Spontaneity from E°_cell and ΔG = −nFE

2.7.5 · Chemistry › Redox & Electrochemistry (Intro)

Core Question

Hum kaise predict karte hain ki koi redox reaction ek electrochemical cell mein spontaneously hogi ya nahi, aur electrochemistry aur thermodynamics ke beech ka bridge kya hai?

[!intuition] The Big Picture

Socho tum ek boulder ko dhakka de rahe ho: agar woh khud neechay ki taraf lud jaaye, toh woh spontaneous hai (energy release hoti hai). Agar tumhe use upar dhakka dena pade, toh woh non-spontaneous hai (energy input chahiye).

Redox reactions mein, electrons flow karte hain reducing agent se oxidizing agent ki taraf. Agar yeh flow energy release karta hai (jaise boulder neechay lud ta hai), toh reaction spontaneous hai. Cell potential E° us "voltage hill" ko measure karta hai — positive E° matlab neechay ki taraf (spontaneous), negative matlab upar (forced). Gibbs free energy ΔG thermodynamics ka spontaneity measure karta hai: negative ΔG = spontaneous, positive = non-spontaneous.

Equation ΔG = −nFE in dono worlds ko link karti hai: electrochemistry ka voltage aur thermodynamics ki free energy.

[!definition] Key Terms

• E°_cell (Standard Cell Potential): Cathode aur anode ke beech voltage difference, standard conditions (1 M, 1 atm, 25°C) par, measured in volts (V).
• ΔG (Gibbs Free Energy Change): Kaam karne ke liye available energy; spontaneity batata hai. Units: joules (J) per mole of reaction as written (ya kJ divided by 1000). Kyunki n "moles of electrons per mole of reaction" hai, result energy per mole of reaction hoti hai.
• n: Balanced redox equation mein transfer hone wale moles of electrons ki sankhya.
• F (Faraday's Constant): 1 mole electrons ka charge 96,485 C/mol ≈ 96,500 C/mol.
• Spontaneous Reaction: Bina bahari energy ke khud hoti hai (ΔG < 0, E° > 0).

[!formula] The Bridge Equation — Derived from First Principles

WHY does this equation exist?

Kyunki electrical work aur chemical free energy ek hi energy change ke do alag views hain.

DERIVATION (with consistent sign conventions):

Step 1: Electrical work done BY the cell Jab cell spontaneously operate karta hai, toh woh charge ko external circuit mein push karta hai, surroundings par useful electrical work karta hai.

• Charge transported = q = nF (n moles of e⁻, har mole charge F coulombs carry karta hai)
• Cell dwara surroundings par potential E ke through kiya gaya work: $w_{bycell}=qE=nFE$

Why positive? Ek spontaneous cell ke liye, E > 0 hota hai aur cell surroundings ko energy deliver karta hai, isliye work output positive hoti hai.

Step 2: Connect to thermodynamics Constant temperature aur pressure par, system dwara surroundings par ki ja sakne wali maximum non-PV (useful) work Gibbs free energy ki decrease ke barabar hoti hai: $w_{bysystem,max}=-\Delta G$

Why? ΔG "useful" energy budget represent karta hai. Jab reaction spontaneously hoti hai, ΔG decrease karta hai (ΔG < 0), aur woh lost free energy usable work ban jaati hai. Toh system jo work deliver karta hai woh +(−ΔG) = −ΔG hai.

Step 3: Equate the two (both are "work done BY the system") Cell dwara deliver ki gayi electrical work = thermodynamics ki allowed maximum useful work: $nFE=-\Delta G$

Dono sides ko −1 se multiply karo: $\boxed{\Delta G=-nFE}$

Key yeh tha ki dono quantities ko by the system on the surroundings kiya gaya work define kiya, isliye algebra consistent hai.

Under Standard Conditions:

$\boxed{\Delta G°=-nFE{°}_{cell}}$

Jahaan:

• ΔG° = standard free energy change (per mole of reaction as written)
• E°_cell = standard cell potential

[!example] Example 1: Predicting Spontaneity from E°

Given: $Zn(s)+C{u}^{2+}(aq)\to Z{n}^{2+}(aq)+Cu(s)$ Standard reduction potentials:

• E°(Cu²⁺/Cu) = +0.34 V
• E°(Zn²⁺/Zn) = −0.76 V

Find: Kya yeh reaction spontaneous hai? ΔG° calculate karo.

Solution:

Step 1: E°_cell calculate karo $E{°}_{cell}=E{°}_{cathode}-E{°}_{anode}$ Cu²⁺ reduce ho raha hai (cathode), Zn oxidize ho raha hai (anode): $E{°}_{cell}=(+0.34)-(-0.76)=+1.10\text{ V}$

Why this step? E°_cell > 0 matlab reaction spontaneous hai.

Step 2: n dhundo Balanced equation dikhata hai ki Zn 2e⁻ kho deta hai, Cu²⁺ 2e⁻ gain karta hai: $n=2$

Step 3: ΔG° calculate karo $\Delta G°=-nFE{°}_{cell}$ $\Delta G°=-(2)(96,485)(1.10)$ $\Delta G°=-212,267\text{ J}=-212.3\text{ kJ (per mole of reaction)}$

Why this step? ΔG° < 0 thermodynamically spontaneity confirm karta hai.

Answer: Reaction spontaneous hai; ΔG° = −212.3 kJ per mole of reaction.

[!example] Example 2: Non-Spontaneous Reaction

Given: ${\text{Cu(s) + Zn}}^{2+}{\text{(aq) → Cu}}^{2+}\text{(aq) + Zn(s)}$

Find: Kya yeh spontaneous hai? ΔG° calculate karo.

Solution:

Step 1: E°_cell calculate karo Ab Zn²⁺ reduce ho raha hai (cathode), Cu oxidize ho raha hai (anode): $E{°}_{cell}=(-0.76)-(+0.34)=-1.10\text{ V}$

Why this step? E°_cell < 0 seedha batata hai: non-spontaneous.

Step 2: n = 2 (same electron transfer)

Step 3: ΔG° $\Delta G°=-(2)(96,485)(-1.10)=+212.3\text{ kJ (per mole of reaction)}$

Answer: Non-spontaneous; proceed karne ke liye external energy chahiye (jaise ek battery).

KEY INSIGHT: Reaction ko reverse karne par E° aur ΔG° dono ke signs flip ho jaate hain.

[!example] Example 3: Multi-Electron Transfer

Given: $2\text{Al}(s)+3{\text{Cu}}^{2+}(aq)\to 2{\text{Al}}^{3+}(aq)+3\text{Cu}(s)$

• E°(Al³⁺/Al) = −1.66 V
• E°(Cu²⁺/Cu) = +0.34 V

Find: ΔG°.

Solution:

Step 1: E°_cell $E{°}_{cell}=(+0.34)-(-1.66)=+2.00\text{ V}$

Step 2: n dhyaan se dhundo

• Har Al 3e⁻ kho deta hai → 2 Al atoms 6e⁻ kho dete hain
• Har Cu²⁺ 2e⁻ gain karta hai → 3 Cu²⁺ 6e⁻ gain karte hain $n=6$

Why this step? Balanced equation mein transfer hone wale total electrons count karne chahiye.

Step 3: ΔG° $\Delta G°=-(6)(96,485)(2.00)=-1,157,820\text{ J}\approx -1158\text{ kJ (per mole of reaction)}$

Answer: Highly spontaneous (large negative ΔG°).

[!mistake] Common Mistakes & How to Steel-Man Them

Mistake 1: "Maine n = 1 use kiya jab equation mein 2e⁻ transfer ho rahe the"

Why it feels right: Tumne "2 electrons" dekha aur socha "n = 2" bina balanced equation ki stoichiometry check kiye.

Steel-man: Tum electron flow identify karne ki koshish kar rahe ho, jo sahi hai. Error yeh hai ki tumne saare stoichiometric coefficients ke across total electrons track nahi kiye.

Fix: Half-reactions balance karne ke baad, ek half-reaction mein electrons count karo (woh barabar hote hain). Agar equation atoms balance karne ke liye multiply ki gayi hai, toh electrons bhi multiply ho jaate hain. 2Al → 2Al³⁺ + 6e⁻ mein, n = 6 hai, 3 nahi.

Mistake 2: "E° = +0.5 V tha lekin mujhe ΔG° positive mila"

Why it feels right: Tumne ΔG = −nFE mein negative sign bhool gaye, toh ΔG = +nFE calculate kar liya.

Steel-man: Tumne formula structure sahi use kiya lekin yeh miss kar gaye ki spontaneous E° ko zaroor negative ΔG dena chahiye, jo minus sign ke bina nahi hota.

Fix: Sign memorize karo: ==ΔG = **−**nFE==. Positive E° → negative ΔG → spontaneous.

Mistake 3: "Maine F = 96.5 use kiya 96,485 ki jagah"

Why it feels right: "96,500" rounded dekha aur units drop kar diye, socha F ≈ 96.5 kJ hai.

Steel-man: Tumne simplify karne ki koshish ki, jo achha instinct hai, lekin magnitude confuse ho gayi.

Fix: F = 96,485 C/mol, hamesha coulombs mein. Agar kJ chahiye, toh convert karna padega: 1 V·C = 1 J, isliye final answer ko 1000 se divide karo kJ paane ke liye.

Mistake 4: "E°_anode − E°_cathode use kiya"

Why it feels right: E°_cell ka formula confuse ho gaya.

Fix: ==E°_cell = E°_cathode − E°_anode==. Cathode reduction hai (electron gain), hamesha anode (oxidation) potential subtract karo. Mnemonic: Cathode Comes first, Reduction is Right.

Mistake 5: "Derivation mein sign confusion"

Why it feels right: Log kabhi kabhi electrical work ko −nFE likhte hain (work done on the system) aur phir use w_max = −ΔG (work done by the system) ke saath equate karte hain, do alag sign conventions mix kar ke.

Fix: Dono quantities ko ek hi ledger side par rakho. Dono ko work done BY the system define karo: electrical work delivered = nFE, aur max useful work = −ΔG. Phir nFE = −ΔG cleanly ΔG = −nFE deta hai.

[!recall]- Feynman Explanation (Explain to a 12-year-old)

Theek hai, socho tumhare paas do metals hain, jaise zinc aur copper, apne ions ke saath paani mein baithe hain. Zinc sachchi electrons dena chahta hai, aur copper ions sachchi electrons lena chahte hain.

Jab tum unhe ek wire se connect karte ho, electrons zinc se copper ki taraf flow karte hain — woh electricity hai! Voltage (E°) waise hai jaise "electron slide" kitni steep hai. Agar slide neechay ki taraf jaati hai (positive E°), toh electrons khud slide karte hain — woh spontaneous hai. Agar upar ki taraf hai (negative E°), toh tumhe unhe push karna padega (ek battery use karni padegi).

Ab, scientists energy "Gibbs free energy" (ΔG) se bhi measure karte hain — socho ise "hume kitni useful energy milegi" ki tarah. Magic formula ΔG = −nFE dono ko connect karta hai: voltage (E) batata hai exactly kitni energy (ΔG) milti hai har flow karne wale electron par. "n" kitne electrons move karte hain woh hai, aur "F" bas ek number hai jo sab kuch sahi units mein convert karta hai.

Bottom line: Positive voltage = electrons freely flow karte hain = reaction hoti hai = negative ΔG. Yeh same cheez hai, bas alag alag tarike se measure ki gayi!

[!mnemonic] Memory Aid

"Negative G, Positive E, Reaction Runs Free"

• ΔG < 0 → spontaneous
• E° > 0 → spontaneous
• Formula: ΔG = −nFE → negative sign flip karta hai relationship ko

n ke liye: Balanced half-reaction mein electrons count karo.

F = 96,485: Socho "F for Faraday, ek mole charge ke liye 96 hazaar coulombs."

Connections

• Standard Electrode Potentials — E° values kahaan se aate hain
• Nernst Equation — concentration ke saath E aur ΔG kaise change hote hain (non-standard)
• Gibbs Free Energy Fundamentals — spontaneity ka thermodynamic basis
• Galvanic vs Electrolytic Cells — spontaneous (ΔG < 0) vs driven (ΔG > 0)
• Relationship between K_eq and ΔG° — ΔG° = −RT ln K equilibrium se link karta hai
• Faraday's Laws of Electrolysis — F use karke quantitative electrochemistry

Flashcards

Concept Map