Electrochemistry

For a strong electrolyte, a plot of molar conductivity against (concentration)1/2 is a straight line, with a negative slope, the correct unit for the slope is

1. S cm2 mol-3/2 L
2. S cm2 mol-3/2 L-1/2
3. S cm2 mol-1 L1/2
4. S cm2 mol-3/2 L1/2

The potential for the given half cell at 298 K is (-) 10-2 V 2H+(aq) + 2e- H2(g) [H+] = 1 M, PH2 = 2 atm (Given : 2.303 RT/F = 0.06 V, 2 = 0.3)

The amount of electricity in Coulomb required for the oxidation of 1 mol of H2O to O2 is 105 C.

Consider the following redox reaction : MnO4- + H+ + H2C2O4 Mn2+ + H2O + CO2 The standard reduction potentials are given as below ( E0red ) : E MnO4-/Mn2+ = +1.51 V E CO2/H2C2O4 = -0.49 V If the equilibrium constant of the above reaction is given as Keq = 10x, then the value of x = (nearest integer)

Identify the factor from the following that does not affect electrolytic conductance of a solution.

1. The nature of solvent used.
2. The nature of the electrolyte added.
3. The nature of the electrode used.
4. Concentration of the electrolyte.

One Faraday of electricity liberates x 10-1 gram atom of copper from copper sulphate. x is .

The values of conductivity of some materials at 298.15 K in S m-1 are 2.1 103, 1.0 10-16, 1.2 10, 3.91, 1.5 10-2, 1 10-7, 1.0 103. The number of conductors among the materials is .

Reduction potential of ions are given below: [figure] The correct order of their oxidising power is :

1. IO4- > BrO4- > ClO4-
2. BrO4- > ClO4- > IO4-
3. ClO4- > IO4- > BrO4-
4. BrO4- > IO4- > ClO4-

The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is 10-4 g. (Atomic mass of zinc = 65.4 amu)

A constant current was passed through a solution of AuCl4- ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is 10-2 F. (Given atomic mass of Au = 197)