#electrochemical cell

Electrolytic Cells and Electrolysis and Faraday's Laws

What is the standard Gibbs energy change for a cell with E°cell = 0.44 V and 2 electrons transferred? (F = 96500 C/mol)

Δr G° = -n F E°cell = -2 × 96500 × 0.44 = -84920 J/mol = -84.92 kJ/mol .

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

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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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Corrosion and Applications of Electrochemistry

In a mercury cell, what is the reducing agent?

Reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l) . Zn is oxidized, so it’s the reducing agent.

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

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Corrosion and Applications of Electrochemistry

What is the emf of the cell Pb(s) | Pb²⁺(0.005 M) || Ag⁺(0.05 M) | Ag(s) at 298 K? (Given: E°Pb²⁺/Pb = -0.13 V , E°Ag⁺/A

E°cell = 0.80 - (-0.13) = 0.93 V . Ecell = 0.93 - (0.059/2) log (0.005/0.05²) = 0.93 - 0.0295 = 0.9005 V .

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

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Corrosion and Applications of Electrochemistry

What is the standard Gibbs energy change for a cell with E°cell = 0.59 V and 1 electron transferred? (F = 96500 C/mol)

Δr G° = -n F E°cell = -1 × 96500 × 0.59 = -56935 J/mol = -56.935 kJ/mol .

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

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Corrosion and Applications of Electrochemistry

What is the standard Gibbs energy change for a cell with E°cell = 0.59 V and 1 electron transferred? (F = 96500 C/mol)

Δr G° = -n F E°cell = -1 × 96500 × 0.59 = -56935 J/mol = -56.935 kJ/mol .

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

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Corrosion and Applications of Electrochemistry

In a mercury cell, what is the overall cell reaction?

Anode: Zn + 2OH⁻ → ZnO + H₂O + 2e⁻ . Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ . Overall: Zn + HgO → ZnO + Hg .

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

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Corrosion and Applications of Electrochemistry

In a mercury cell, what is the overall cell reaction?

Anode: Zn + 2OH⁻ → ZnO + H₂O + 2e⁻ . Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ . Overall: Zn + HgO → ZnO + Hg .

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

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