# Important Questions for Class 12 Chapter 3 - Electrochemistry

**Important questions based on NCERT syllabus for Chapter 3 - Electrochemistry:**

*Question-1*: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

*Solution*: Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.

Molar conductivity of a solution at a dilution (V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of cross-section of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by Λm.

The conductivity of a solution (both for strong and weak electrolytes) decreases with decrease in concentration of the electrolyte, i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on ‘ dilution. The molar conductivity of a solution increases with decrease in concentration of the electrolyte. This is because both number of ions as well as mobility of ions increases with dilution. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

*Question-2*: Standard electrode potentials given as,

Mg2+/Mg = −2.37 V, Hg2+/Hg = 0.79V, Cr3+/Cr = − 0.74V, Ag+/Ag = 0.80V, K+/K = −2.93V

In the order of increasing of reducing power arrange the given metals accordingly.

*Solution*:

The reducing power increases with the lowering of reduction potential. In order of given standard electrode potential (increasing order) : K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag

Thus, in the order of reducing power, we can arrange the given metals as : Ag< Hg < Cr < Mg < K

*Question-3*: Considering the case of a conductivity cell having 0.001 M KCl solution at 298 K is 1500 Ω. If given, conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S, find the cell constant?

*Solution*:

Given,

Conductivity, k = 0.146 × 10−3 S cm−1

Resistance, R = 1500 Ω

Cell constant = k × R

= 0.146 × 10−3 × 1500

= 0.219 cm−1