#Nernst equation

CHEMISTRY

What is the potential of a hydrogen electrode in a solution with pH = 5 at 298 K, given E⁰ = 0 V ?

Given: What is the potential of a hydrogen electrode in a solution with pH = 5 at 298 K, given E⁰ = 0 V ? Formula: For H+ + e- -> 1/2 H₂, E = E⁰ - 0.059/1 log frac1[H^+]. Substitution & Calculation: [H^+] = 10⁻⁵, E = 0 - 0.059 × 5 = -0.295 V . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

What is the EMF of a ll with Ni(s) | Ni²⁺(0.005 M) || Ag⁺(0.001 M) | Ag(s) at 298 K, given E_{Ni^{2+/Ni⁰ = -0.25 V and E

Given: What is the EMF of a ll with Ni(s) | Ni²⁺(0.005 M) || Ag⁺(0.001 M) | Ag(s) at 298 K, given E_{Ni^{2+/Ni⁰ = -0.25 V and E_{Ag^{+/Ag⁰ = 0.80 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.80 - (-0.25) = 1.05 V, n = 2. This is standard NCERT relation. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Ni^{2+][Ag^{+]² . Q = 0.005/(0.001)² = 5000, log Q = 3.699 . E_{ll = 1.05 - 0.059/2 × 3.699 = 1.05 - 0.109 = 0.941 V approx 0.94 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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Variation of Conductivity with Concentration and Measurement

What is the emf of a cell Zn(s) | Zn²⁺(0.001 M) || H⁺(1 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Zn²⁺/Zn = -0.76 V

E°cell = 0.00 - (-0.76) = 0.76 V . Ecell = 0.76 - (0.059/2) log (0.001/1²) = 0.76 + 0.0885 = 0.8485 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

A cell operates with Ecell = 1.50 V at 298 K when [Anode] = 0.001 M and [Cathode] = 0.1 M . If E°cell = 1.44 V , what is

Ecell = E°cell - (0.059/n) log Q , 1.50 = 1.44 - (0.059/2) log Q . 0.06 = -0.0295 log Q , log Q = -(0.06/0.0295) = -2.034, Q = 10⁻².⁰³⁴ ≈ 0.0092 . Since Q = ([Anode]/[Cathode]) = (0.001/0.1) = 0.01 , matches approximately.

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

What is the equilibrium constant for a cell reaction with E°cell = 0.236 V at 298 K involving 2 electrons?

E°cell = (0.059/n) log Kc . 0.236 = (0.059/2) log Kc , log Kc = 8, Kc = 10⁸ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

What is the emf of the cell Co(s) | Co²⁺(0.01 M) || Ag⁺(0.001 M) | Ag(s) at 298 K? (Given: E°Co²⁺/Co = -0.28 V , E°Ag⁺/A

E°cell = 0.80 - (-0.28) = 1.08 V . Ecell = 1.08 - (0.059/2) log ([Co²⁺]/[Ag⁺]²) = 1.08 - 0.0295 log (0.01/0.000001) = 1.08 - 0.177 = 0.903 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

A hydrogen electrode operates at 298 K with PH_₂ = 0.5 atm and pH = 3. What is its potential? (Given: E°H⁺/H_₂ = 0.00 V

E = E° - (0.059/2) log (PH_₂/[H⁺]²) , [H⁺] = 10⁻³ . E = 0 - 0.0295 log (0.5/10⁻⁶) = 0 - 0.0295 × 5.301 = -0.1564 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

What is the emf of a cell Ni(s) | Ni²⁺(0.1 M) || Cu²⁺(0.001 M) | Cu(s) at 298 K? (Given: E°Ni²⁺/Ni = -0.25 V , E°Cu²⁺/Cu

E°cell = 0.34 - (-0.25) = 0.59 V . Ecell = 0.59 - (0.059/2) log (0.1/0.001) = 0.59 - 0.059 = 0.531 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

A cell has E°cell = 0.59 V and an equilibrium constant of 10²⁰ at 298 K. How many electrons are transferred?

E°cell = (0.059/n) log Kc , 0.59 = (0.059/n) × 20 , n = (1.18/0.59) = 2 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

What is the potential of a hydrogen electrode in a solution with pH = 2 at 298 K? (Given: E°H⁺/H_₂ = 0.00 V )

E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻² , n = 1 . E = 0 - 0.059 × 2 = -0.118 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

A cell has E°cell = 0.52 V and an equilibrium constant of 10¹⁷ at 298 K. How many electrons are transferred?

E°cell = (0.059/n) log Kc , 0.52 = (0.059/n) × 17 , n = (1.003/0.52) ≈ 2 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Nernst Equation and Gibbs Energy and Equilibrium Constant

What is the cell potential at 298 K for Zn(s) | Zn²⁺(0.01 M) || Cu²⁺(0.1 M) | Cu(s) ? (Given: E°cell = 1.10 V )

Ecell = E°cell - (0.059/2) log ([Zn²⁺]/[Cu²⁺]) = 1.10 - (0.059/2) log (0.01/0.1) = 1.129 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant

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