Elektrolisis bagian 2 -HUKUM FARADAY 1
Summary
TLDRThis 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
🔬 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.
🔍 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.
🔋 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.
Mindmap
Keywords
💡Electrolysis
💡Faraday's Laws
💡Electrode
💡Charge
💡Copper Sulfate (CuSO4)
💡Valency
💡Current (I)
💡Time (t)
💡Mass (m)
💡Mole (mol)
💡Faraday's Constant (F)
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
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