Q26 of 27 Page 1

(a) A cell is prepared by dipping a zinc rod in 1M zinc sulfate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential is given:
E0Zn2+/Zn = -0.76 V,
 E0Ag2+/Ag = +0.80 V,

What is the effect of an increase in the concentration of Zn2 + on the Ecell?


(b) Write the products of electrolysis of aqueous solution of NaCl with platinum electrodes.


(c) Calculate e. m. f. of the following cell at 298 K:


Ni(s) / Ni2 + (0. 01 M) // Cu2 + (0. 1M) / Cu (s)



Write the overall cell reaction.


OR


(a) Apply Kohlrausch law of independent migration of ions, write the expression to determine the limiting molar conductivity of calcium chloride.


(b) Given are the conductivity and molar conductivity of NaCl solutions at 298K at different concentrations:



Compare the variation of conductivity and molar conductivity of NaCl solutions on dilution. Give reason.


(c) 0. 1 M KCl solution offered a resistance of 100 ohms in a conductivity cell at 298 K. If the cell constant of the cell is 1. 29 cm - 1, calculate the molar conductivity of KCl solution.

(a) The Ecell decreases.


Explanation:


As we know the Ecell of a cell is calculated by the Nerst equation:
E(Zn2+/Zn) = E0(Zn2+/Zn) -


This is the equation for the cell given above:


Therefore ecell here is inversely proportional to the concentration of Zn2 + ions as we see.


Clearly, if the concentration is increased, Ecell will decrease.


(b) At the Anode: Chlorine gas is evolved.


At the Cathode: Hydrogen gas is evolved


Explanation:


in the electrolysis of aqueous NaCl,


the ions produced will be –


Na+,Cl- ,H+ and OH- released by the dissociation of NaCl and H2O respectively.


At the Cathode:
Na+ (aq) + e-→ Na (s), E0(cell) = - 2.71V
H+ (aq) + e-→ 1/2 H2 (s), E0(cell) = 0.00 V


As the reaction with high Ecell is preferred,His released at cathode


At the anode:
Cl- (aq) → 1/2 Cl2 (g), E0(cell) = - 2.71V


Chlorine gets oxidized at anode to give chlorine at the anode.


(c) The overall cell - reaction is –


Cu2 + (aq) + Ni (s) → Ni2 + (aq) + Cu(s)


FORMULA TO FIND E0CELL


E0cell = E0cathode - E0anode


E0cell = 0. 34 - ( - 0. 25)


E0 = 0. 59V


FORMULA TO FIND ECELL


Ecell = E0cell - log OR Ecell = E0cell - log


Where,


n = the no. of exchanged electrons


r = gas constant


T = absolute temperature 298K


F = Faradays’s constant


Ecell = 0. 59 - log [n = 2]


Ecell = 0. 6195V


CONCLUSION


Ecell of the given cell is = 0. 6195V


OR


(a) If an electrolyte on dissociation gives cations and anions,then its limiting molar conductivity is given by,



For calcium chloride,



(b) The conductivity of the NaCl solution decreases on dilution as the number of ions per unit volume decreases.


The molar conductivity of NaCl increases on dilution as on dilution the inter - ionic interactions are overcome and ions are free to move about. This is also clearly seen in the data given above.


(c) Given,


Resistance = 100 ohms


C = 0. 1M


Cell constant G* = 1. 29 cm - 1


G* = K . R


K = = 0. 0129 S cm - 1


Formula to calculate limiting molar conductivity is =



= 129 S cm2 mol – 1


Conclusion:


The molar conductivity of the KCl solution is 129 S cm2 mol - 1


More from this chapter

All 27 →
23

Explain the following:

(a) Out of Sc3 + , Co2 + and Cr3 + ions, Sc3 + only is colorless in aqueous solutions. (Atomic no. : Co = 27; Sc = 21 and C = 24)


(b) The E°Cu2 + /Cu for copper metal is positive ( + 0. 34), unlike the remaining members of the first transition series


(c) La(OH)3 is more basic than Lu(OH)3.

 24

A metal complex having composition Cr(NH3)4Cl2Br has been isolated in two forms A and B. form A reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.

(i) Write the formulae of isomers A and B.


(ii) State the hybridization of chromium in each of them.


(iii) Calculate the magnetic moment (spin only value) of the isomer.

 25

(a) Identify A – D


(b) Distinguish between the following pair of compounds:


(i) Aniline and Benzylamine.


(ii) Methylamine and Dimethylamine


(c) Complete the following:



OR


(a) Account for the following:


(i) Direct nitration of aniline yields a significant amount of meta derivative.


(ii) Primary aromatic amines cannot be prepared by Gabriel phthalimide synthesis.


(b) Carry out the following conversions:


(i) Ethanoic acid into methenamine.


(ii) Aniline to p - Bromoaniline.


(c) Arrange the following in increasing order of basic strength: Aniline, p - nitroaniline, and p - toluene.

 27

(a) Account for the following observations:

(i) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed


(ii) Chlorine water is a powerful bleaching agent.


(iii) is a stronger oxidizing agent than Sb(V)


(b) What happens when


(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2.


(ii) XeF6 undergoes partial hydrolysis.


(Give the chemical equations involved).


OR


(a) What inspired N. Bartlett for carrying out a reaction between Xe and PtF6?


(b) Arrange the following in the order of property indicated against each set:


(i) F2, I2, Br2, Cl2(increasing bond dissociation enthalpy)


(ii) NH3, AsH3, SbH3, BiH3, PH3(decreasing base strength)


(c) Complete the following equations:


(i)


(ii)