Author: DiplomaMitra

 Practical No. 08: Determination of Equivalent Weight of Metal using Faraday’s Second Law

• Conclusion: Same quantity of electricity deposits weights proportional to equivalent weights, verifying Faraday’s second law.
• Recommendation: Clean electrodes carefully, maintain constant current, and accurately measure weights for reliable results.

• Answer: Cleaning removes impurities, oxide layers, or grease from electrode surfaces. This ensures proper adhesion of deposited metal and accurate weight measurement. Clean surfaces provide better electrical contact and uniform deposition. It prevents contamination of the electrolyte and ensures the reaction occurs properly at the electrode surface.

• Answer: During electrolysis, the anodes (positive electrodes) undergo oxidation and dissolve into the electrolyte as metal ions. The copper anode dissolves as Cu²⁺ ions, and the Zinc anode dissolves as Zn²⁺ ions. This dissolution causes a decrease in the weight of both anodes. The mass lost from the anode equals the mass deposited at the cathode.

• Answer: According to Faraday’s laws, the amount of substance deposited is directly proportional to the time for which current is passed when current is constant. W ∝ t. Therefore, longer electrolysis time results in greater deposition of metal on the cathode, provided current remains constant throughout the process.

• Answer: The increase in cathode weight indicates successful electrodeposition of metal on the cathode. It confirms that the ions from the electrolyte are being reduced and deposited on the electrode. This increase is used to calculate the electrochemical equivalent and equivalent weight of the metal, making it essential for verifying Faraday’s laws.

 Practical No. 07: Electrochemical Equivalent of Copper using Faraday’s First Law

• Conclusion: Weight of substance deposited is directly proportional to quantity of electricity passed.
• Recommendation: Clean cathode thoroughly, maintain constant current, and accurately record time for precise results.

• Answer: Chemical equivalence refers to the equivalent weight of a substance (atomic weight / valency). Electrochemical equivalent is the weight of substance deposited by 1 coulomb of electricity. They are related by Faraday’s laws: Electrochemical equivalent (Z) = Equivalent weight / 96500. Thus, electrochemical equivalent is directly proportional to chemical equivalent.

• Answer: According to Faraday’s first law, the weight of substance deposited (W) is directly proportional to the time (t) for which current is passed, provided current (I) is constant. W ∝ t, or W = Z × I × t, where Z is electrochemical equivalent. Thus, longer time results in greater deposition of substance.

• Answer: The copper anode is a consumable/sacrificial anode in this electrolysis process. As current passes, the copper anode dissolves into Cu²⁺ ions, which migrate to the cathode where they get deposited. The anode loses mass, and this type of anode is called a “soluble anode.”

 Practical No. 06: Determination of Voltage Generated from Chemical Reaction using Daniel Cell

• Conclusion: Voltage depends on electrolyte concentrations, with maximum voltage at maximum concentrations.
• Recommendation: Clean electrodes, use fresh salt bridge, and ensure tight connections for accurate voltage measurement.

• Answer: In a Daniel cell, Zinc electrode (Zn) acts as the Anode (negative terminal) where oxidation occurs. Copper electrode (Cu) acts as the Cathode (positive terminal) where reduction occurs. Electrons flow from zinc anode through the external circuit to copper cathode.

• Answer:
  • At Anode (Zinc): Zn → Zn²⁺ + 2e⁻ (Oxidation).
  • At Cathode (Copper): Cu²⁺ + 2e⁻ → Cu (Reduction).
  • Net reaction: Zn + Cu²⁺ → Zn²⁺ + Cu.
  • Observation: The zinc dissolves, and copper deposits on the copper electrode.

• Answer: A half cell is a single electrode immersed in a solution of its own ions. It consists of a metal electrode and an electrolyte solution containing the corresponding metal ions. When two half cells are connected, they form an electrochemical cell. Each half cell has its own electrode potential, and the combination of two half cells gives the total cell potential.

• Answer: The salt bridge maintains electrical neutrality by allowing the flow of ions between the two half cells. It prevents the accumulation of excess positive charges near the anode and negative charges near the cathode. It completes the electrical circuit and allows the continuous flow of electrons in the external circuit, maintaining the cell potential.

• Answer: Suitable electrolytes for a salt bridge include:
  • (i) KCl (Potassium chloride)
  • (ii) KNO₃ (Potassium nitrate)
  • (iii) Na₂SO₄ (Sodium sulphate)
  • (iv) NH₄NO₃ (Ammonium nitrate)
  • Note: The ions should not react with the electrode solutions and should have similar ionic mobilities.

 Practical No. 05: Determination of Electrode Potential of Iron

• Conclusion: Zinc is more electropositive than iron (-0.76 V vs -0.44 V). Therefore zinc acts as sacrificial anode protecting iron from corrosion.
• Recommendation: Use clean electrodes and fresh electrolyte solutions for accurate measurements.

• Answer: De-electronation is the process of removal of electrons from an atom or ion, i.e., oxidation (loss of electrons). Electronation is the process of addition of electrons to an atom or ion, i.e., reduction (gain of electrons). At anode, de-electronation occurs, and at cathode, electronation occurs in an electrochemical cell.

• Answer: Factors influencing electrode potential of iron include:
  • (i) Nature of the metal and its ions.
  • (ii) Concentration of metal ions in solution.
  • (iii) Temperature of the solution.
  • (iv) Pressure (for gaseous electrodes).
  • (v) Nature of the electrolyte.
  • (vi) Surface condition of the electrode.

• Answer: Applications include:
  • (i) Designing corrosion protection systems like sacrificial anodes and cathodic protection.
  • (ii) Electroplating and galvanization of iron surfaces.
  • (iii) Manufacturing of batteries and electrochemical cells.
  • (iv) Determining the feasibility of redox reactions.
  • (v) Metallurgy and metal extraction processes.
  • (vi) Sensor and analytical applications.

 Practical No. 04: Determination of Electrode Potential of Copper

• Conclusion: Zinc is more electropositive than copper (lower reduction potential -0.76 V vs +0.34 V). Hence zinc undergoes corrosion in preference to copper when in contact.
• Recommendation: Clean electrode surfaces, use fresh solutions, and ensure proper connections for accurate readings.

• Answer:
  • At Anode (Zinc electrode – oxidation): Zn → Zn²⁺ + 2e⁻ (Zinc loses electrons and dissolves as Zn²⁺ ions).
  • At Cathode (Copper electrode – reduction): Cu²⁺ + 2e⁻ → Cu (Copper ions gain electrons and deposit as copper metal on the electrode).
  • Net Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu.

• Answer: Lower reduction potential indicates a higher tendency to undergo oxidation (corrosion). Metals with more negative reduction potentials (like Zn, Fe) have a higher tendency to lose electrons and corrode. Metals with positive reduction potentials (like Cu, Ag) have a lower tendency to corrode. Thus, more electropositive metals corrode preferentially.

• Answer: Zinc electrode acts as the Anode (negative terminal) where oxidation occurs. Copper electrode acts as the Cathode (positive terminal) where reduction occurs. Electrons flow from zinc (anode) through the external circuit to copper (cathode), and conventional current flows from copper to zinc.

 Practical No. 03: Identification of States of Matter

• Conclusion: Matter exists in solid, liquid, and gas states depending on temperature and pressure.
• Recommendation: Use simulation carefully, record observations systematically, and plot accurate graphs for precise determination of phase transition temperatures.

• Answer: Particles of matter have the following characteristics:
  • (i) They are continuously moving.
  • (ii) They have spaces between them.
  • (iii) They attract each other (intermolecular forces).
  • (iv) They are very small in size.
  • (v) The kinetic energy of particles increases with an increase in temperature.
  • (vi) The movement of particles decreases with a decrease in temperature.

• Answer: The physical state of a substance is determined by:
  • (i) Temperature: Increasing temperature changes solid to liquid and liquid to gas.
  • (ii) Pressure: Increasing pressure can change gas to liquid and liquid to solid.
  • (iii) Intermolecular forces: Stronger forces favour the solid state, while weaker forces favour the gaseous state.
  • (iv) Kinetic energy of particles: Higher kinetic energy favours the gaseous state.

• Answer:
  • At 0°C: Water exists in two states — solid (ice) and liquid (water) at the melting point (both coexist).
  • At 100°C: Water exists in two states — liquid (water) and gas (steam/water vapour) at the boiling point (both coexist).
  • Note: Above 0°C and below 100°C, water is purely in a liquid state.

 Practical No. 02: Identification of Anions

• Conclusion: Systematic anion analysis with group reagents effectively identifies acidic radicals.
• Recommendation: Use freshly prepared reagents, avoid contamination, and perform tests in proper sequence for accurate results.

• Answer: The acidic radical is CO₃²⁻ (Carbonate). When carbonate reacts with dilute HNO₃, it liberates CO₂ gas which turns lime water milky due to formation of CaCO₃. The effervescence observed confirms the presence of carbonate ion.

• Answer: To separate halides, add dilute HNO₃ followed by AgNO₃ solution. White precipitate indicates Cl⁻, pale yellow precipitate indicates Br⁻, and yellow precipitate indicates I⁻.
• Confirmation: For further confirmation, add chloroform and chlorine water — chlorine water liberates halogens which dissolve in chloroform giving characteristic colours: Cl⁻ gives colourless layer, Br⁻ gives yellowish brown layer, and I⁻ gives violet layer.

• Answer: The anion is SO₄²⁻ (Sulphate). Barium nitrate gives a white precipitate of BaSO₄ which is insoluble in dilute HNO₃. This confirms the presence of sulphate ion. The confirmatory test can also be performed using BaCl₂ solution which gives the same white precipitate.