Calculate the E.M.F. of the cell, Mg s | Mg 2 + ( 0 . 1 M ) | | Ag + 1 × 10 - 3 | Ag s E Ag + / Ag ° = + 0 . 8 V , E Mg 2 + / Mg ° = - 2 . 37 V What happens to the E.M.F. if the concentration of Ag + is decreased to 1 × 10 - 4 M ? [given log 5 = 0.6990]

E Cell ° = E Cathode ° − E Anode ° = 0 . 80 − ( − 2 . 37 ) = 3 . 17 volt Cell reaction , Mg + 2 Ag + → 2 Ag + Mg 2 + E cell = E cell ° − 0 . 0591 n log Mg 2 + Ag + 2 = 3 . 17 − 0 . 0591 2 log 0 . 1 1 × 10 - 3 2 = 3 . 17 − 0 . 1478 = 3 . 02225 volt when Ag + = 1 × 10 - 4 M E cell = E cell ° − 0 . 0591 2 log 0 . 1 1 × 10 - 4 2 = 3 . 17 − 0 . 20685 volt = 2 . 963 volt If the concentration of Ag + is decreased to 1 × 10 - 4 M the EMF is 2.963 volt.

State Kohlrausch law of independent migration of ions. How is the law used to determine the degree of dissociation of a weak electrolyte?

Kohlrausch's law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, if λ ° Na + and λ ° Cl - are limiting molar conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation : Λ ° m ( NaCl ) = λ ° ( Na + ) + λ ° ( Cl - ) . With increase in degree of dissociation due to dilution more number of ions are furnished and equivalent conductance increases. The degree of dissociation is calculated from the conductance ratio i.e., the ratio of equivalent conductance at any concentration to that at infinite dilution. α = Λ c Λ 0

Write the Nernst equation and calculate e.m.f. of the cell, Mg s ∣ Mg 2 + ( 0 . 001 M ) ∥ Cu 2 + 0 . 0001 M ∣ Cu s at 298 K . Given E ° Cell = 2 . 71 volt .

Mg s ∣ Mg 2 + ( 0 . 001 M ) ∥ Cu 2 + 0 . 0001 M ∣ Cu s Given, E ° Cell = 2 . 71 volt , Mg 2 + = 0 . 001 M Cu 2 + = 0 . 0001 M and E cell = ? Cell reactions from the given representation can be written as, Anode: Mg → Mg 2 + + 2 e - ⋯ ⋯ ⋯ ( i ) Cathode: Cu 2 + + 2 e - → Cu ⋯ ⋯ ⋯ ( ii ) Equation i + ii , Mg + Cu 2 + → Mg 2 + + Cu Nernst equation for the above reaction at 298 K as follows, E Cell = E ° Cell - 0 . 0591 n log Mg 2 + Cu 2 + ⋯ ⋯ ⋯ ⋯ ( iii ) Put all the given values in the equation iii , E Cell = 2 . 71 - 0 . 0591 2 log 0 . 001 0 . 0001 E Cell = 2 . 71 - 0 . 02955 E Cell = 2 . 6805 volt Therefore, the calculated e.m.f. of the given cell is 2 . 6805 volt .

Calculate the mass of hydrogen evolved by passing a current of 0 . 5 ampere for 40 minutes through acidified water.

The following reaction involve in this process, 2 H + + 2 e - → H 2 g Given, Current I = 0 . 5 ampere , time t = 40 minutes = 40 × 60 s Mass of hydrogen W = ? According to Faraday's first law, W = ZIt ⋯ ⋯ ( i ) Where, Electrochemical equivalent of a substance, Z = Equivalent weight of a substance 96500 = M . W Valency or charge n × 96500 W = M . W × I × t n × 96500 W = 2 × 0 . 5 × 40 × 60 2 × 96500 W = 0 . 0124 gram Therefore, the mass of hydrogen evolved by passing a current through acidified water for the given question, is 0 . 0124 gram .

The resistivity of a 0 . 8 M solution of an electrolyte is 5 × 10 - 3 ohm - cm . Calculate its molar conductivity.

Given, Concentration ( C ) = 0 . 8 M Resistivity ρ = 5 × 10 - 3 ohm - cm Molar conductivity ( λ m ) = ? As we know, Molar conductivity, λ m = 1000 × k C ⋯ ⋯ ⋯ ⋯ ( i ) also, k = 1 ρ ∴ λ m = 1000 C × ρ ⋯ ⋯ ⋯ ( ii ) Put all the values in equation ii , = 1000 0 . 8 × 5 × 10 - 3 = 2 . 5 × 10 5 ohm - 1 cm 2 mol - 1 Therefore, the molar conductivity of the electrolyte is 2 . 5 × 10 5 ohm - 1 cm 2 mol - 1 .

What is meant by 'limiting molar conductivity'

The limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte. Mathematically, Λ ° m for A x B y = xλ ° + + yλ ° - where Λ ° m is the limiting molar conductivity of the electrolyte; λ ° + and λ ° - are the limiting molar conductivities of the cation A y + and the anion B x - respectively. For example, Λ ° m for NaCl = λ ° Na + + λ ° Cl - , Λ ° m for Al 2 SO 4 3 = 2 λ ° Al 3 + + 3 λ ° SO 4 2 - .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Electrochemistry>Manipur Board-2019>Q 4

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12th Manipur Board

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In the electrode of the first kind ${Cl}_{2} / {Cl}^{-}$ , the electrode material ${Cl}_{2}$ is a non-conductor. How can the electron transfer be carried out with the ion $({Cl}^{-})$ .

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