8 public questions tagged with this topic.
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
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
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
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
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
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
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
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
