Elektrolisis bagian 2 -HUKUM FARADAY 1

Summary

This educational video explores the quantitative relationship between the amount of substance reacting and the electric charge involved in electrochemistry, focusing on Faraday's First Law. The law, named after English scientist Michael Faraday, establishes that the mass of a substance deposited at an electrode is proportional to the electric charge passed through the cell. The video uses the formula m = e * I * t / (n * F) to calculate the mass of metals produced at the cathode during electrolysis, where m is mass, e is the equivalent mass, I is current, t is time, n is the number of electrons transferred, and F is Faraday's constant. Practical examples, including the electrolysis of CuSO4 and AgNO3 solutions, demonstrate how to apply this law to determine the mass of metals produced.

Takeaways

• 🔬 The video discusses the quantitative relationship between the amount of substance reacting and the electric charge involved, known as Faraday's laws of electrolysis.
• 🌐 Faraday's laws are named after the English scientist Michael Faraday, a physicist and chemist who formulated these principles.
• ⚡ The first law of Faraday states that the mass of a substance deposited at an electrode is directly proportional to the quantity of electric charge passed through the solution.
• 🔋 The formula m = e * I * t / (n * F) is used to calculate the mass of the substance produced, where m is mass, e is the equivalent mass of the substance, I is current, t is time, n is the number of electrons transferred, and F is Faraday's constant.
• 🔢 Faraday's constant (F) is approximately 96,500 C/mol, representing the charge of one mole of electrons.
• 🏗️ The video provides examples of how to calculate the mass of metals deposited at the cathode during electrolysis, using the formula and Faraday's constant.
• ⚖️ The script includes a problem-solving session where the audience is guided through calculating the mass of copper, silver, and magnesium produced during electrolysis.
• 🕒 Time is a crucial factor in these calculations, and it must be converted into seconds when used in the formula.
• 🔋 The valency of the metal ion in the electrolyte solution is essential for determining the mass of the metal deposited, as it affects the number of electrons involved in the reaction.
• 🔌 The video also covers the calculation of the electric current required to produce a specific mass of a metal during electrolysis, demonstrating the application of Faraday's laws in practical scenarios.
• 📚 Understanding Faraday's laws is fundamental for solving problems related to electrolysis and is emphasized as a key learning outcome of the video.

Q & A

What is the relationship between the amount of substance reacting and the electric charge involved according to Faraday's laws?

Faraday's laws establish a quantitative relationship between the amount of substance reacting and the electric charge involved, allowing us to determine the quantity of a substance reacting based on the amount of electric charge passed through an electrolyte in a given time.

Why is the law named Faraday's law?

The law is named Faraday's law because it was formulated by the English scientist Michael Faraday, a physicist and chemist known for his work in electromagnetism and electrochemistry.

What is the formula used to calculate the mass of a substance produced at an electrode according to Faraday's first law?

The formula used to calculate the mass of a substance produced at an electrode is m = e * I / (n * F), where m is the mass, e is the equivalent mass of the substance, I is the current in amperes, n is the number of electrons involved in the redox reaction, and F is Faraday's constant.

What is the value of Faraday's constant and what does it represent?

Faraday's constant is approximately 96,500 coulombs per mole of electrons. It represents the amount of electric charge carried by one mole of electrons.

How can you calculate the mass of copper deposited at the cathode during electrolysis of a CuSO4 solution with a given current?

To calculate the mass of copper deposited, you would use the formula m = (M * I * t) / (n * F), where M is the molar mass of copper, I is the current in amperes, t is the time in seconds, n is the valency of copper, and F is Faraday's constant.

What is the molar mass of copper and how is it used in Faraday's law calculations?

The molar mass of copper is 63.5 g/mol. It is used in Faraday's law calculations to determine the mass of copper deposited at the cathode during electrolysis by multiplying it with the current, time, and valency, and then dividing by Faraday's constant.

How do you determine the mass of silver produced by passing a current through a AgNO3 solution?

The mass of silver produced can be determined using the formula m = (M * I * t) / (n * F), where M is the molar mass of silver, I is the current in amperes, t is the time in seconds, n is the valency of silver, and F is Faraday's constant.

What is the valency of silver in the context of the electrolysis of AgNO3?

In the context of the electrolysis of AgNO3, silver has a valency of one, as it forms Ag+ ions in the solution.

How can you calculate the current required to produce a specific mass of magnesium through the electrolysis of MgCl2?

The current required to produce a specific mass of magnesium can be calculated using the formula I = (m * n * F) / (M * t), where m is the mass of magnesium, n is the valency of magnesium, F is Faraday's constant, and t is the time in seconds.

What is the relative atomic mass of a metal produced at the cathode with a known current, time, and mass of the metal?

The relative atomic mass of the metal can be calculated using the formula M = (m * F) / (n * I * t), where m is the mass of the metal, F is Faraday's constant, n is the valency of the metal, I is the current in amperes, and t is the time in seconds.

How does the valency of a metal ion affect the amount of metal deposited during electrolysis?

The valency of a metal ion affects the amount of metal deposited during electrolysis because it determines the number of electrons required to reduce one mole of the metal ions to metal atoms, which in turn affects the amount of charge needed for the deposition process.

Outlines

00:00

🔬 Introduction to Faraday's Laws in Electrochemistry

This paragraph introduces the concept of Faraday's Laws in the context of electrochemistry. It explains that these laws, formulated by the English scientist Michael Faraday, relate the amount of substance reacting in an electrochemical cell to the charge of electricity involved. The law is fundamental for determining the mass of a substance produced at an electrode during electrolysis, which is directly proportional to the charge passed through the cell. The formula m = e * (I * t) / (n * F) is introduced, where m is the mass of the substance, e is the equivalent mass of the substance, I is the current in amperes, t is the time in seconds, n is the number of electrons involved in the redox reaction, and F is Faraday's constant (approximately 96,500 C/mol). The paragraph also discusses how to calculate the mass of a substance produced at an electrode using this formula.

05:00

🔍 Calculating Mass of Metals in Electrolysis

The second paragraph delves into practical applications of Faraday's Laws by calculating the mass of metals produced during electrolysis. It presents a problem-solving approach where the current (I), time (t), and valency (n) of the metal are known, and the goal is to find the mass (m) of the metal produced. The paragraph uses specific examples, such as the electrolysis of CuSO4 and AgNO3 solutions, to demonstrate the calculations. It explains the process of determining the molar mass of the metal ions involved, applying the charge of the current, and using Faraday's constant to find the mass of the metal deposited at the cathode. The examples illustrate how to convert the given values into the appropriate units and apply them to the formula to get the final mass of the metal produced.

10:04

🔋 Further Applications of Faraday's Laws in Electrolysis

The third paragraph continues the exploration of Faraday's Laws with additional examples, focusing on the electrolysis of MgCl2 to produce magnesium. It discusses the process of calculating the required electrical current to produce a specific mass of magnesium within a given time frame. The paragraph emphasizes the importance of understanding the valency of the metal ion and the stoichiometry of the electrolysis reaction. It provides a step-by-step calculation, showing how to use the formula m = (I * t) / (n * F) to determine the current needed for electrolysis. The examples highlight the practical use of Faraday's Laws in predicting the outcomes of electrochemical processes and underscore the importance of unit consistency and correct application of the formula.

Keywords

💡Electrolysis

Electrolysis is a chemical process that uses an electric current to drive a non-spontaneous chemical reaction. In the context of the video, electrolysis is used to discuss the deposition of metals from their ionic solutions onto an electrode. The script mentions electrolysis in relation to the deposition of copper from a CuSO4 solution, illustrating how the process is fundamental to understanding the principles of electrochemistry.

💡Faraday's Laws

Faraday's Laws describe the relationship between the amount of substance produced or consumed at an electrode during electrolysis and the quantity of electric charge passed through the electrolyte. The video script refers to Faraday's First Law, which is crucial for calculating the mass of substances involved in electrochemical reactions. It is used to explain how the mass of a substance deposited at an electrode is directly proportional to the amount of charge passed.

💡Electrode

An electrode is a conductor through which electric current enters or leaves an electrolyte. In the video, the script discusses the deposition of metals at the cathode during electrolysis, highlighting the role of electrodes in facilitating electrochemical reactions. The cathode is where reduction occurs, and metals are deposited from their ionic form.

💡Charge

In electrochemistry, charge refers to the quantity of electricity that is transferred during an electrochemical reaction. The script mentions the charge in the context of calculating the mass of metals deposited during electrolysis using Faraday's Laws. The charge is a critical factor in determining the amount of substance that will be deposited or dissolved at an electrode.

💡Copper Sulfate (CuSO4)

Copper sulfate is a chemical compound that serves as an electrolyte in the video's discussion of electrolysis. The script uses CuSO4 as an example to explain how copper metal is deposited at the cathode during the electrolysis process. It is a common electrolyte used in educational demonstrations of electrochemistry.

💡Valency

Valency, also known as oxidation state, refers to the combining power of an element for other atoms, which is often indicated by the number of hydrogen atoms it can displace or the number of oxygen atoms with which it can combine. In the script, valency is used to calculate the amount of charge required for the deposition of a metal from its compound during electrolysis, as seen in the calculation for copper deposition.

💡Current (I)

Current in the context of the video refers to the flow of electric charge in a circuit, measured in amperes (A). The script discusses the current in relation to the amount of time it takes to deposit a certain mass of metal during electrolysis. Current is a key variable in the equations used to calculate the mass of substances deposited or dissolved at electrodes.

💡Time (t)

Time is a critical factor in electrochemical reactions, as it determines the total charge passed through the electrolyte. In the video, time is mentioned in the context of the duration for which a current is applied during electrolysis, which directly affects the amount of substance deposited at the electrode. The script uses time in seconds to calculate the total charge and subsequently the mass of the deposited metal.

💡Mass (m)

Mass in the video refers to the amount of substance deposited or dissolved at an electrode during an electrochemical reaction. The script uses the mass of metals like copper and silver as examples to demonstrate the application of Faraday's Laws in calculating the outcome of electrolysis. The mass is calculated based on the charge passed and the valency of the metal ions.

💡Mole (mol)

A mole is a unit of measurement for the amount of substance, defined as the amount of a chemical substance that contains as many entities (such as atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12. In the video, the script uses moles to explain the relationship between the number of electrons transferred and the amount of substance deposited during electrolysis, as seen in the discussion of Faraday's constant.

💡Faraday's Constant (F)

Faraday's constant is approximately 96,485 C/mol and represents the amount of electric charge carried by one mole of electrons. The script mentions Faraday's constant in the context of calculating the mass of substances deposited during electrolysis. It is used to relate the charge passed through the electrolyte to the amount of substance deposited at the electrode.

Highlights

Introduction to the concept of electrolysis cells and the focus on discussing Faraday's Laws.

Explanation of why it is called Faraday's Law, named after the English scientist Michael Faraday.

Quantitative relationship between the amount of substance reacting and the electric charge involved.

How to determine the amount of substance reacting based on the electric charge used in a specific time frame during electrolysis.

Faraday's First Law states that the mass of a substance deposited on an electrode is directly proportional to the charge involved in the cell.

The formula m = x * f is introduced to calculate the mass of the substance produced.

The significance of the number 96,500 in the formula, representing the charge of one mole of electrons.

The formula m = e * (I * t) / (n * F) is explained, where m is mass, e is the equivalent mass of the substance, I is current, t is time, n is the valency, and F is Faraday's constant.

Practical application of Faraday's Law in determining the mass of metal deposited at the cathode during electrolysis.

Example problem: Calculating the mass of copper deposited at the cathode with a given current and valency.

Explanation of how to calculate the mass of a substance using the formula m = (I * t) / (n * F).

Example problem: Determining the mass of silver produced by a given current over a specific time in a silver nitrate solution.

Calculation of the required electrical current to produce a specific mass of magnesium from the electrolysis of molten MgCl2.

Example problem: Determining the relative atomic mass of a metal produced by electrolysis with given current, time, and mass of the metal.

Emphasis on understanding the application of Faraday's Law in solving electrolysis problems.

Encouragement for viewers to continue learning and exploring further videos on the topic.

Transcripts

Halo assalamualaikum warahmatullahi

wabarakatuh kembali lagi di channel

cerdas kimia teman-teman di video kali

ini kita akan melanjutkan pembahasan

kita mengenai sel elektrolisis dan kali

ini kita akan fokus membahas hukum

Faraday 1 nah Mengapa disebut hukum

Faraday karena hukum ini dicetuskan oleh

seorang ilmuwan Inggris ya seorang

ilmuwan Kimia Fisika Inggris yang

bernama Michael Faraday sehingga disebut

dengan hukum Faraday jadi teman-teman di

dalam sel elektrokimia yang kemarin

sudah kita pelajari bersama itu ternyata

Terdapat hubungan kuantitatif antara

jumlah zat yang bereaksi dengan muatan

listrik yang terlibat nah sehingga kita

itu bisa menentukan banyaknya zat yang

bereaksi

akan jumlah muatan listrik yang

digunakan dalam rentang waktu tertentu

Ya seperti kalau pada elektrolisis

kemarin itu kan di katoda terdapat

endapan logam ya dari proses

elektrolisis itu nah kita bisa

menentukan nih massa logam yang

terbentuk itu dari jumlah muatan listrik

yang digunakan dalam rentang waktu

tertentu pada saat kita melakukan

elektrolisis itu nah sehingga

berdasarkan hukum satu Faraday itu

parade mengemukakan bahwa dalam sel

elektrokimia massa zat yang diendapkan

pada suatu elektrode itu sebanding

dengan besarnya muatan listrik yang

terlibat di dalam sel jadi massa zat

yang dihasilkan tadi sebanding dengan

listrik yang digunakan atau listrik yang

terlibat sehingga ini bisa kita rumuskan

sebagai m = x * f

Hai atau ini kalau diuraikan lagi

menjadi m = e * high kali tv96 500 nah

angka 96 500 ini itu tidak muncul dengan

tiba-tiba ya teman-teman Wih angka 96

500 ini muncul karena hasil kali antara

muatan elektron dengan satu Mal

elektronnya 1 mol elektron muatan

elektron itu oleh milikan itu kan

besarnya 1,60 dua kali 10 pangkat min 19

sementara satu Mal itu kalian ingat ya

jadi 6602 kali 10 pangkat 23na hasil

kalinya ini itu dapatnya adalah sekitar

96500 jadi ini Angka 96 500 ini tidak

lain dan tidak bukan adalah menyatakan

satu mal

Hai elektron atau biasa disebut dengan

Molek yang Molek Throne oke nah kemudian

m itu adalah massa zat yang

dihasilkannya dalam gram nya kemudian e

e itu adalah massa ekivalen zat atau ini

bisa kita rumuskan dengan arper bilangan

oksidasi atau er prevalensi dari logam

itu kemudian ini adalah kuat arus

satuannya ampere dan t adalah waktu

satuannya detik jadi ketika kita nanti

mengerjakan soal-soal elektrolisis maka

aturannya untuk satuannya harus

disesuaikan ya Oke Supaya kita lebih

paham mengenai hukum Faraday 1 ini kita

langsung ke latihan soalnya Oke latihan

soal yang pertama arus listrik sebesar

satu Faraday dialirkan kedalam larutan

CuSO4 Tentukan massa tembaga yang di

akan pada katoda jika diketahui Ar Cu =

6 3,5 Nah dari sini ini kan diketahui

muatannya berarti efeknya ya F = 1

Faraday kemudian Arnya RC = 6 3,5

berarti kalau yang ditanya adalah masa

ya jadi masanya berapa maka jawabannya

masa sama dengan menggunakan rumus yang

tadi Edi kali F ya karena f atau

muatannya sudah apa muatan Listriknya

sudah diketahui berarti ini Eh ini sama

dengan arper belok dikalikan dengan F

Jadi airnya itu 63,5 beloknya nah gimana

cara kita tahu pilox jadi ini kan CuSO4

nah CuSO4 ini kalau diuraikan itu akan

menjadi cu2 plus-plus

Hai so4 2 minus sehingga pilot CEO di

dalam dewasa 4 adalah nyanyian muatannya

12 atau valensinya 12 sehingga ini kita

bagi dengan dua gitu kemudian dikali

dengan FBnya fb-nya tadi satu ya lalu

ketemu 63 setengah bagi2 Berarti 31,2 25

apa nih satuannya satuannya yaitu G ya

jadi massa tembaga yang dihasilkan pada

katoda adalah 31 koma tujuh puluh lima

gram kita ke soal yang kedua Berapakah

Massa perak yang dihasilkan oleh arus 10

ampere yang dilewatkan selama 10 menit

ke dalam larutan AgNO3 Oke jadi Disini

Hai diketahui adalah arosya arus

10ampere berarti ih sama dengan 10

ampere kemudian dilewatkan selama 10

menit berarti ini TT sama dengan 10

menit ingat teh itu harus dalam detik

berarti ini kalau dijadikan detik dikali

60 ya jadinya 600 Detik Nah air AG 108

oke nah kalo ada yg diketahui teh

diketahui Maka kalau disuruh mencari

masanya ya m ditanyakan berarti cara

mengerjakannya adalah masa sama dengan

Edi kali-kali TP 965 ratus nah enya itu

= Ar per bilok atau air Valencia

sekali-kali pepper 965

200 Ok saya lanjutkan di sebelah sini

berarti m = airnya 108 beloknya Oke

kalau AgNO3 ini berarti kan AgNO3 ini

menjadi AG + itu ya + no3 minus jadi

plus satu batia sudah bagi satu dia Iya

10 kemudian tehnya 600 per 965 ratus

jadi ini boleh dicoret dengan ini

sehingga ketemunya m yaitu = 6 koma 715

satuannya dan Oke jadi Massa perak yang

dihasilkan dari proses elektrolisis itu

sama dengan 6,7 15 G spam ya teman-teman

kita lanjut ke soal yang ketiga

Hai berapakah arus listrik yang

diperlukan untuk menghasilkan 0,05 kg

magnesium dari proses elektrolisis

leburan MgCl2 selama satu jam oke yang

diketahui dulu nih Apa nih berarti ini

akan Mas saya massa zat yang dihasilkan

berarti m-nya = 0,0 5 kg ingat Masa

harus dalam G berarti ini kalau

diizinkan g g kan kali 1000 berarti sama

dengan 50gram lalu tanya ya Tanyain ya

waktunya tv-nya adalah satu jam teh itu

harus dalam detik berarti ini 3600 Detik

oke yang ditanya adalah i ya yang

ditanyain berarti untuk menjawabnya kita

masukkan aja ke rumus SN = a

er pernah valensi di kali ini digali

t-perpus 96500 Gear nya berapa nih 24

beloknya berapa nih dari MgCl2 berarti

ini MgCl2 itu kan menjadi MG 2 + + 2hcl

minus berarti bloknya akan nyapres buaya

berarti ini per 2 dikali itu yang

ditanya-tanya 3600 kemudian dibagi

dengan 965 ratus Oh ya Emangnya tadi 50

Oke sehingga ini bisa kita coret ya Oke

selanjutnya di sebelah sampingnya ini

jadi 50 kali 965 jutaan kali silangkan

berarti 4

28 250 sama dengan ini 24 bagi2 12 i&d

kali 36 ya ketemunya kayak gini aja

pelan-pelan ya = 12 kali 3643 dua i-iii

nya sama dengan 482 50/4 32 jadi Isinya

= 111 koma enam puluh sembilan satunya

adalah ampere kejadian arus listrik yang

dibutuhkan untuk menghasilkan 50 gram

magnesium dari elektrolisis leburan

MgCl2 selama satu jam adalah 111.com

a609 ampere key yang keempat suatu

elektrolisis leburan garam sulfat

bervalensi dua dengan arus listrik

sebesar 9000

2656 peer selama 10 detik menghasilkan

endapan logam di katoda dengan massa 20

gram berapakah massa atom relatif dari

logam tersebut Oke jadi ini yang

ditanyakan adalah ardarini logam itu ya

gimana dalam soal ini itu diketahui

diantaranya adalah logamnya bervalensi

dua Berarti tadi ya valensinya dua

kemudian arus listriknya atau I sebesar

9650 ampere waktunya adalah 10 detik

lalu massa logam yang dihasilkan adalah

20gram di soal ini yang ditanya adalah

aer sehingga kita bisa menghitungnya

dengan mudah menggunakan rumus yang

Nadia bahwa m = e t

super 96 500m nya itu 20 jam

hai eh itu kan arper valensi valensinya

disitu ketahui dua ya jadi arper dua

kali Inya 9650 c-nya 10 detik per 965

ratus Nah ini kan dicoret ya berarti

Bisa langsung kita hitung bahwa airnya

sama dengan ya ini ya tinggal 20 kali

dua ya 20 kali gua berarti airnya sama

dengan 40 jadi ardarini logam tersebut

adalah 40 Oke jadi seperti itu

teman-teman mudah sekali ya teman-teman

jadi asalkan teman-teman Nah tadi Paham

bagaimana cara menggunakan rumus dari

hukum Faraday 1 ok teman-teman sampai

disini dulu video kita cowok nanti kita

lanjutkan ke video video selanjutnya

Terima kasih tetap semangat dan

warahmatullahi wabarakatuh