Standard electrode potentials — SHE reference, electrochemical series

2.7.2 · Chemistry › Redox & Electrochemistry (Intro)

Overview

Different half-reactions ki "driving force" compare karne ke liye, humein ek universal reference electrode ke against measure kiya gaya standard electrode potential ($E°$) chahiye. Standard Hydrogen Electrode (SHE) zero-reference point ka kaam karta hai, aur baaki saare electrode potentials isi ke relative measure kiye jaate hain. Electrochemical series in potentials ko most negative (strongest reducing agents) se most positive (strongest oxidizing agents) tak rank karti hai.

[!intuition] Hume Reference Electrode Ki Zaroorat Kyun Hai

Problem: $C{u}^{2+}+2e^{-}\to Cu$ jaisi ek half-reaction mein inherent "electron-pulling power" hoti hai, lekin hum ise isolation mein measure nahi kar sakte—complete circuit chahiye. Har measurement mein DO half-cells hote hain.

Solution: EK half-reaction ko universal reference ke roop mein chuno aur uska potential exactly zero define karo. Phir baaki sabko isi reference ke relative measure karo. Yeh bilkul waise hai jaise altitude ke liye sea level ko zero-point set karte hain—saari heights usi standard ke relative measure hoti hain.

SHE kyun?: Hydrogen electrode yeh hai:

• Reproducible: Pure H₂ gas, unit activity par H⁺ ions ke saath set up karna aasaan hai
• Chemically simple: Sirf H⁺/H₂ involve hai, koi complex ions nahi
• Historically established: 20th century ki shuruaat se universal convention hai

[!definition] Standard Hydrogen Electrode (SHE)

Standard Hydrogen Electrode mein yeh hota hai:

• Ek platinum electrode (inert, reaction ko catalyze karta hai)
• Pure H₂ gas 1 bar pressure par electrode ke upar bubbling karti hui
• Aqueous solution jisme $[{\text{H}}^{+}]=1\text{ M}$ ho (unit activity)
• Temperature: 25°C (298 K)

Half-reaction: ${\text{2H}}^{+}\text{ (aq, 1M)}+2e^{-}\rightleftharpoons {\text{H}}_2\text{ (g, 1 bar)}$

Defined potential: $E{°}_{{\text{H}}^{+}/{\text{H}}_2}=0.00\text{ V (by convention)}$

Yeh ek arbitrary zero-reference hai, koi physical measurement nahi. Baaki saare electrode potentials ek half-cell ko SHE se couple karke measure kiye jaate hain.

[!formula] Standard Electrode Potentials Measure Karna

Setup

${\text{Zn}}^{2+}+2e^{-}\to \text{Zn}$ jaisi half-reaction ka $E°$ find karne ke liye:

1. Left half-cell: SHE (reference, $E°=0.00\text{ V}$)
2. Right half-cell: Zn electrode $[{\text{Zn}}^{2+}]=1\text{ M}$ mein
3. Connect karo salt bridge se aur cell potential $E{°}_{\text{cell}}$ measure karo

Cell notation: ${\text{Pt(s) | H}}_2{\text{(g, 1bar) | H}}^{+}{\text{(1M) || Zn}}^{2+}\text{(1M) | Zn(s)}$

First Principles Se Derivation

Cell potential cathode (reduction) aur anode (oxidation) potentials ka difference hai:

$E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$

Zn/SHE cell ke liye:

• Observation: Zn dissolve ho raha hai (oxidation), H₂ ban rahi hai (reduction)
• To: Cathode = SHE, Anode = Zn

$E{°}_{\text{cell}}=E{°}_{{\text{H}}^{+}/{\text{H}}_2}-E{°}_{{\text{Zn}}^{2+}/\text{Zn}}$

Kyunki $E{°}_{{\text{H}}^{+}/{\text{H}}_2}=0.00\text{ V}$:

$E{°}_{\text{cell}}=0.00-E{°}_{{\text{Zn}}^{2+}/\text{Zn}}$

Measurement se $E{°}_{\text{cell}}=+0.76\text{ V}$ milta hai, isliye:

$E{°}_{{\text{Zn}}^{2+}/\text{Zn}}=-0.76\text{ V}$

Negative kyun?: Zn half-cell anode ki tarah kaam karta hai (electrons khota hai). Negative $E°$ ka matlab hai reduction non-spontaneous hai SHE ke relative—Zn oxidize hona prefer karta hai.

General Formula

Kisi bhi half-cell X ke liye jo SHE se coupled ho:

$E{°}_{\text{X}}=E{°}_{\text{cell}}\text{ (agar X cathode hai)}$ $E{°}_{\text{X}}=-E{°}_{\text{cell}}\text{ (agar X anode hai)}$

Key insight: $E{°}_{\text{cell}}$ ka sign batata hai kaun sa electrode oxidize hota hai. Positive $E{°}_{\text{cell}}$ → jaise likha hai waise spontaneous; negative → reverse spontaneous hai.

[!example] Worked Example 1: Copper Ka $E°$ Measure Karna

Setup: ${\text{Pt | H}}_2{\text{(1 bar) | H}}^{+}{\text{(1M) || Cu}}^{2+}\text{(1M) | Cu}$

Observation: H₂ ke bubbles gayab ho jaate hain, Cu electrode par Cu deposit ho jaata hai.

• Oxidation (anode): ${\text{H}}_2\to 2{\text{H}}^{+}+2e^{-}$ Pt par
• Reduction (cathode): $C{u}^{2+}+2e^{-}\to Cu$ Cu par

Measured cell potential: $E{°}_{\text{cell}}=+0.34\text{ V}$

Calculation: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$ $+0.34=E{°}_{{\text{Cu}}^{2+}/\text{Cu}}-0.00$ $E{°}_{{\text{Cu}}^{2+}/\text{Cu}}=+0.34\text{ V}$

Yeh step kyun? Cu electrode cathode hai (wahan reduction hoti hai), isliye $E{°}_{\text{cathode}}=E{°}_{{\text{Cu}}^{2+}/\text{Cu}}$. Positive value ka matlab hai Cu²⁺ reduction H⁺ reduction ke relative spontaneous hai.

[!example] Worked Example 2: Spontaneous Direction Predict Karna

Question: Kya ${\text{Fe}}^{3+}$, ${\text{I}}^{-}$ ko standard conditions mein ${\text{I}}_2$ mein oxidize karega?

Given: $E{°}_{{\text{Fe}}^{3+}/{\text{Fe}}^{2+}}=+0.77\text{ V}$, $E{°}_{{\text{I}}_2/{\text{I}}^{-}}=+0.54\text{ V}$

Proposed reaction: $2{\text{Fe}}^{3+}+2{\text{I}}^{-}\to 2{\text{Fe}}^{2+}+{\text{I}}_2$

Step 1: Half-reactions identify karo

• Reduction: ${\text{Fe}}^{3+}+e^{-}\to {\text{Fe}}^{2+}$ ($E°=+0.77\text{ V}$)
• Oxidation: $2{\text{I}}^{-}\to {\text{I}}_2+2{\text{e}}^{-}$ ($E°=+0.54\text{ V}$ ka reverse)

Step 2: $E{°}_{\text{cell}}$ calculate karo $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$ $E{°}_{\text{cell}}=0.77-0.54=+0.23\text{ V}$

Yeh step kyun? Cathode woh hai jahan reduction hoti hai (${\text{Fe}}^{3+}$), anode woh hai jahan oxidation hoti hai (${\text{I}}^{-}$). Hum cathode potential mein se anode potential subtract karte hain.

Step 3: Spontaneity determine karo $\Delta G°=-nFE{°}_{\text{cell}}=-2\times 96485\times 0.23<0$

Conclusion: $E{°}_{\text{cell}}>0$ → Spontaneous. ${\mathrm{Fe}}^{3+}$, ${\text{I}}^{-}$ ko oxidize karega.

Physical interpretation: Fe³⁺ I₂ se zyada strong oxidizing agent hai (higher $E°$), isliye woh "jeet ta" hai aur reduce ho jaata hai, aur I⁻ ko oxidize hone par majboor karta hai.

[!definition] Electrochemical Series

Electrochemical series standard electrode potentials ki ek ranked list hai, jo most negative (upar) se most positive (neeche) order mein hoti hai.

Standard Reduction Potentials (excerpt):

Series Padhna

Reducing agents (electron donors, oxidize ho jaate hain):

• Series ke upar (negative $E°$): Strong reducers (Li, K, Zn)
• Yeh species reduction resist karti hain lekin aasaani se oxidation undergo karti hain

Oxidizing agents (electron acceptors, reduce ho jaate hain):

• Series ke neeche (positive $E°$): Strong oxidizers (F₂, Cl₂, Ag⁺)
• Yeh species aasaani se reduction undergo karti hain lekin oxidation resist karti hain

Rule of thumb:

• Higher $E°$ (more positive) → stronger oxidizing agent (oxidized form)
• Lower $E°$ (more negative) → stronger reducing agent (reduced form)

[!formula] Series Se Redox Reactions Predict Karna

Principle: Ek spontaneous redox reaction tab hoti hai jab stronger oxidizing agent (higher $E°$) stronger reducing agent (lower $E°$) ko oxidize karta hai.

Algorithmic approach:

1. Series se do half-reactions likho
2. Jiska higher $E°$ ho woh reduction ke roop mein proceed karta hai (cathode)
3. Jiska lower $E°$ ho woh oxidation ke roop mein proceed karta hai (anode, half-reaction reverse karo)
4. Calculate karo $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$
5. Agar $E{°}_{\text{cell}}>0$ → spontaneous; agar $E{°}_{\text{cell}}<0$ → non-spontaneous

Rule ki derivation:

Thermodynamics se: $\Delta G°=-nFE{°}_{\text{cell}}$

Spontaneity ke liye, $\Delta G°<0$, isliye: $E{°}_{\text{cell}}>0$

Expand karne par: $E{°}_{\text{cathode}}-E{°}_{\text{anode}}>0$ $E{°}_{\text{cathode}}>E{°}_{\text{anode}}$

Yeh kyun kaam karta hai: Jis species ka higher $E°$ hota hai woh electrons zyada strongly "pull" karti hai, isliye woh cathode ki tarah kaam karti hai. Jiska lower $E°$ hota hai woh electrons zyada aasaani se release karta hai, isliye woh anode ki tarah kaam karta hai.

[!example] Worked Example 3: Kya Zinc Copper Ko Reduce Karega?

Question: Kya Zn metal ${\text{Cu}}^{2+}$ ions ko reduce karega?

Half-reactions:

• $Z{n}^{2+}+2e^{-}\to Zn$, $E°=-0.76\text{ V}$
• $C{u}^{2+}+2e^{-}\to Cu$, $E°=+0.34\text{ V}$

Cathode/anode identify karo:

• Higher $E°$: Cu²⁺/Cu → cathode (reduction)
• Lower $E°$: Zn²⁺/Zn → anode (oxidation, reaction reverse karo)

Overall reaction: $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$

$E{°}_{\text{cell}}$ calculate karo: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=0.34-(-0.76)=+1.10\text{ V}$

Yeh step kyun? Hum cathode potential (Cu) mein se anode potential (Zn) subtract karte hain. Double negative positive ban jaata hai.

Conclusion: $E{°}_{\text{cell}}=+1.10\text{ V}>0$ → Spontaneous. Zn, Cu²⁺ ko reduce karega.

Physical observation: Zn strip ko ${\text{CuSO}}_4$ solution mein daalo → Zn par Cu deposit ho jaata hai, blue colour fade ho jaata hai.

[!mistake] Common Mistake 1: Half-Reactions Reverse Karte Waqt Signs Flip Karna

Galat idea: "Agar main $C{u}^{2+}+2e^{-}\to Cu$ ($E°=+0.34\text{ V}$) ko reverse karke $\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$ banaaun, toh potential $E°=-0.34\text{ V}$ ho jaayega."

Yeh sahi kyun lagta hai: Reaction reverse karne se $\Delta G°$ reverse ho jaata hai, aur kyunki $\Delta G°=-nFE°$ hai, toh kya $E°$ ka sign nahi badalna chahiye?

Sach: Electrochemical series mein $E°$ values hamesha reduction half-reaction ke liye hoti hain. Jab tum kisi half-reaction ko anode ke roop mein use ke liye reverse karte ho, tum $E°$ value nahi badlte—tum ise $E{°}_{\text{cell}}$ formula mein $E{°}_{\text{anode}}$ ke roop mein use karte ho jisme minus sign pehle se formula mein built-in hai.

Sahi approach:

• Dono half-reactions ka $E°$ reductions ke roop mein find karo
• $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$ use karo
• Subtraction reversal ka dhyaan rakhti hai

Example: $\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$ ko anode ke roop mein use karne par: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{{\text{Cu}}^{2+}/\text{Cu}}=E{°}_{\text{cathode}}-0.34$

$-0.34$ ka effect subtraction se aata hai, $E°$ ka sign flip karne se nahi.

[!mistake] Common Mistake 2: Standard Conditions Bhool Jaana

Galat idea: "Cu²⁺/Cu ke liye $E°=+0.34\text{ V}$ hai, toh reaction hamesha spontaneous hai."

Yeh sahi kyun lagta hai: $E°>0$ spontaneity suggest karta hai.

Sach: $E°$ sirf standard conditions mein apply hota hai:

• 25°C
• 1 M concentrations (unit activity)
• 1 bar gas pressures

Non-standard conditions mein, Nernst equation use karo: $E=E°-\frac{RT}{nF}\ln Q=E°-\frac{0.0592}{n}\log Q\text{ (at 25°C)}$

Example: $C{u}^{2+}+2e^{-}\to Cu$ ke liye, agar $[{\text{Cu}}^{2+}]=0.001\text{ M}$ ho: $E=0.34-\frac{0.0592}{2}\log \frac{1}{0.001}=0.34-0.089=+0.25\text{ V}$

Abhi bhi positive hai, lekin driving force kam ho gayi. Bahut low $[{\text{Cu}}^{2+}]$ par, $E$ negative bhi ho sakta hai.

Fix: Hamesha check karo ki conditions standard hain ya nahi. Agar nahi, toh Nernst equation apply karo (covered in 2.7.03-Nernst-equation-and-concentration-effects).

[!mistake] Common Mistake 3: $E°$ Ko Stoichiometric Coefficients Se Multiply Karna

Galat idea: "${\text{2Ag}}^{+}+{\text{2e}}^{-}\to \text{2Ag}$ reaction mein twice electrons hain, toh $E°=2\times 0.80=1.60\text{ V}$."

Yeh sahi kyun lagta hai: $\Delta G°=-nFE°$, $n$ par depend karta hai, toh kya $E°$ scale nahi hona chahiye?

Sach: $E°$ ek intensive property hai (temperature ya pressure ki tarah)—yeh amount ke saath nahi badlta. Jab tum stoichiometry double karte ho:

• $\Delta G°$ double ho jaata hai ($n$ double hota hai)
• $E°$ same rehta hai ($\Delta G°=-nFE°$ mein $n$ compensate karta hai)

Proof: $\Delta G°=-nFE°$

${\text{Ag}}^{+}+e^{-}\to \text{Ag}$ ke liye: $\Delta G{°}_1=-(1)FE°$

${\text{2Ag}}^{+}+{\text{2e}}^{-}\to \text{2Ag}$ ke liye: $\Delta G{°}_2=-(2)FE°$

Kyunki $\Delta G{°}_2=2\Delta G{°}_1$ (extensive property): $-(2)FE°=2[-(1)FE°]$ $E°=E°$

Physical analogy: Battery ki "voltage" nahi badlti agar tum do identical batteries parallel mein connect karo (same voltage, double current). $E°$ "voltage per electron" hai.

Fix: Hamesha tabulated $E°$ value use karo, chahe equation kaise bhi balance karo.

[!recall]- Ek 12-Saal Ke Bachche Ko Explain Karo

Socho ek playground mein bahut saare bachche hain, aur kuch REALLY ek toy (electrons) ke saath khelna chahte hain. Maan lo:

• Strong oxidizers (jaise F₂, Cl₂) woh pushy bachche hain jo REALLY toy chahte hain aur kisi se bhi le lete hain.
• Strong reducers (jaise Li, Zn) woh generous bachche hain jo khushi se toy de dete hain.

Ab, agar tum ek pushy bachche (F₂) ko ek generous bachche (Li) ke paas rakhte ho, toh pushy wala toy le lega—yeh ek spontaneous reaction hai!

Lekin hum "pushiness" measure kaise karein? Hum har bachche se akele nahi pooch sakte—humein unhe compete karna hoga! Toh hum ek bachche ko reference choose karte hain (Standard Hydrogen Electrode, SHE) aur kehte hain, "Is bachche ki pushiness zero hai by definition." Phir hum baaki sabko measure karte hain ki woh reference bachche se kitna zyada ladte hain.

• Agar koi reference bachche se toy le le, woh zyada pushy hai (positive $E°$, achha oxidizer).
• Agar koi reference bachche ko toy de de, woh zyada generous hai (negative $E°$, achha reducer).

Hum saari "pushiness scores" ek list mein likhte hain—yeh electrochemical series hai! Sabse pushy bachcha (F₂, $E°=+2.87\text{ V}$) neeche hai, aur sabse generous bachcha (Li, $E°=-3.04\text{ V}$) upar hai.

Jab tum do bachhon ko (do half-cells ko) milate ho, toh zyada pushy wala (higher $E°$) toy le lega (reduce ho jaayega), aur zyada generous wala (lower $E°$) toy de dega (oxidize ho jaayega). Agar pushiness ka difference ($E{°}_{\text{cell}}$) positive hai, toh yeh "trade" apne aap ho jaata hai—spontaneous!

[!mnemonic] Electrochemical Series Yaad Karne Ka Tarika

"Lucky Lions Can Zoom High, Courageously Fighting Bears After Fierce Competition"

• Li ($-3.04$ V)
• Li (wait, that's K for Potassium!) → K ($-2.93$ V)
• Ca ($-2.87$ V)
• Zn ($-0.76$ V)
• H₂ (0.00 V, reference)
• Cu ($+0.34$ V)
• Fe³⁺ ($+0.77$ V)
• Ag ($+0.80$ V) (Bears? close enough!)
• Br₂ ($+1.07$ V)
• Cl₂ ($+1.36$ V) (After)
• F₂ ($+2.87$ V) (Fierce Competition)

Oxidizer strength ke liye alternate mnemonic: "Fine Chefs Bake Awesome Cookies Hot Zesty Creations Keep Lads" (bottom to top, strongest to weakest oxidizers).

Connections

• Previous: 2.7.01-Oxidation-states-and-balancing-redox-equations — Oxidation states assign karna aur redox reactions balance karna
• Next: 2.7.03-Nernst-equation-and-concentration-effects — Concentration ke saath $E$ kaise badlta hai
• Relates to: 2.7.04-Electrochemical-cellsand-cell-potential — $E°$ ko galvanic cells mein apply karna
• Relates to: 2.7.05-Gibs-free-energy-and-equilibrium-constant — $E°$ ko $\Delta G°$ aur $K$ se connect karna
• Foundation: 1.5.02-Thermodynamics-spontaneity-and-Gibbs-energy — $\Delta G°<0$ ka matlab spontaneous kyun hota hai

#flashcards/chemistry

Standard Hydrogen Electrode (SHE) kya hai, aur iska defined potential kya hai? :: SHE mein ek Pt electrode, 1 M H⁺ solution, 1 bar par H₂ gas, aur 25°C hota hai. Half-reaction hai $2H^{+}+2e^{-}\rightleftharpoons H_2$. Iska potential convention se exactly 0.00 V define kiya jaata hai, jo baaki saare electrode potentials measure karne ke liye universal reference hai.

Negative standard electrode potential ($E°<0$) kya indicate karta hai? :: Negative $E°$ ka matlab hai reduction half-reaction SHE ke relative non-spontaneous hai—species oxidize hona prefer karti hai (reducing agent ki tarah kaam karti hai). Example: $E{°}_{{\text{Zn}}^{2+}/\text{Zn}}=-0.76\text{ V}$, isliye Zn aasaani se electrons khota hai.

Concept Map