13 public questions tagged with this topic.
Anode: Zn(Hg) + 2OH- -> ZnO(s) + Hâ‚‚O + 2e- . Cathode: HgO + Hâ‚‚O + 2e- -> Hg(l) + 2OH- . Overall: Zn(Hg) + HgO(s) -> ZnO(s) + Hg(l) . This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.
The mercury cell has a potential of approximately 1.35 V, which remains constant.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ . HgO is the cathode material.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻ . Product is ZnO.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Anode: Zn → Zn²⁺ + 2e⁻ , Moles = (0.34/68) = 0.005 mol , Charge = 0.005 × 2 × 96500 = 965 C . Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ , Moles Hg = 0.005 mol , Mass = 0.005 × 200 = 1.0 g .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Zn + 2OH⁻ → ZnO + H₂O + 2e⁻ . 1 mol Zn (68 g) requires 2F. Moles = (0.68/68) = 0.01 mol , Charge = 0.01 × 2 = 0.02 F .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Zn + 2OH⁻ → ZnO + H₂O + 2e⁻ . 1 mol Zn (68 g) requires 2F. Moles = (0.68/68) = 0.01 mol , Charge = 0.01 × 2 = 0.02 F .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
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
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
Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ , Moles = (0.5/200) = 0.0025 mol , Charge = 0.0025 × 2 × 96500 = 482.5 C . Anode: Zn → Zn²⁺ + 2e⁻ , Moles Zn = 0.0025 mol , Mass = 0.0025 × 68 = 0.17 g .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ , Moles = (0.5/200) = 0.0025 mol , Charge = 0.0025 × 2 × 96500 = 482.5 C . Anode: Zn → Zn²⁺ + 2e⁻ , Moles Zn = 0.0025 mol , Mass = 0.0025 × 68 = 0.17 g .
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
