Berry Tokoh Guru No.1 – Encik Arbain bin Miswan

September 1, 2010

Berry Berry Easy is pleased to start a new section to honour and highlight excellent teachers who have gone beyond the scope of duty to guide students to greater heights. The greater heights mentioned are not confined to academic excellence, but also in the form of extra-curricular events, competitions and other ways which touches a student or community in general. A teacher which fits into this profile and selected as our first Berry Tokoh Guru is the very experienced (21 years in service) Encik Arbain bin Miswan. He guided his student Jeep Wai Hong to a second place nationwide in the Essay Writing competition organised by the Persatuan Sejarah Malaysia (History Society of Malaysia). For this, we congratulate both this pair of teacher-student combination. We are fortunate enough to be able to share with all Berry Readers his story and give insights into the mind and thoughts of a great teacher.

Berry Tokoh Guru #1 – Encik Arbain bin Miswan – Second Place in National Level Essay Writing Competition by Persatuan Sejarah Malaysia – Mentor/Guru Pembimbing

We’ll start of with the biodata of Mr Arbain. The subsequent interview will be conducted in Malay in conjunction with Merdeka Day 2010 and also as a tribute to the specialisation/option of Mr Arbain as a Malay language teacher.

Biodata Encik Arbain bin Miswan

Encik Arbain B. Miswan dan Jeep Wai Hong menerima Sijil Pemenang.

Encik Arbain B. Miswan dan Jeep Wai Hong menerima Sijil Pemenang.

  • Ijazah Kepengetuaan (Universiti Malaya) NPQH (National Professional Qualification Of Headship) dalam bahasa Melayu Sijil Kelayakan Profesional Kepengetuaan Kebangsaan.
  • Ijazah Sarjana Muda Sastera (Universiti Malaya) Sejarah dan Pengajian Bahasa Melayu
  • Berkhidmat sebagai guru sejak 01/08/1989
  • Sekarang memegang jawatan Penolong Kanan 1 di SMK Dato’ Jaafar Johor Bahru.
  • Antara karya-karya yang dihasilkan:
    Merangsang Pelajar Berfikir dalam Mata Pelajaran BM dan Pengetua Unggul (Pendidik)
    Memartabatkan Profesion Perguruan (Dewan Masyarakat)
    Beberapa ulasan buku ilmiah (Dewan Masyarakat & Pelita Bahasa)
Encik Arbain B. Miswan dan Jeep Wai Hong menang dalam Pertandingan Menulis Esei Sejarah Malaysia

Encik Arbain B. Miswan dan Jeep Wai Hong menang dalam Pertandingan Menulis Esei Sejarah Malaysia

The Interview with Encik Arbain bin Miswan

1. Bilakah pengumuman bahawa SMK Dato’ Jaafar telah merangkul tempat kedua dalam pertandingan Menulis Sejarah Malaysia peringkat kebangsaan (anjuran Persatuan Sejarah Malaysia) diumumkan? Apakah perasaan cikgu pada masa itu?

> Kita mengetahui secara rasmi keputusan tersebut pada 17 Julai 2010. Perasaan saya gembira dan bersyukur. Hal ini sebab apabila terlibat dengan pengurusan sekolah, memang agar sukar untuk membimbing pelajar dalam kegiatan seumpama ini.

2. Cikgu merupakan guru mata pelajaran Bahasa Melayu, bagaimanakah cikgu dapat membimbing pelajar ini untuk menghasilkan karya/esei Sejarah yang mendapat pencapaian yang begitu cemerlang sekali sehingga menjadi pemenang peringkat kebangsaan?

> Penjelasan saya tentang perkara ini agak panjang. Sebenarnya opsyen saya adalah Bahasa Melayu dan Sejarah. Namun semenjak saya menjawat Guru Kanan Mata Pelajaran Bahasa sejak 1997 saya tidak lagi mengajarkan Sejarah, kerana tuntutan dan keperluan saya menumpukan kepada Bahasa Melayu.

Sebelum saya menjawat GKMP Bahasa, di sekolah pertama saya bertugas, saya selama 6 tahun membimbing pelajar dalam pertandingan esei, pantun, bahas, forum dan beberapa lagi aktiviti dalam bahasa Melayu. Pencapaian tertinggi yang diperoleh adalah johan peringkat negeri. Hal ini memang membantu saya dalam membimbing Jeep Wai Hong.

Tentang persoalan membimbing pelajar menghasilkan karya sejarah saya perlu kembali kepada asas penulisan sesuatu karya. Sebenarnya semasa saya mengajarkan Bahasa Melayu kepada para pelajar, saya menekankan aspek pemahaman terhadap apa-apa bahan yang dibaca. Bukti yang menunjukkan seseorang itu memahami apa-apa yang dibaca adalah seseorang itu dapat mengenal pasti isu, maklumat-maklumat, fakta-fakta, huraian dan contoh-contoh yang terkandung dalam sesuatu bahan bacaan, sama ada satu halaman atau sebuah buku.

Bertolak daripada menguasai kemahiran di atas, seseorang yang hendak menghasilkan sesuatu karya perlu banyak membuat rujukan. Maka ia perlu menentukan perkara yang hendak diutarakan dalam penulisannya. Seterusnya menentukan kerangka yang hendak dikembangkan dari segi isu, maklumat dan fakta-fakta yang berkaitan. Oleh itu, apabila seseorang itu membuat rujukan daripada pelbagai sumber seperti buku, internet dan dokumen, ia dapat menentukan isu, maklumat dan fakta-fakta yang berkaitan dengan tugasan yang dilakukannya.

Hal ini mengambarkan bahawa pertandingan penulisan yang disertai oleh Jeep Wai Hong adalah pemantapan kepada proses pembelajaran yang dialaminya di dalam kelas. Aspek-aspek yang saya bimbing adalah dari segi pengolahan maklumat dan fakta. Serta beberapa dimensi pemikiran yang kritikal dan anatikal. Hal ini sebab Jeep Wai Hong telah dapat menguasai kemahiran membaca dan pemahaman untuk mengenal pasti, membanding-membeza serta menilai aspek isi, maklumat, fakta, huraian dan contoh.

3. Bagaimana encik membuat pilihan pelajar? Apakah sifat-sifat pelajar yang dicarikan semasa pemilihan dibuat ? Bagaimanakah encik melatih pelajar berkenaan?
Apakah teknik-teknik yang digunakan untuk memberikan motivasi kepada pelajar tersebut?

> Penjelasan yang akan saya beri, berbeza daripada tujuan saudari mengemukakan soalan ini. Di sinilah keunikan, penghasilan karya tersebut.

Sebenarnya Jeep Wai Hong telah mengambil inisiatif sendiri untuk menyertai pertandingan ini. Keunikan juga berlaku apabila pelajar (peserta) yang memilih guru pembimbing. Oleh itu, saya berpendapat, mungkin saudari boleh mengemukakan pertanyaan kepada Jeep Wai Hong mengapakah dia menyertai pertandingan tersebut dan memilih saya sebagai guru pembimbing? Sehubungan dengan itu saya rasa, saya tidak perlu memberi respon dahulu terhadap aspek sifat-sifat pelajar.

Menyentuh tentang cara saya memberikan bimbingan pula; pertama saya meminta Jeep menulis terdahulu esei yang berkaitan. Kedua, saya membaca esei tersebut (saya anggap sebagai draf pertama). Tujuan saya membaca keseluruhan draf tersebut, untuk memahami kerangka idea yang hendak disampaikan oleh Jeep. Ketiga, saya menganalisis kekuatan dan kelemahan draf tersebut dan menganalisis perkara-perkara yang perlu ditambah baik. Keempat saya mengadakan perbincangan dengan Jeep, memintanya mencari maklumat dan fakta yang berkaitan dan perlu ada di dalam esei tersebut. Di samping meminta dia menyusun semula kedudukan beberapa fakta dan maklumat. Juga mengugurkan apa-apa yang tidak relevan. Proses ini berlaku sebanyak tidak kurang tiga kali. Bentuk perbincangan saya dengan Jeep seperti bentuk perbincangan saya dengan guru.

Selepas saya berpuas hati dengan esei yang dihasilkan barulah kami membincangkan tajuk yang seseuai. Hal ini sebab dalam pertandingan ini, pihak penganjur hanya memberikan tema sahaja.

4. Apakah bantuan yang telah diberi kepada pelajar berkenaan?

> Oleh itu, bantuan yang saya berikan adalah dalam bentuk idea, teknik penulisan dan teori pensejarahan.

Bagi soalan yang lain saya berpendapat elok kiranya saya memberikannya selepas saudari memperoleh jawapan daripada Jeep terhadap dua soalan yang telah saya kemukakan tadi.

Encik Arbain B. Miswan mengajar dalam kelas

Encik Arbain B. Miswan mengajar dalam kelas

5. Apakah hasrat cikgu pada masa yang akan datang?

> Walaupun saya terlibat dalam menguruskan sekolah, jika ada pelajar yang berinisiatif untuk menyertai sesuatu pertandingan dan bersedia dibimbing mengikut apa-apa yang saya rancangkan, saya akan memperuntukan masa dan fikiran. Prinsip saya, apabila membimbing pelajar sejak 20 tahun lalu adalah untuk menjadi johan.

6. Apa kata-kata yang ingin cikgu sampaikan kepada pelajar-pelajar?

> Hayati falsafah, konsep dan objektif kegiatan kokurikulum yang sebenar. Ingat kokurikulum adalah pelengkap kepada pembelajaran di dalam kelas. Jangan tinggalkan kelas saja-saja untuk menjayakan kegiatan kokurikulum, itu adalah konsep yang salah. Sebenarnya, sebarang kegiatan kokurikulum bermula dari pembelajaran di dalam kelas. Jika anda mengabaikan pelajaran dan tiada disiplin kendiri, jangan harap anda akan berjaya dalam pelajaran. Rekod saya mengajar selama 21 tahun membuktikan keyakinan saya ini.

7. Apakah nasihat yang ingin cikgu sampaikan kepada pelajar-pelajar yang mahu mencapai kejayaan dalam hidup ?

> Bagi mencapai kejayaan dalam peperiksaan kuasailah ilmu, bukan sekadar ingin lulus cemerlang dalam peperiksaan. Dua tahun lepas seorang pensyarah UTM Skudai, mencari dan mengubungi saya, menyatakan hasratnya untuk bertemu dengan saya. Akhirnya apabila bertemu beliau menyatakan bahawa apa-apa yang saya ajarkan semasa di tingkatan 4 (1996), beliau boleh gunakan sehingga peringkat PhD di Amerika Syarikat. Saya mengajarkan Bahasa Melayu, pengajian beliau peringkat PhD dalam bahasa Inggeris. Subjek yang beliau ajar di UTM adalah Keselamatan Komputer. Itulah ilmu boleh digunakan dalam apa-apa bahasa pun.

Inilah citra sebahagian daripada bekas pelajar-pelajar saya. Ramai bekas pelajar saya yang menceburi pelbagai bidang kerjaya menyatakan cara saya mengajar mereka. Inilah yang menyebabkan saya terus meneroka idea baru dalam pengajaran saya dan cara saya menguruskan bidang tugas saya di sekolah.

Cemerlang dalam peperiksaan sangat penting, tetapi itu bukan matlamat, sebenarnya hanya alat. Aspek-aspek lain yang penting untuk berjaya dalam hidup antaranya berdisiplin, perasaan kemanusiaan yang tinggi, gigih dan sentiasa kuat semangat apabila menghadapi cabaran dalam kerjaya serta kehidupan.

Terima kasih kerana sudi bertukar fikiran dengan saya, semoga apa-apa yang saya nyatakan bermanfaat kepada anda semua.

— End of Interview —

We thank Mr Arbain for his time and sharing his experience on the competition and also general life lessons. So do remember what was said, especially the fact that “academic excellence is important but not the aim, it should be the instrument to success instead. Other aspects such as discipline, human compassion, determination and spirit/morale are also important in future endeavors and in life“.

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STPM Form 6 Academic Grading System (Johor Bahru District)

August 27, 2010

Seeing that STPM examinations for the academic year of 2010 is quickly approaching, there are three questions/requests that are in abundance by our Berry Readers. They are namely:

  1. STPM 2010 Tips
  2. STPM 2010 Trial papers/ or other past year papers
  3. STPM 2010 Grading system

The first request will come soon enough but will not be released first as Berry Berry Easy hopes that all students will study all chapters and only look for tips when you have studied everything one round. So tips will come either in September or October, but not in August. The second request would require some time before it can materialise.

So this post focuses on the third request, which is the grading system. The following grading system showed is for the JB district’s STPM Form 6 Academic grading system. It shall not be taken as representing the real STPM, but should give students some idea on the scoring system used.

STPM Form 6 Academic Grading System (Johor Bahru District)

This is a grading system for Form 6 schools in Johor Bahru District.

A (4.00) = 75-100 marks
A- (3.67) = 70-74 marks
B+ (3.33) = 60-69 marks
B (3.00) = 50-59 marks
B- (2.67) = 45 – 49 marks
C+ (2.33) = 40-44 marks
C (2.00) = 35-39 marks
C- (1.67) = 30-34 marks
D+ (1.33) = 25-29 marks
D (1.00) = 20-24 marks
F (0.00) = 0-19 marks

Marks below 35 is consider failed.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 11)

August 26, 2010

What is the differences between an electrolytic cell and a chemical cell? Most student will answer that electrolytic cells involves electrical supply while a chemical cell do not. While structurally that is the correct answer but most students do not remember the other differences or similarities between the two. So,the Berry Berry Teacher summarises everything for our Berry Readers, in turn creating Part 11 of the SPM Form 5 Chemistry notes on “Oxidation and Reduction” – redox reactions in electrolytic cell and chemical cell. This subtopic is a popular topic in SPM Chemistry exams, so students should understand and memorise the tables of differences given below. Do also understand and memorise the similarities between the two different type of cells. Examples of redox reactions in the electrolytic cell and chemical cell are given after the table which is useful for elaboration of points in an essay or strutural question.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 11)

Redox Reactions in Electrolytic Cell and Chemical Cell

Similarities

  • redox reaction.
  • Anode: oxidation
  • Cathode: reduction
  • Electrons flow from anode to cathode in the external circuit

Differences

Differences Electrolytic Cell (Electrolysis) Chemical Cell / Voltaic Cell
Structure With electrical supply. No electrical supply.
Electrodes Can be the same or difference metal (graphite or platinum). Must be two different metals.
Flows of electrons From anode to cathode through external circuit. From more electropositive metal to less electropositive metal through external circuit.
Transformation of energy Electrical energy to chemical energy. Chemical energy to electrical energy.
At positive terminal Anode.Oxidation occurs. Anions release electrons at the anode. Cathode.Reduction occurs.

Oxidising agent gain electrons.

At negative terminal Cathode.Reduction occurs.

Cations gain electrons from the cathode.

Anode.Oxidation occurs.

Reducing agent releases electrons.

1) Redox Reactions in Electrolytic Cell

Example 1: Electrolysis of molten zinc chloride

  • Electrodes: Carbon
  • Ions present: Cl- and Zn2+
  • Anode: Oxidation / 2Cl-(l) –> Cl2(g) + 2e / Cl- ions act as reducing agent.
  • Cathode: Reduction / Zn2+(l) + 2e –> Zn(s) / Zn2+ ions act as oxidising agent.

Example 2: Electrolysis of copper(II) sulphate solution.

  • Electrodes: Carbon
  • Ions present: Cu2+, SO42-, H+, OH-
  • OH- ions are discharged because OH- ion is below SO42- ion in electrochemistry series.
    Anode
    : Oxidation / 4OH-(aq) –> O2(g) + 2H2O(l) + 4e / Oxygen gas is liberated.
  • Cu2+ ions are discharged because Cu2+ ion is below H+ ion in electrochemistry series.
    Cathode
    : Reduction / Cu2+(aq) + 2e –> Cu(s) / Cu2+ ions are reduced to copper metal (brown layer formed).
  • –> Overall equation: Cu2+(aq) + 4OH-(aq) –> O2(g) + 2H2O(l) + Cu(s)

Example 3: Electrolysis of copper(II) sulphate solution.

  • Electrodes: Copper
  • Ions present: Cu2+, SO42-, H+, OH-
  • OH- ions  and SO42- ion are not discharged.
    Anode
    : Oxidation / Cu(s) –> Cu2+(aq) + 2e / Copper anode (electrode) dissolves.
  • Cu2+ ions are discharged because Cu2+ ion is below H+ ion in electrochemistry series.
    Cathode
    : Reduction / Cu2+(aq) + 2e –> Cu(s) / Cu2+ ions are reduced to copper metal.

Example 4: Electrolysis of concentrated sodium chloride solution.

  • Electrodes: Carbon
  • Ions present: Na+, Cl-, H+, OH-
  • Cl- ions are discharged because of the higher concentration. (Concentration of Cl- ion is high, the ion is selectively discharged rather than the OH- ion, the one that is placed below the electrochemical series.)
    Anode
    : Oxidation / 2Cl-(aq) –> Cl2(g) + 2e / Chlorine gas (green gas with choking smell) is liberated.
  • H+ ions are discharged because Na+ ion is below H+ ion in electrochemistry series. (H+ ions and Na+ ion are placed very far apart in the electrochemical series, the concentration factor becomes unimportant.)
    Cathode
    : Reduction / 2H+(aq) + 2e –> H2(g) / Hydrogen gas is liberated.
  • –> Overall equation: 2Cl-(aq) + 2H+(aq) –> Cl2(g) + H2(g)

2) Redox Reactions in Chemical Cell

Example 1: Daniel cell

  • Anode (negative terminal): Oxidation / Zinc strip immerses in zinc sulphate solution.
    Zn(s) –> Zn2+(aq) + 2e / Zinc strip becomes thinner.
  • Cathode (positive terminal): Reduction / Copper strip immerses in copper(II) sulphate solution.
    Cu2+(aq) + 2e –> Cu(s) / A brown layer formed around copper strip. / Concentration Cu2+ ions decreases cause the intensity blue colour of solution decreases.
  • Zinc is more electropositive than copper. Electrons are flowed from zinc strip to copper strip through the external circuit. (Note: Conventionally, electrons flow in the opposite direction of electrical current).
  • –> Overall equation:  Zn(s) + Cu2+(aq) –> Zn2+(aq) + Cu(s)

Good news to all Berry Readers, as the series involving “Oxidation and Reduction” will end with the next post with the various chemical cells that you would encounter in daily life.


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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 10)

August 25, 2010

As continuation to Part 9 featuring yet again the reactivity series, this part focuses on the position of hydrogen in the reactivity series of metal and what is the significance of the position. So the Berry Berry Teacher would like to highlight to all Berry Readers, Part 10 of the SPM Form 5 Chemistry notes on “Oxidation and Reduction” – position of hydrogen. Please read up Part 9 of this series of notes first if you have not done so, as you’ll find yourself missing a lot of crucial points if you were to start directly on this part. The tips to this SPM Chemistry Form 5 subchapter is to always remember the relative position of hydrogen on the reactivity series which is in between Zinc (Zn) and Iron (Fe). After that, try and understand the examples. If you can fully understand the examples, it means that you have grasp the concept and significance of the position of hydrogen in the series.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 10)

Position of Hydrogen in the Reactivity Series of Metals

Reactivity Series
K, Na, Ca, Mg, Al, C, Zn, H, Fe, Sn, Pb, Cu, Hg, Ag, Au
<——– increase in reactivity

Metal oxide + Hydrogen –> Metal + Water

Any metal below hydrogen in the reactivity series, hydrogen will reduce the oxide of metal to metal.

Example 1:

  • CuO(s) + H2(g) –> Cu(s) + H2O(l)
  • Observation: Burns quickly with a bright flame. The black solid turns brown solid.
  • H2: Reducing agent
  • CuO: Oxidising agent
  • Hydrogen is more reactive than copper.

Example 2:

  • PbO(s) + H2(g) –> Pb(s) + H2O(l)
  • Observation: Burns with a bright flame. The yellow solid turns grey solid.
  • H2: Reducing agent
  • PbO: Oxidising agent
  • Hydrogen is more reactive than lead.

Example 3:

  • Fe2O3(s) + 3H2(g) –> 2Fe(s) + 3H2O(l)
  • Observation: Glows dimly. The reddish-brown solid turns grey solid.
  • H2: Reducing agent
  • Fe2O3: Oxidising agent
  • Hydrogen is more reactive than iron.

Example 4:

  • ZnO(s) + H2(g) –> no reaction
  • Observation: No glow is observed. It turns yellow when hot and white when cold.
  • Hydrogen is unable to reduce zinc oxide. Hydrogen is less reactive than zinc.

Next up in Part 11 of this series is the slightly (only marginally) difficult subtopic of redox reactions in an electrolytic cells and chemical cells.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 9)

August 20, 2010

The reactivity of metals and its application in SPM Form 5 Chemistry level is usually regarded as a difficult topic by half of the students and easy topic by the other half. So today, the Berry Berry Teacher will like to share the secret to this subtopic with all Berry Readers, Part 9 of the SPM Form 5 Chemistry notes on “Oxidation and Reduction”. For those who find this subtopic easy, you won’t need this secret. For those who find it hard, then the secret is so simple that you would be disappointed with the revelation of it. To master this topic, all you really need is to just understand it rather than memorising it. With understanding, you can easily memorise this part because they all have logical links. So, before you start to read this post, it will be good to refresh the whole topic from Part 1-8.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 9)

The Reactivity Series of Metals and Its Application

1. Metal form metal oxides when burnt in air (excess).
Metal + Oxygen –> Metal oxide
Example: 2Zn(s) + O2(g) –> 2ZnO(s)

2. The more reactive a metal is, the more vigorously it burns in oxygen.

Reactivity of Metals
K, Na, Ca, Mg, Al, Zn, Fe, Sn, Pb, Cu, Hg, Ag, Au
<——– increase in reactivity of metals

3. Reactivity of Metals with Oxygen.

Metal Observation Inference Equation
Magnesium (Mg) Burns vigorously with a very brilliant white flame.The residue is white when hot and cold. The reactivity of Mg towards O2 is very high.Magnesium oxide is formed. 2Mg(s) + O2(g)–> 2MgO(s)
Zinc (Zn) Burns quickly with a bright flame.The residue is yellow when hot and white when cold. The reactivity of Zn towards O2 is high.Zinc oxide is formed. 2Zn(s) + O2(g)–> 2ZnO(s)
Iron (Fe) Glows very brightly.The residue is reddish-brown when hot and cold. The reactivity of Fe towards O2 is medium.Iron(III) oxide is formed. 2Fe(s) + O2(g)–> 2Fe2O3(s)
Lead (Pb) Glows brightly.The residue is brown when hot and yellow when cold. The reactivity of Pb towards O2 is low.Lead(II) oxide is formed. 2Pb(s) + O2(g)–> 2PbO(s)
Copper (Cu) Glows faintly.The residue is black when hot and cold. The reactivity of Cu towards O2 is very low.Copper(II) oxide is formed. 2Cu(s) + O2(g)–> 2CuO(s)
  • Glass wool – prevents metal powder mixes with potassium manganate(VII)
  • Solid potassium manganate(VII) – liberates oxygen gas when it is heated / decomposed.

2KMnO4(s) —-> K2MnO4(s) + MnO2(s) + O2(g)
heat

  • Other than potassium manganate(VII),

- solid potassium chlorate(V) with manganese(IV) oxide as a catalyst.

MnO2
2KClO3(s) —-> KCl(s) + 3O2(g)
heat

- solid potassium nitrate

2KNO3(s) —-> KNO2(s) + O2(g)
heat

4. Position of Carbon in the Reactivity Series of Metals

Reactivity Series
K, Na, Ca, Mg, Al, C, Zn, Fe, Sn, Pb, Cu, Hg, Ag, Au
<——– increase in reactivity

a) Metal + Carbon dioxide –> Metal oxide + Carbon

Any metal above carbon in the reactivity series can displace oxygen from carbon dioxide.

Example: 2Mg(s) + CO2(g) –> 2MgO(s) + C(s)

Mg: Reducing agent
CO2: Oxidising agent
MgO: White residue
C: Black spots

–> Therefore, magnesium is more reactive than carbon.
(If the metal is less reactive than carbon, the metal is unable to remove oxygen from carbon dioxide.)

b) Carbon + Metal oxide –> Carbon dioxide + Metal

Any metal below carbon in the reactivity series can displace carbon from its oxide.

Example: C(s) + 2ZnO(s) –> 2Zn(s) + CO2(g)

C: Reducing agent
ZnO: Oxidising agent
Zn: Grey residue

–> Therefore, zinc is less reactive than carbon.
(If carbon is less reactive than the metal, the carbon is unable to remove oxygen from metal oxide.)

Next up in this series, is the position of hydrogen and how it is used as a baseline for comparison. So stay tune and log in frequently to BerryBerryEasy

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 8)

August 19, 2010

Berry Readers will definitely be familiar with the concept of rusting. But do you know that rusting is also a redox reaction? Presented in this post is Part 8 of SPM Chemistry Form 5 notes on “Oxidation and Reduction” from Berry Berry Easy. In this post, you will get to read about the simplified 4-step rust process. Also important are the ways to prevent rust. Ways to prevent rust is typically found in exams on multiple choice questions (especially those that group answers together) and also in the structured paper. It’s popularity in exams is cyclical, so do read up on them.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 8 )

Rusting as a Redox Reaction

  • Rust / hydrated iron(III) oxide, Fe2O3•xH2O – formed slowly at the surface of iron when it exposed to the damp air.
  • Rusting – a redox reaction that take places between iron and oxygen to form hydrated iron(III) oxide and this is a slow reaction.
    4Fe(s) + 3O2(g) + 2xH2O(l) –> Fe2O3•xH2O(s)

Corrosion – a redox reaction that take places between a metal and the gases in air. Metal is oxidised to form an oxide layer on the surface. Metal atoms lose electrons to form positive ions.

  1. Group 1 metals are very reactive.
  2. Metals are exposed to air will corrode rapidly and become tarnished.
  3. Aluminium, lead and zinc corrode rapidly in the air and forms an oxide layer. The oxide layer is hard, non-porous, impermeable and difficult to crack. This protects the aluminium, lead and zinc below it from further corrosion.

Example: Corrosion of metal.
Zn(s) –> Zn2+(aq) + 2e
Cu(s) –> Cu2+(aq) + 2e

K, Na, Ca, Mg, Al, Zn, Fe, Sn, Pb, Cu, Hg, Ag, Au
<————Tendency of metal to corrode increases.

Process of Rusting of Iron

  1. Anode (negative terminal) – concentration of oxygen is lower and iron rust (oxidation process) to form iron(II) ions:
    Fe(s) –> Fe2+(aq) + 2e
  2. Cathode (positive terminal) – concentration of oxygen is higher and oxygen gains electrons that reduced to hydroxide ions:
    O2(g) + 2H2O(l) + 4e –> 4OH-(aq)
  3. Fe2+ ions and OH- ions combine to form iron(II) hydroxide, Fe(OH)2
  4. Oxygen further oxidises iron(II) hydroxide, Fe(OH)2 to hydrated iron(III) oxide, Fe2O3•xH2O.
    2Fe(OH)2(s) –> Fe2O3•xH2O(s)

The Effect of Other Metals on the Rusting of Iron

Potassium hexacyanoferrate(III), K3Fe(NO)6 is used to detect Fe2+ ions
(produces dark blue colour in the presence of Fe2+).

Phenolphthalein is used to detect OH- ions
(produces pink colour in the presence of OH-).

Test tube Observation Reaction
Fe only Intensity of blue colour is low. Oxidation:Fe(s) –> Fe2+(aq) + 2e
Control Pink colour is not present. Reduction:O2(g) + 2H2O(l) + e –> 4OH-(aq)OH- ions react with Fe2+ ions to form Fe(OH)2.
Inference Fe nail rusts a little.
Test tube Observation Reaction
Fe & Mg Blue colour is not present. Oxidation:Mg(s) –> Mg2+(aq) + 2e
Intensity of pink colour is very high. Reduction:O2(g) + 2H2O(l) + e –> 4OH-(aq)
Inference Mg is corroded and Fe nail does not rust.

i) Fe act as the (+) terminal (cathode)
ii) Mg act as the (-) terminal (anode)

Test tube Observation Reaction
Fe & Zn Blue colour is not present. Oxidation:Zn(s) –> Zn2+(aq) + 2e
Intensity of pink colour is high. Reduction:O2(g) + 2H2O(l) + e –> 4OH-(aq)
Inference Zn is corroded and Fe nail does not rust.

i) Fe act as the (+) terminal (cathode)
ii) Zn act as the (-) terminal (anode)

Test tube Observation Reaction
Fe & Sn Intensity of blue colour is high. Oxidation:Fe (s) –> Fe2+(aq) + 2e
Pink colour is not present. Reduction:O2(g) + 2H2O(l) + e –> 4OH-(aq)OH- ions react with Fe2+ ions to form Fe(OH)2.
Inference Fe nail rusts quickly (high rate).

i) Sn act as the (+) terminal (cathode)
ii) Fe act as the (-) terminal (anode)

Test tube Observation Reaction
Fe & Cu Intensity of blue colour is very high. Oxidation:Fe (s) –> Fe2+(aq) + 2e
Pink colour is not present. Reduction:O2(g) + 2H2O(l) + e –> 4OH-(aq)OH- ions react with Fe2+ ions to form Fe(OH)2.
Inference Fe nail rusts very quickly (the highest rate).

i) Cu act as the (+) terminal (cathode)
ii) Fe act as the (-) terminal (anode)

  • Iron nail does not rust if it has contacted with more electropositive metals (Mg and Zn).
  • Iron nail rusts quickly if it has contacted with less electropositive metals (Sn and Cu).

Prevention of Rusting of Iron

The rate of rusting of iron decreases if the iron (Fe) in contact with any of these metals: K, Na, Ca, Mg, Al and Zn.
The rate of rusting of iron increases if the iron (Fe) in contact

  • with any of these metals: Sn, Pb, Cu, Hg, Ag and Au.
  • a strong electrolyte (salt and acid) is present.

Ways Used for Prevention of Rusting

  1. Painting – Protect iron surface (prevent from contacting with air and water)
  2. Coat with plastic – Used in metal netting
  3. Apply oil and grease – Protective coating for machine part
  4. Alloying the iron – Alloying the iron with 18% chromium and 8% nickel that provide a protective oxide coating.
  5. a) Tin plating (less electropositive metal) – Cans of food (iron) is covered with a thin layer of tin to provide a protective oxide coating to the cans.
    b) Chrome plating
  6. Cathodic protection / Electrical protection (more electropositive metal)
    a) Galvanising (coat with zinc metal) – Zinc layer provides a protective oxide coating and zinc is oxidized instead of iron. Iron cannot form ions, so it will not rust.
    b) Sacrificial protection – Blocks of magnesium are attached at the intervals of the water piping system & zinc bars are attached to the part of the ship submerged in sea water.

Stay tune and log in back again for Part 9 of this series on reactivity series of metal and their applications only at BerryBerryEasy

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 7)

August 15, 2010

The topic of Oxidation and Reduction is indeed a very long chapter in SPM Chemistry Form 5. Presented in this post is the Part 7 on the “Redox Reactions by the Transfer of Electrons as a Distance” from Berry Berry Easy. This is by far one of the most popular topics in exams for the written paper. Also one of the most hated topic as it is deemed too difficult by many students as there are a lot of things to be memorised. Berry Readers need not fear this topic as you should understand the concept of this ‘experiment’ and how electrons move across the salt bridge.

Six Easy Steps to Understand this SubTopic:

  • First, try to understand why the apparatus are set up this way. Is it for isolation? If so, why? (Answer: So that redox reaction can occur through the transfer of electrons without contact but through a conduit.
  • Secondly, try to understand the nature of both the reducing and oxidising agents. Why would certain combinations be chosen?
  • Thirdly, remember what is added onto the solution in the reducing agent arm of the U-tube (U-shaped glass, not the video site).
  • Fourthly, try to understand why is it tested on the reducing agent arm only.
  • Fifth, try to logically visualise the observation, visualising it is sometimes and important way to understand things.
  • Finally, use all your chemistry knowledge to infer your observations.

A summary of the most important part of this subtopic is summarised in the tables below.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 7)

Redox Reactions by the Transfer of Electrons at a Distance

Set I

Reducing agent Oxidising agent Test on the solution in the reducing agent arm of U-tube
Iron(II) sulphate, FeSO4 solution Acidified potassium dichromate(VI), K2Cr2O7 solution Add a few drops of potassium thiocyanate, KSCN solution
Observation Inference
The electrode in the iron(II) sulphate, FeSO4 solution acts as the negative terminal, whereas the electrode in the acidified potassium dichromate(VI), K2Cr2O7 solution acts as the positive terminal. Electrons flow from iron(II) sulphate, FeSO4 solution to acidified potassium dichromate(VI), K2Cr2O7 solution
Iron(II) sulphate solution changes from pale green to yellow/brown. It gives blood-red colouration with potassium thiocyanate solution (KSCN) Iron(III) ions are present. Iron(II) ions are oxidised to  iron(III) ions.
Acidified potassium dichromate(VI), K2Cr2O7 solution changes colour from orange to green. Dichromate(VI) ions are reduced to chromium(III) ions.
  • Oxidation half-equation: Fe2+(aq) –> Fe3+(aq) + e
  • Reduction half-equation: Cr2O72-(aq) + 14H+(aq) + 6e –> 2Cr3+(aq) + 7H2O(l)
  • Overall reaction: Cr2O72-(aq) + 6Fe2+(aq) 14H+(aq) –> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

Set II

Reducing agent Oxidising agent Test on the solution in the reducing agent arm of U-tube
Iron(II) sulphate, FeSO4 solution Acidified manganate(VII), KMnO4 solution Add sodium hydroxide, NaOH solution
Observation Inference
The electrode in the iron(II) sulphate, FeSO4 solution acts as the negative terminal, whereas the electrode in the acidified potassium dichromate(VI), K2Cr2O7 solution acts as the positive terminal. Electrons flow from iron(II) sulphate, FeSO4 solution to acidified potassium dichromate(VI), K2Cr2O7 solution
Iron(II) sulphate solution changes from pale green to yellow/brown. It formed a brown precipitate when the brown solution is tested with sodium hydroxide solution (NaOH) Iron(III) ions are present.Iron(II) ions are oxidised to  iron(III) ions.
Purple acidified manganate(VII), KMnO4 solution turns colourless. Manganate(VII) ions are reduced to manganese(II) ion.
  • Oxidation half-equation: Fe2+(aq) –> Fe3+(aq) + e
  • Reduction half-equation: MnO4-(aq) + 8H+(aq) + 5e –> Mn2+(aq) + 4H2O(l)
  • Overall reaction: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) –> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Set III

Reducing agent Oxidising agent Test on the solution in the reducing agent arm of U-tube
Potassium iodide, KI solution Bromine water, Br2 Add a few drops of starch solution
Observation Inference
The electrode in the potassium iodide, KI solution acts as the negative terminal, whereas the electrode in the bromine water acts as the positive terminal. Electrons flow from potassium iodide, KI solution to bromine water, Br2 (aq).
Colourless potassium iodide solution turns brown. It formed a dark blue colouration when the brown solution is tested with starch solution. Iodine is present.Iodide ions have oxidised to iodine.
Brown bromine water turns colourless. Bromines are reduced to bromide ion.
  • Oxidation half-equation: 2I-(aq) –> I2(aq) + 2e
  • Reduction half-equation: Br2(aq) + 2e –> 2Br-(aq)
  • Overall reaction: Br2(aq) + 2I-(aq) –> 2Br-(aq) + I2(aq)

Other pairs of reducing agent and oxidising agent

Reducing agent Oxidising agent
Potassium iodide,KI solution Iron(III) sulphate,Fe2(SO4)3 solution
Potassium iodide,KI solution Acidified potassium dichromate(VI),K2Cr2O7 solution
Potassium bromide,KBr solution Chlorine, Cl2 water

Do log in back to Berry Berry Easy for Part 8 in the series focusing on “rusting

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 6)

August 12, 2010

Halogens are non-metal elements listed under Group 17 in the periodic table. The halogen elements contained within Group 17 are fluorine, chlorine, bromine, iodine and astatine. (Within the SPM Chemistry context, you may just focus on the first four halogens mentioned) Hence, you may guessed it already, this post is a continuation Berry Berry Easy‘s SPM Chemistry Form 5 notes on “Oxidation and Reduction” – Part 6. It’s the only group in the periodic table to consist of elements in liquid, solid and gaseous form at S.T.P. Unlike the elements from other groups, halogens typically is not found in its free form in nature, due to its high reactivity. In all, it is rather easy to understand halogens in SPM level and fortunately this short sub-topic is frequently asked in exams of all levels.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 6)

Displacement of Halogen (Group 17) from Its Halide Solution

Halogen – elements in Group 17 of the Periodic Table

Example: chlorine, bromine and iodine.

Halogen Identification

Halogen Colour in (conc.) aq. solution Colour in (dilute) aq. solution Colour in 1,1,1-trichloroethane
Iodine Reddish-brown Yellow Purple
Bromine Brown Yellow Brown
Chlorine Pale yellow Colourless Colourless

Strength of oxidising agent in halogen

Cl2, Br2. I2

<——— Oxidising power increases

Halide / Halogen Chlorine Bromine Iodine
Potassium chloride - No changes No changes
Potassium bromide Chlorine displace bromine from KBr solution - No changes
Potassium iodide Chlorine displace iodine from KI solution Bromine displace iodine from KI solution -

So stay tune for the next part of this series of “Oxidation and Reduction” on redox reaction by the transfer of electrons from a distance.

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