Q35 of 37 Page 1

(a) The e.m.f. of the following cell at 298 K is 0.1745 V

Fe (s) / Fe2+(0.1 M)//H+(xM)/ H2 (g) (1 bar)/ Pt (s)


Given:


= -0.44V


Calculate the H+ ions concentration of the solution at the electrode where hydrogen is being produced.


(b)Aqueous solution of copper sulphate and silver nitrate are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.


OR


(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10-5 S cm-1.


Given: = 349.6 S cm2 mol-1; = 40.9 S cm2 mol-1


(b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases to a smaller extent while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer.


(a)The electrode reaction is given as


Fe(s) + 2H+→ Fe2+(aq) + 2e- + H2(g)


Fe → Fe2+ + 2e-


Applying Nernst equation,


Ecell = E0cell -


0.175 = -0.44V -


0.615 = -(0.0591/2)log(0.1/[H+])


-20.795 = log(0.1/[H])


-20.795 = log0.1 – logH+


Log[H+] = -19.795


Taking antilogs,


[H+] = 6.24 x 1019


(b)Given,


t = 10 mins = 600s


charge = current x time = 1A x 600s = 600C


The reaction is given by Cu2+(aq) + 2e-→ Cu(s)


We need 2 x 96500C to deposit 1 mol or 63g of copper on the electrode, for 900C the amount of copper deposited will be 0.29g.


The amount of silver and copper deposited will be different because Ag has a different molar mass and equivalent mass.


OR


(a)κ = 8.0 × 10-5 S cm-1


C = M = 0.0024M


Molar conductivity = 40.9 S cm2 mol-1


Degree of dissociation is given by,


α = / = 40.9 S cm2mol-1/349.6 S cm2mol-1


= 0.11


(b)‘B’ is the strong electrolyte because on dilution of a strong electrolyte, number of ions almost remain the same. Only the interionic attraction decreases and therefore the increase in limiting molar conductivity of ‘B’ is small. This is because on dilution for both strong and weak electrolyte, there is increase in molar conductivity. But it has been observed that there is a large increase in molar conductivity of weak electrolyte on dilution than strong electrolyte. Because, weak electrolytes dissociate very less at higher concentration. On dilution, degree of dissociation of weak electrolyte increases means there is increase in number of ions per unit volume. Hence, ‘A’ is the weaker electrolyte. On the other hand, for strong electrolyte, degree of dissociation increases slowly on dilution. Thus, ‘B’ whose limiting molar conductivity increases to a lesser extent is strong electrolyte.


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33

(a) Give chemical tests to distinguish between the following pairs of compounds:

(i) Ethanal and Propanone.


(ii) Pentan-2-one and Pentan-3-one.(b) Arrange the following compounds in increasing order of their acid strength: Benzoic acid, 4- Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4- Methoxybenzoic acid.


OR


Compare the reactivity of benzaldehyde and ethanal towards nucleophilic addition reactions. Write the cross aldol condensation product between benzaldehyde and ethanal.


 34

Define and write an example for the following:

(a) Broad-spectrum antibiotics.


(b) Analgesics


 36

An organic compound A with molecular formula C7H7NO reacts with Br2/aq. KOH to give compound B, which upon reaction with NaNO2 & HCl at 0°C gives C. Compound C on heating with CH3CH2OH gives a hydrocarbon D. Compound B on further reaction with Br2 water gives white precipitate of compound E. Identify the compound A, B, C, D & E; also justify your answer by giving relevant chemical equations.

OR


(a) How will you convert:


(i) Aniline into Fluorobenzene.


(ii) Benzamide into Benzylamine.


(iii) Ethanamine to N,N-Diethylethanamine.


(b)Write the structures of A and B in the following:


(i)


(ii)


37

(a) When a chromite ore (A) is fused with an aqueous solution of sodium carbonate in free excess of air, a yellow solution of compound (B) is obtained. This solution is filtered and acidified with sulphuric acid to form compound (C). Compound (C) on treatment with solution of KCl gives orange crystals of compound (D). Write the chemical formulae of compounds A to D.

(b) Describe the cause of the following variations with respect to lanthanoids and actinoids:


(i) Greater range of oxidation states of actinoids as compared to lanthanoids.


(ii) Greater actinoid contraction as compared to lanthanoid contraction.


(iii) Lower ionisation enthalpy of early actinoids as compared to the early lanthanoids.


OR


(a) What happens when:


(i) Manganate ions (MnO42-) undergoes disproportionation reaction in acidic medium?


(ii) Lanthanum is heated with Sulphur?


(b) Explain the following trends in the properties of the members of the First series of transition elements:


(i) E0(M2+|M) value for copper is positive(+0.34 V) in contrast to the other members of the series.


(ii) Cr2+ is reducing while Mn3+ is oxidising, though both have d4 configuration.


(iii) The oxidising power in the series increases in the order


VO2+< Cr2O72-< MnO4-.