Answer :

Given:

E° = 2.71 V

Solution:

For the given reaction the Nernst equation is written as,

For the given conditions,

(i) E_{cell} > 2.71 V, the current flows from cathode to anode (conventional direction)

(ii) E_{cell} < 2.71 V, i.e. the external voltage is greater than the emf of the cell, the direction of the flow is from anode to cathode (direction is reversed)

**OR**

(i) As the conductivity is linearly dependent on the square root of concentration the line on Λ v/s c^{1/2} suggests that A is a strong electrolyte.

When initially when the concentration is increases the conductivity drops rapidly but after certain point there are enough ions and no more addition of B will decrease the conductivity significantly, thus B is a weak electrolyte.

(ii) Extrapolation of values of Λ_{m} gives the maximum values of conductivity i.e. Λ°_{m} for strong electrolytes at zero concentration. This cannot be found for Weak electrolytes as the relation is non-linear curvature which does not meet the conductivity.

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