Electrolysis — Faraday's laws (m = ZIt), industrial electrolysis (NaCl, Al)

2.7.10 · Chemistry › Redox & Electrochemistry (Intro)

YEH KYUN ZAROORI HAI: Isi tarah hum aluminum ko ore (Al₂O₃) se nikaalte hain, brine se chlorine aur NaOH banate hain, metals ko electroplate karte hain, aur copper refine karte hain. Electrolysis ke bina, modern metallurgy aur chemical industry exist hi nahi karti.

Faraday's First Law — Quantitative Link

$m \propto Q$

Kyunki $Q = I \cdot t$ (current × time), toh hume milta hai:

$m = Z \cdot I \cdot t$

jahan:

$m$ = deposited/liberated mass (grams)
$Z$ electrochemical equivalent (g/C) — ek coulomb mein kitni mass deposit hoti hai
$I$ = current (amperes)
$t$ = time (seconds)

Step 1: Current kya measure karta hai? Current $I$ charge flow per second hai: $I = \frac{Q}{t}$ , isliye $Q = I \cdot t$ coulombs.

Step 2: Charge ka moles of electrons se kya relation hai? Electrons ka ek mole Faraday's constant $F = 96, 500$ C/mol (approximately). Agar charge $Q$ pass hota hai, toh electrons ke moles = $\frac{Q}{F} = \frac{I \cdot t}{F}$

Step 3: Electrons ka mass se kya relation hai? Ek metal ion $M^{n +} + n e^{-} \to M$ consider karo

$n$ electrons, $M$ ka 1 atom deposit karte hain
$M$ ke 1 mole ke liye $n$ moles of electrons chahiye
$M$ ke deposited moles = $\frac{moles of e ^{-}}{n} = \frac{I \cdot t}{n \cdot F}$

Step 4: Moles ko mass mein convert karo $m = moles \times M_{molar} = \frac{I \cdot t}{n \cdot F} \times M$

Step 5: Electrochemical equivalent define karo $Z = \frac{M}{n \cdot F}$

Yeh mass hai jo ek coulomb mein deposit hoti hai. Isliye:

$m = Z \cdot I \cdot t = \frac{M}{n \cdot F} \cdot I \cdot t$

YEH FORMULA KYUN KAAM KARTA HAI: Jitne bhi electrons flow karte hain, woh ek ion ka ek unit charge reduce/oxidize karte hain. Zyada current → zyada electrons → zyada atoms deposited. Factor $\frac{M}{n F}$ "coulombs" ko "is specific substance ke grams" mein convert karta hai.

Faraday's Second Law — Different Substances ki Comparison

$\frac{m _{1}}{m _{2}} = \frac{M _{1} / n _{1}}{M _{2} / n _{2}} = \frac{Z _{1}}{Z _{2}}$

KYUN: Utne hi electrons se alag-alag substances ki proportional amounts banani padti hain, jo scale hoti hain is baat se ki har ek ko kitne electrons chahiye aur uski atomic mass kitni hai.

Solution:

$I = 2.5$ A
$t = 40 \times 60 = 2400$ s
$F = 96, 500$ C/mol
$n = 2$ (kyunki $Cu^{2 +} + 2 e^{-} \to Cu$ )

$m = \frac{M \cdot I \cdot t}{n \cdot F} = \frac{63.5 \times 2.5 \times 2400}{2 \times 96500}$

$m = \frac{381 , 000}{193 , 000} = 1.97 g$

Yeh step kyun? Hum count kar rahe hain: (2.5 C/s)(2400 s) = 6000 C total. Yeh hai $\frac{6000}{96500} = 0.0622$ mol $e^{-}$ . Kyunki 2 electrons se 1 Cu atom banta hai, hume milta hai $\frac{0.0622}{2} = 0.0311$ mol Cu = $0.0311 \times 63.5 = 1.97$ g.

Solution: $\frac{m _{Cu}}{m _{Ag}} = \frac{M _{Cu} / n _{Cu}}{M _{Ag} / n _{Ag}} = \frac{63.5/2}{108/1} = \frac{31.75}{108}$

$m_{Cu} = 1.08 \times \frac{31.75}{108} = 0.317 g$

Kyun? Same electrons pass hue. Ag ko ek electron chahiye per atom (electrons ke hisaab se halka), Cu ko 2 electrons chahiye per atom (electrons ke hisaab se bhaari lekin zyada electrons maangta hai), isliye Cu ki kam mass deposit hoti hai.

Industrial Electrolysis ke Applications

Figure — Electrolysis — Faraday's laws (m = ZIt), industrial electrolysis (NaCl, Al)

1. Chlor-Alkali Process (Brine ka Electrolysis)

Setup: Brine, inert electrodes (aksar Ti/graphite) wale electrolytic cell mein se flow karti hai.

Anode par (oxidation): $2 Cl^{-} \to Cl_{2} + 2 e^{-}$ Chloride ions electrons chodti hain → chlorine gas bubbles nikalti hai.

Cathode par (reduction): $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}$ Paani reduce hota hai (Na⁺ nahi, kyunki paani reduce karna zyada aasaan hai) → hydrogen gas + hydroxide ions.

Overall Reaction: $2 NaCl + 2 H_{2} O electrolysis Cl_{2} + H_{2} + 2 NaOH$

YEH PRODUCTS KYUN?

Cl⁻ ko anode par paani se zyada aasaan oxidize kiya ja sakta hai (lower oxidation potential)
Cathode par Na⁺ se paani ko reduce karna zyada aasaan hai (Na⁺ ke liye bahut zyada voltage chahiye)
Peeche bache Na⁺ aur OH⁻ milke NaOH solution banate hain

Industrial Importance:

Chlorine: PVC plastics, disinfectants, solvents
Sodium hydroxide: Paper, soap, chemical manufacturing
Hydrogen: Ammonia synthesis, fuel

Cells ke Types:

Diaphragm cell: Porous barrier Cl₂ aur NaOH ko mix hone se rokta hai (warna NaOCl ban jaata)
Membrane cell: Ion-exchange membrane (modern, zyada efficient)
Mercury cell: Hg cathode use karta hai (toxicity ki wajah se phase out ho raha hai)

Solution: Reaction: $2 Cl^{-} \to Cl_{2} + 2 e^{-}$ Har Cl₂ molecule ke liye 2 electrons chahiye, isliye $n = 2$ .

$t = \frac{m \cdot n \cdot F}{M \cdot I} = \frac{10 , 000 \times 2 \times 96 , 500}{71 \times 50}$

$t = \frac{1 , 930 , 000 , 000}{3 , 550} = 543, 661 s = 151 hours$

Kyun? 10 kg = 10,000/71 = 140.85 mol Cl₂. Har mol ke liye 2 mol electrons chahiye = 281.7 mol $e^{-}$ . Yeh hai $281.7 \times 96, 500 = 27, 184, 050$ C. 50 A par, time = 27,184,050/50 = 543,681 s ≈ 151 hours.

2. Aluminum ka Extraction (Hall-Héroult Process)

AL₂O₃ ko sirf heat kyun nahi karte? Aluminum bahut zyada reactive hai—yeh oxygen ko bahut mazbooti se pakde rehta hai. Chemical reduction ke liye ek aur bhi zyada reactive metal chahiye. Electrolysis hi ek practical method hai.

Setup:

Electrolyte: Al₂O₃ jo molten cryolite mein dissolve hai (~1000°C)
- Pure Al₂O₃ 2072°C par melt hota hai (bahut zyada, bahut expensive)
- Cryolite melting point ~1000°C tak ghata deta hai aur electricity conduct karta hai
Cathode: Carbon-lined steel cell (lining cathode ki tarah kaam karti hai)
Anode: Multiple carbon (graphite) rods jo bath mein dip hain

Cathode par (reduction): $Al^{3 +} + 3 e^{-} \to Al$ Molten aluminum neeche collect hota hai (cryolite se bhaari, metal hone ke bawajood).

Anode par (oxidation): $2 O^{2 -} \to O_{2} + 4 e^{-}$

Oxygen turant carbon anode ke saath react karti hai: $C + O_{2} \to CO_{2}$

CARBON ANODES KYUN? Yeh saste hain, acche conductor hain, aur CO₂ reaction se anode dheere-dheere "consume" hota hai, isliye anodes periodically replace karne padte hain—lekin isse O₂ accumulation nahi hoti.

Overall Reaction: $2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}$

Energy Cost: Aluminum production bahut zyada energy-intensive hai—lagbhag 15 kWh per kg Al. Isliye Al smelters sasti hydroelectric power ke paas banaye jaate hain.

Solution: $m = \frac{M \cdot I \cdot t}{n \cdot F} = \frac{27 \times 150 , 000 \times ( 24 \times 3600 )}{3 \times 96 , 500}$

$m = \frac{27 \times 150 , 000 \times 86 , 400}{289 , 500} = \frac{349 , 920 , 000 , 000}{289 , 500} = 1, 208, 839 g \approx 1209 kg$

Kyun? Charge = $150, 000 \times 86, 400 = 1.296 \times 1 0^{10}$ C. Yeh hai $\frac{1.296 \times 1 0 ^{10}}{96 , 500} = 134, 300$ mol $e^{-}$ . Kyunki 3 electrons se 1 Al atom banta hai, hume milta hai $\frac{134 , 300}{3} = 44, 767$ mol Al = $44, 767 \times 27 = 1, 209$ kg.

Common Mistakes aur Unhe Theek Karne ke Tarike

Kyun sahi lagta hai: 2+ charge aur 2 electrons same number hain.

Asal mein kya ho raha hai: $n$ half-reaction mein transfer hone wale electrons ki sankhya hai, sirf charge nahi. $Cu^{2 +} + 2 e^{-} \to Cu$ ke liye, hume 2 electrons chahiye, isliye $n = 2$ . Lekin $Fe^{3 +} + 3 e^{-} \to Fe$ ke liye, $n = 3$ . Electrons count karne ke liye hamesha half-reaction likhein.

Fix: Pehle reduction/oxidation half-equation likhein, electrons explicitly count karein.

Kyun sahi lagta hai: Practice mein hum aksar time minutes ya hours mein measure karte hain.

Asal mein kya ho raha hai: Formula mein $I$ amperes (C/s) mein aur $t$ seconds mein chahiye. Agar minutes use karein, toh answer 60 ke factor se galat ho jaayega.

Fix: Calculation se pehle hamesha time ko seconds mein convert karein. $t_{seconds} = t_{minutes} \times 60 = t_{hours} \times 3600$ .

Kyun sahi lagta hai: Same charge se lagta hai same kaam hona chahiye.

Asal mein kya ho raha hai: Faraday's Second Law kehta hai ratio $M / n$ par depend karta hai. Halke metals with high charge (jaise Al, $M / n = 27/3 = 9$ ) same charge ke liye heavy metals with low charge (jaise Ag, $M / n = 108/1 = 108$ ) se kam mass deposit karte hain.

Fix: Alag substances compare karte waqt hamesha molar mass aur valency dono account karo.

Kyun sahi lagta hai: Hum ek sodium salt electrolyze kar rahe hain, sodium baahir aana chahiye.

Asal mein kya ho raha hai: Electrode potential decide karta hai ki asal mein kya reduce hoga. Paani ka reduction potential Na⁺ se kam negative hai, matlab paani ko Na⁺ se zyada aasaan reduce kiya ja sakta hai. Cell easier path leta hai jab tak bahut zyada voltage force na kare.

$Na^{+} + e^{-} \to Na E^{\circ} = - 2.71 V (bahut negative—mushkil)$ $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-} E^{\circ} = - 0.83 V (aasaan)$

Fix: Yaad rakho ki aqueous solution mein cathode par woh species pehle reduce hoti hai jiska reduction potential kam negative (zyada positive) ho.

Connections & Extensions

Galvanic Cells and Standard Electrode Potentials — Electrolysis, galvanic cells ka ulta hai; hum spontaneous reaction ko reverse karne ke liye voltage apply karte hain
Nernst Equation — Electrode potentials decide karte hain ki har electrode par asal mein kaunsi species react karti hai
Thermodynamics of Electrochemical Cells — Electrolysis ke liye minimum voltage = reverse galvanic reaction ka $- E_{cell}^{\circ}$
Redox Reactions and Balancing — Har electrolysis ek redox reaction hai jo external voltage se force ki jaati hai
Industrial Chemistry — Chlor-alkali aur aluminum extraction multi-billion dollar industries hain
Corrosion and Electrochemical Protection — Sacrificial anodes, structures protect karne ke liye electrolysis principles ko ulta use karte hain

Recall Ek 12-saal ke bachche ko samjhao

Socho bijli bahut saare chhote workers (electrons) ki tarah hai jo wire mein daud rahe hain. Jab woh liquid mein metal tak pahunchte hain, toh woh atoms ko sath chipkane ya todhne mein madad kar sakte hain. Faraday ne ek simple rule pata kiya: jitne zyada workers bhejo (current) aur jitni der woh kaam karein (time), utna hi zyada kuch banega ya tootega (mass). Yeh LEGO jaisa hai—agar 100 workers 1 ghante ke liye bana rahe hain, toh ek certain size ka castle milega. Bada castle chahiye toh 200 workers ko 1 ghante ke liye bhejo, ya 100 workers ko 2 ghante ke liye. Formula $m = Z I t$ bas yahi rule maths mein likha hua hai.

Ab, alag-alag metals ko ek atom ke liye alag-alag workers chahiye. Aluminum zidd hai—ise 3 workers chahiye per atom. Silver aasaan hai—sirf 1 worker per atom. Isliye same kaam se silver zyada banta hai aluminum se.

Factories mein, hum yeh trick use karte hain aluminum banane ke liye (soda cans aur airplanes ke liye) aur chlorine banane ke liye (pools saaf karne ke liye). Hum aluminum wali chattan dissolve karte hain, bahut zyada garam karte hain, aur BAHUT BADI amount of electricity se zap karte hain. Aluminum pighal kar neeche baith jaata hai jahan hum ise collect karte hain. Isme itni zyada electricity lagti hai ki yeh factories bade baandhon ke paas banai jaati hain!

$Z = M / (n F)$ ke liye: "My Naughty Friend" (Molar mass, valency, Faraday constant)

Chlor-alkali ke products ke liye: "Children Hate Naughty behavior" → Cl₂, H₂, NaOH

Aluminum extraction ke liye: "Carbon Anodes Consume Oxygen, Creating CO₂" → Anode par, C + O₂ → CO₂

#flashcards/chemistry

Electrolysis kya hai? :: Electrical energy use karke ek non-spontaneous redox reaction ko force karna.

Faraday's First Law words mein batao.

Electrode par deposited ya liberated substance ki mass seedha proportional hoti hai electrolyte mein se pass hone wali electricity ki quantity (charge) ke.

Electrolysis mein deposited mass ka formula likhein sab terms define karke.

m = Z \cdot I \cdot t = \frac{M}{n F} \cdot I \cdot t

jahan

m

= mass (g),

Z

= electrochemical equivalent (g/C),

I

= current (A),

t

= time (s),

M

= molar mass (g/mol),

n

= electrons transferred,

F

= Faraday constant (96,500 C/mol).

Electrochemical equivalent

Z

kya hai?

Z = \frac{M}{n F}

— pass hone wale charge ke ek coulomb mein deposit hone wali substance ki mass, jo molar mass, valency, aur Faraday's constant par depend karti hai.

Faraday's Second Law batao.

Jab same quantity of electricity alag-alag electrolytes mein se pass hoti hai, toh deposited masses unke chemical equivalents (

M / n

) ke ratio mein hoti hain.

Faraday's constant

F

kya hai aur uski value kya hai?

Electrons ke ek mole par charge, lagbhag 96,500 coulombs per mole.

Chlor-alkali process mein kaun se teen products bante hain?

Chlorine gas (Cl₂), hydrogen gas (H₂), aur sodium hydroxide (NaOH).

Chlor-alkali electrolysis mein anode reaction likhein.

2 Cl^{-} \to Cl_{2} + 2 e^{-}

(chloride ions ka chlorine gas mein oxidation).

Chlor-alkali electrolysis mein cathode reaction likhein.

2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}

(paani ka hydrogen gas aur hydroxide ions mein reduction).

Brine electrolysis mein cathode par Na⁺ ki jagah paani kyun reduce hota hai?

Paani ka reduction potential kam negative (zyada positive) hai (

- 0.83

V) Na⁺ (

- 2.71

V) se, isliye paani reduce karna zyada aasaan hai aur preferentially hota hai.

Chlor-alkali electrolysis ki overall reaction kya hai?

2 NaCl + 2 H_{2} O electrolysis Cl_{2} + H_{2} + 2 NaOH

Hall-Héroult process kya hai? :: Alumina (Al₂O₃) jo molten cryolite mein dissolve hai, usse aluminum nikaalने ka industrial electrolytic process.

Aluminum extraction mein cryolite kyun use kiya jaata hai?

Pure Al₂O₃ 2072°C par melt hota hai (bahut zyada); cryolite melting point ~1000°C tak ghata deta hai aur mixture ko electrically conductive banata hai.

Hall-Héroult process mein cathode reaction likhein.

Al^{3 +} + 3 e^{-} \to Al

(aluminum ions ka liquid aluminum metal mein reduction).

Hall-Héroult process mein anode reaction likhein.

2 O^{2 -} \to O_{2} + 4 e^{-}

ke baad

C + O_{2} \to CO_{2}

(oxide ions ka oxidation aur carbon anode se reaction).

Aluminum extraction mein carbon anodes kyun use kiye jaate hain?

Yeh saste hain, bijli acche se conduct karte hain, aur produce hone wali oxygen ke saath react karke CO₂ banate hain, oxygen buildup rok kar (haalaanki anodes periodically replace karne padte hain).

Hall-Héroult process mein overall reaction kya hai?

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

Aluminum extraction itna energy-intensive kyun hai?

Aluminum ko ek atom ke liye 3 electrons chahiye aur zyada temperature bhi; ek kg Al produce karne mein lagbhag 15 kWh lagti hai.

CuSO₄ mein se 5 A current 1930 seconds ke liye pass karne par copper ke kitne grams deposit honge? (

M_{Cu} = 63.5

n = 2

)

m = \frac{63.5 \times 5 \times 1930}{2 \times 96500} = 3.175

Agar 10.8 g Ag deposit hota hai, toh same charge se kitna Al deposit hoga? (

M_{Ag} = 108

n_{Ag} = 1

;

M_{Al} = 27

n_{Al} = 3

)

\frac{m _{Al}}{m _{Ag}} = \frac{M _{Al} / n _{Al}}{M _{Ag} / n _{Ag}} = \frac{27/3}{108/1} = \frac{9}{108} = \frac{1}{12}

use karke,

m_{Al} = \frac{10.8}{12} = 0.9