#EMF

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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Nernst Equation and Gibbs Energy and Equilibrium Constant

The standard emf of a cell is 0.96 V, and its equilibrium constant at 298 K is 10³² . How many electrons are transferred

E°cell = (0.059/n) log Kc , 0.96 = (0.059/n) × 32 , n = (1.888/0.96) ≈ 2 .

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

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Nernst Equation and Gibbs Energy and Equilibrium Constant

What is the emf of the cell Sn(s) | Sn²⁺(0.002 M) || Pb²⁺(0.02 M) | Pb(s) at 298 K? (Given: E°Sn²⁺/Sn = -0.14 V , E°Pb²⁺

E°cell = -0.13 - (-0.14) = 0.01 V . Ecell = 0.01 - (0.059/2) log (0.002/0.02) = 0.01 + 0.0295 = 0.0395 V .

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

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Nernst Equation and Gibbs Energy and Equilibrium Constant

What is the emf of the cell Mn(s) | Mn²⁺(0.05 M) || H⁺(0.005 M) | H₂(g)(1 bar) | Pt(s) at 298 K? (Given: E°Mn²⁺/Mn = -1.

E°cell = 0.00 - (-1.18) = 1.18 V . Ecell = 1.18 - (0.059/2) log (0.05/0.005²) = 1.18 - 0.07375 = 1.10625 V .

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

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Electrolytic Cells and Electrolysis and Faraday's Laws

What is the emf of the cell Cr(s) | Cr³⁺(0.001 M) || Cu²⁺(0.01 M) | Cu(s) at 298 K? (Given: E°Cr³⁺/Cr = -0.74 V , E°Cu²⁺

E°cell = 0.34 - (-0.74) = 1.08 V . Ecell = 1.08 - (0.059/6) log ((0.001)²/0.01) = 1.08 - 0.01967 = 1.06033 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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Electrochemical Cells - Galvanic Cells and Electrode Potential

The emf of a cell Fe(s) | Fe²⁺(0.001 M) || Cu²⁺(0.1 M) | Cu(s) at 298 K is (Given: E°Fe²⁺/Fe = -0.44 V , E°Cu²⁺/Cu = 0.3

E°cell = 0.34 - (-0.44) = 0.78 V . Ecell = 0.78 - (0.059/2) log (0.001/0.1) = 0.78 + 0.059 = 0.839 V .

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

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Corrosion and Applications of Electrochemistry

What is the emf of the cell Cd(s) | Cd²⁺(0.01 M) || Ni²⁺(0.1 M) | Ni(s) at 298 K? (Given: E°Cd²⁺/Cd = -0.40 V , E°Ni²⁺/N

E°cell = -0.25 - (-0.40) = 0.15 V . Ecell = 0.15 - (0.059/2) log (0.01/0.1) = 0.15 + 0.0295 = 0.1795 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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Corrosion and Applications of Electrochemistry

The emf of a cell Ni(s) | Ni²⁺(0.001 M) || Ag⁺(0.002 M) | Ag(s) at 298 K is (Given: E°Ni²⁺/Ni = -0.25 V , E°Ag⁺/Ag = 0.8

E°cell = 0.80 - (-0.25) = 1.05 V . Ecell = E°cell - (0.059/2) log ([Ni²⁺]/[Ag⁺]²) = 1.05 - (0.059/2) log (0.001/(0.002)²) = 0.989 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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Corrosion and Applications of Electrochemistry

The emf of a cell Ni(s) | Ni²⁺(0.001 M) || Ag⁺(0.002 M) | Ag(s) at 298 K is (Given: E°Ni²⁺/Ni = -0.25 V , E°Ag⁺/Ag = 0.8

E°cell = 0.80 - (-0.25) = 1.05 V . Ecell = E°cell - (0.059/2) log ([Ni²⁺]/[Ag⁺]²) = 1.05 - (0.059/2) log (0.001/(0.002)²) = 0.989 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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