The resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 . If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 , calculate the conductivity and molar conductivity of 0.02 mol L-1 KCl solution. The conductivity of 0.1 mol L-1 KCl solution is 1.29x 10-2
-1cm-1
State Faraday's first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
Calculate emf of the following cell at 298 K: Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s)
[Given E0 cell = +2.71 V, 1 F = 96500 C mol-1]
Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(ii) The enthalpy of atomization is lowest for Zn in 3d series of the transition elements.
(iii) Actinoid elements show a wide range of oxidation states.
i) Electronic configuration of Mn2+ is [Ar]183d5
Electronic configuration of Fe2+ is [Ar]18 3d6
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in +2 state has a stable d5 configuration. Therefore, Mn2+ shows resistance to oxidation to Mn3+. Also,
e2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration.
Therefore, Fe2+ gets oxidised to Fe3+ easily.
ii) The extent of metallic bonding an element undergoes, decides the enthalpy of atomisation.
The more extensive the metallic bonding of an element, the more will be its enthalpy of atomisation. In all transition metals (except Zn, electronic configuration: 3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomisation.
iii) Actinides exhibit larger oxidation states because of very small energy gap between 5f, 6d and 7s sub-shells. The energies are calculated on the basis of (n+l) rule. The (n+l) values of the three orbitals are:
5 f = 5 + 3 = 8
6 d = 6 + 2 = 8
7 s = 7 + 0 = 7
Since, all the values are almost same, therefore all orbitals can involve in bonding resulting in larger oxidation number for actinoids.
(i) Name the elements of 3d transition series that show a maximum number of oxidation states. Why does this happen?
(ii) Which transition metal of 3d series has positive E0 (M2+/M) value and why?
(iii) Out of Cr3+ and Mn3+, which is a stronger oxidising agent and why?
(iv) Name a member of the lanthanoid series that is well-known to exhibit +2 oxidation state.
(v) Complete the following equation: MnO4- + 8H+ + 5e- -->