#hydrogen electrode

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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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 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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Electrochemical Cells - Galvanic Cells and Electrode Potential

A hydrogen electrode operates at 298 K with PH_₂ = 1.5 atm and [H⁺] = 0.0001 M . What is its potential? (Given: E°H⁺/H_₂

E = E° - (0.059/2) log (PH_₂/[H⁺]²) . E = 0 - 0.0295 log (1.5/10⁻⁸) = 0 - 0.0295 × 8.176 = -0.2412 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the potential of a hydrogen electrode in a solution with pH = 7 at 298 K? (Given: E°H⁺/H_₂ = 0.00 V )

E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻⁷ , n = 1 . E = 0 - 0.059 × 7 = -0.413 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the potential of a hydrogen electrode in a solution with pH = 7 at 298 K? (Given: E°H⁺/H_₂ = 0.00 V )

E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻⁷ , n = 1 . E = 0 - 0.059 × 7 = -0.413 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the emf of a cell Fe(s) | Fe²⁺(0.1 M) || H⁺(0.001 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Fe²⁺/Fe = -0.44

E°cell = 0.00 - (-0.44) = 0.44 V . Ecell = 0.44 - (0.059/2) log (0.1/0.001²) = 0.44 - 0.0885 = 0.3515 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the emf of a cell Fe(s) | Fe²⁺(0.1 M) || H⁺(0.001 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Fe²⁺/Fe = -0.44

E°cell = 0.00 - (-0.44) = 0.44 V . Ecell = 0.44 - (0.059/2) log (0.1/0.001²) = 0.44 - 0.0885 = 0.3515 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the potential of a hydrogen electrode in a solution with [H⁺] = 10⁻⁵ M at 298 K? (Given: E°H⁺/H_₂ = 0.00 V )

E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻⁵ , n = 1 . E = 0 - 0.059 × 5 = -0.295 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the potential of a hydrogen electrode in a solution with [H⁺] = 10⁻⁵ M at 298 K? (Given: E°H⁺/H_₂ = 0.00 V )

E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻⁵ , n = 1 . E = 0 - 0.059 × 5 = -0.295 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A hydrogen electrode operates at 298 K with PH_₂ = 0.1 atm and pH = 4. What is its potential? (Given: E°H⁺/H_₂ = 0.00 V

E = E° - (0.059/2) log (PH_₂/[H⁺]²) , [H⁺] = 10⁻⁴ . E = 0 - 0.0295 log (0.1/10⁻⁸) = 0 - 0.0295 × 7 = -0.2065 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A hydrogen electrode operates at 298 K with PH_₂ = 0.1 atm and pH = 4. What is its potential? (Given: E°H⁺/H_₂ = 0.00 V

E = E° - (0.059/2) log (PH_₂/[H⁺]²) , [H⁺] = 10⁻⁴ . E = 0 - 0.0295 log (0.1/10⁻⁸) = 0 - 0.0295 × 7 = -0.2065 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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