Practice questions on corrosion processes and practical applications of electrochemistry with solutions for competitive and board exams.
Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃ . In MnO₂ , Mn = +4; in MnO(OH) , Mn + 2(-2) + 1 = 0, Mn = +3.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃ . In MnO₂ , Mn = +4; in MnO(OH) , Mn + 2(-2) + 1 = 0, Mn = +3.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) , 0.05 = (19/Lambdam°) , Lambdam° = 380 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) , 0.05 = (19/Lambdam°) , Lambdam° = 380 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Δr G° = -n F E°cell = -2 × 96500 × 1.05 = -202650 J/mol = -202.65 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Δr G° = -n F E°cell = -2 × 96500 × 1.05 = -202650 J/mol = -202.65 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Cathode: Na⁺ + e⁻ → Na . Sodium metal is produced.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Cathode: Na⁺ + e⁻ → Na . Sodium metal is produced.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
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
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
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
