Practice questions on galvanic cells, electrode potential, and electrochemical cells with detailed explanations for competitive and board exams.
Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ . Oxygen gas is produced.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Lambdam = α Lambdam° = 0.02 × 390 = 7.8 S cm² mol⁻¹ . kappa = (Lambdam × c/1000) = (7.8 × 0.1/1000) = 0.00078 S cm⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
kappa = (cell constant/R) , 0.0025 = (cell constant/400) , cell constant = 1.0 cm⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Anode: Zn → Zn²⁺ + 2e⁻ . Zn is oxidized, so it’s the reducing agent.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O . For 1 mol IO₃⁻ , 5 electrons.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Reaction: Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s) . E°cell = E°cathode - E°anode = 0.80 - (-0.44) = 1.24 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E°cell = E°cathode - E°anode = 0.80 - (-0.28) = 1.08 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E = E° - (0.059/2) log (PH_₂/[H⁺]²) . E = 0 - 0.0295 log (1.5/10⁻⁸) = 0 - 0.0295 × 8.176 = -0.2412 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
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
E°cell = 0.85 - (-1.18) = 2.03 V . Ecell = 2.03 - (0.059/2) log (0.005/0.02) = 2.03 + 0.02065 = 2.05065 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E°cell = E°cathode - E°anode = 0.80 - (-0.74) = 1.54 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
The cell reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) . E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 1.10 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
