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.

Kohlrausch gave the following relation for strong electrolyte : Λ = Λₒ − A√C. Which of the following equality holds true ?

Calculate the emf of the following cell at 298 K : Al (s) | Al³⁺ (0·001 M) || Ni²⁺ (0·1 M) | Ni (s). [Given : E°(Al³⁺/Al) = –1·66 V, E°(Ni²⁺/Ni) = –0·25 V, log 10 = 1]