+ + 2e- → 2Na(l)




Power Source - +

PtPt

Gas Br2 is produced at the anode.

Na+

Br-





Power Source - +

PtPt

Liquid Na is produced at the cathode.

Na+

Br-





Power Source - +

PtPt

The potential for each half reaction is calculated and the oxidation sign is reversed

Na+

Br-





Power Source - +

PtPt

The potential for each half reaction is listed and the oxidation sign is reversed

Na+

Br-


-2.71 v


-1.09 v



Power Source - +

PtPt

The overall redox reaction is written.

Na+

Br-


-2.71 v


-1.09 v



Power Source - +

PtPt

The overall redox reaction is written.

Na+

Br-


-2.71 v


-1.09 v

2Na+ + 2Br- → Br2(g)+ 2Na(l) E0 = -3.80 v153 Dr. Elhelece W. A.


Power Source - +

PtPt

Na+

Br-


-2.71 v


-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.



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-


-2.71 v


-1.09 v

2Na+ + 2Br- → Br2(g)+ 2Na(s) E0 = -3.80 v MTV = +3.80 v



Power Source - +

PtPt

Na+

Br-


-2.71 v


-1.09 v


Electrons flow through the wire from anode to cathode.



Power Source - +

PtPt

Electrons flow through the wire from anode to cathode.

Na+

Br-


-2.71 v


-1.09 v


e-

e-


2. Draw and completely analyze a 1.0 M KI electrolytic cell.



Power Source - +

Pt Pt



Power Source - +

Pt Pt

K+

I-

H2O

Add the ions. (aq) or M

or solution means water.



Power Source - +

Pt Pt

K+

I-

H2O

Label the -, +, anode, cathode, oxidation, and reduction.



Power Source - +

Pt Pt

K+

I-

H2O

Label the -, +, anode, cathode, oxidation, and reduction.

-Cathodereduction

+Anodeoxidation



Power Source - +

Pt Pt

K+

I-

H2O

The cation and water migrate to the cathode

-Cathodereduction

+Anodeoxidation



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


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


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


Cl2 + 2e- → 2Cl- 1.36 v


Reduction of water

2H2O + 2e- → 2H2 + 2OH- -0.42 v

Zn2+ + 2e- → Zn(s) -0.76 v

Reduction of K+


Cl2 + 2e- → 2Cl- 1.36 v


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


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.



Power Source - +

Pt Pt

K+

I-

H2O


-CathodeReduction

+Anodeoxidation



Power Source - +

Pt Pt

K+

I-

H2O


-CathodeReduction2H2O+2e- → 2H2+ 2OH--0.41 v

+Anodeoxidation



Power Source - +

Pt Pt

K+

I-

H2O

The anion + water goes to the anode.

+Anodeoxidation




Power Source - +

Pt Pt

K+

I-

H2O

For oxidation the most spontaneous reaction is found on the redox chart and is lowest.

+Anodeoxidation



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


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


Cl2 + 2e- → 2Cl- 1.36 v overpotential effect means water is here



Reduction of water

2H2O + 2e- → 2H2 + 2OH- -0.42 v

Zn2+ + 2e- → Zn(s) -0.76 v


Cl2 + 2e- → 2Cl- 1.36 v overpotential effect means water is here



pick strongest reducing agent- lowerReduction of water

2H2O + 2e- → 2H2 + 2OH- -0.42 v

Zn2+ + 2e- → Zn(s) -0.76 v



Power Source - +

Pt Pt

K+

I-

H2O


+AnodeOxidation

-CathodeReduction

2H2O+2e- → 2H2+ 2OH--0.41 v



Power Source - +

Pt Pt

K+

I-

H2O


+AnodeOxidation2I- → I2(s) + 2e-

-0.54 v

-CathodeReduction

2H2O+2e- → 2H2+ 2OH--0.41 v



Power Source - +

Pt Pt

K+

I-

H2O

Write the overall reaction with the cell potential.


-0.54 v

-CathodeReduction

2H2O +2e- → 2H2+ 2OH-

-0.41 v



Power Source - +

Pt Pt

K+

I-

H2O



-0.54 v

-CathodeReduction

2H2O+2e- → 2H2+ 2OH--0.41 v

2H2O+ 2I- → 2H2+ I2(s) + 2OH- E0 = -0.95 v



Power Source - +

Pt Pt

K+

I-

H2O



-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-


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


1.0 M

Zn+2

e- e-

AnodeCathode

A battery >1.10V

Zn Cu1.0 M

Cu+2


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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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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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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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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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+


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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Dr. Elhelece W. A.

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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Dr. Elhelece W. A.

dr.wael elhelece electrochemistry 331chem

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