Question

Read the passage given below and answer the questions that follow : Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. It is an electrochemical cell that uses spontaneous redox reactions to generate electricity. A salt bridge also connects to the half cells. The reading of the voltmeter gives the cell voltage or cell potential or electromotive force. If E°cell is positive the reaction is spontaneous and if it is negative the reaction is non-spontaneous and is referred to as electrolytic cell. Electrolysis refers to the decomposition of a substance by an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as Cu²⁺. This was first formulated by Faraday in the form of laws of electrolysis. The conductance of material is the property of materials due to which a material allows the flow of ions through itself and thus conducts electricity. Conductivity is represented by k and it depends upon nature and concentration of electrolyte, temperature etc. A more common term molar conductivity of a solution at a given concentration is conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with the unit area of cross-section and distance of unit length. Limiting molar conductivity of weak electrolytes cannot be obtained graphically.

The conductivity of 2·5 × 10⁻⁴ M methanoic acid is 5·25 × 10⁻⁵ S cm⁻¹ and its Λ°ₘ has a value 400 S cm² mol⁻¹. Calculate its molar conductivity and degree of dissociation.

When a certain conductivity cell was filled with 0·05 M KCl solution, it has a resistance of 100 ohm at 25ºC. When the same cell was filled with 0·02 M AgNO₃ solution, the resistance was 90 ohm. Calculate the conductivity and molar conductivity of AgNO₃ solution. (Given : Conductivity of 0·05 M KCl solution = 1·35 × 10⁻² ohm⁻¹cm⁻¹)

Resistance of a conductivity cell filled with 0.2 mol L⁻¹ KCl solution is 200 Ω. If the resistance of the same cell when filled with 0.05 mol L⁻¹ KCl solution is 620 Ω, calculate the conductivity and molar conductivity of 0.05 mol L⁻¹ KCl solution. The conductivity of 0.2 mol L⁻¹ KCl solution is 0.0248 S cm⁻¹.