Good Earth School Naduveerapattu Class XII Chemistry...

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Good Earth School Naduveerapattu Class XII Chemistry Worksheet (1) The resistance of 0.01 M NaCl solution in a conductivity cell was found to be 210Ω. The specific conductance of this solution is 4.5 x 10 –3 S cm –1 . What is the cell constant of the cell? [Ans: 0.945 cm –1 ] (2) The resistance of 0.01 N NaCl solution is 200 Ω at 25°C. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution. [Ans: 500 –1 cm 2 eq –1 ] (3) The specific conductance of 50 solution of KCl at 25°C is 0.002765 mho cm –1 . If the resistance of a cell containing this solution is 400 ohms, find out the cell constant. [Ans: 1.106 cm –1 ] (4) The resistance of a decinormal solution of a salt occupying a volume between two platinum electrodes 1.80cm apart and 5.4 cm 2 in area was found to be 32 ohms. Calculate the specific and equivalent conductance of the solution. [Ans: 0.01042 ohm –1 cm –1 , 104.2 ohm –1 cm 2 eq –1 ] (5) A potential difference of 20 volts applied to the ends of a column of 10 AgNO3 solution, 4cm in diameter and 12cm in length, gives a current of 1.198 amperes. Calculate the specific and equivalent conductivity of the solution. [Ans: 0.057 ohm –1 cm –1 , 570 ohm –1 cm 2 eq –1 ] (6) At 298K, the molar conductivities at infinite dilution of sodium propionate (CH3CH2COONa), HCl and NaCl are 85.9, 426.1 and 126 ohm –1 cm 2 mol –1 respectively. Calculate the molar conductivity of propionic acid at infinite dilution. [Ans: 386.0 ohm –1 cm 2 mol –1 ] (7) The molar conductivity of a weak monobasic acid at infinite dilution is 387 ohm –1 cm 2 mol –1 . The 0.02M solution of the same acid has a specific conductivity of 3.3 x 10 –4 ohm –1 cm 2 mol –1 . Calculate the degree of dissociation of the acid at the given concentration. [Ans: 0.0426] (8) The molar conductivity of CH3COOH at infinite dilution is found to be 387 ohm –1 cm 2 mol –1 but at a dilution of 1g mol in 1000 litres it is 55 ohm –1 cm 2 mol –1 . What is percent dissociation of the acid at the given dilution? [Ans: 14.21%] (9) Calculate the molar conductivity at infinite dilution for CH3COOH if the molar conductivities at infinite dilution for NaCl, HCl and CH3COONa are 126.45, 426.16 and 91.0 ohm –1 cm 2 mol –1 respectively. [Ans: 390.71 ohm –1 cm 2 mol –1 ] (10) A potential difference of 0.25 volts is observed when an electrode system M/M 2+ (1 mole L –1 ) is connected to a standard hydrogen electrode. If the direction of the flow of current is from hydrogen electrode to the metal electrode, calculate the standard potential of the electrode. [Ans: E ° M 2+ /M = - 0.25 volts]

Transcript of Good Earth School Naduveerapattu Class XII Chemistry...

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Good Earth School Naduveerapattu

Class XII Chemistry Worksheet

(1) The resistance of 0.01 M NaCl solution in a conductivity cell was found to be 210Ω.

The specific conductance of this solution is 4.5 x 10–3 S cm–1. What is the cell constant of the cell?

[Ans: 0.945 cm–1] (2) The resistance of 0.01 N NaCl solution is 200 Ω at 25°C. Cell constant of the

conductivity cell is unity. Calculate the equivalent conductance of the solution. [Ans: 500 𝛀–1 cm2 eq–1]

(3) The specific conductance of 𝑁50

solution of KCl at 25°C is 0.002765 mho cm–1. If the resistance of a cell containing this solution is 400 ohms, find out the cell constant.

[Ans: 1.106 cm–1] (4) The resistance of a decinormal solution of a salt occupying a volume between two

platinum electrodes 1.80cm apart and 5.4 cm2 in area was found to be 32 ohms. Calculate the specific and equivalent conductance of the solution.

[Ans: 0.01042 ohm–1 cm–1, 104.2 ohm–1 cm2 eq–1] (5) A potential difference of 20 volts applied to the ends of a column of 𝑁

10 AgNO3

solution, 4cm in diameter and 12cm in length, gives a current of 1.198 amperes. Calculate the specific and equivalent conductivity of the solution.

[Ans: 0.057 ohm–1 cm–1, 570 ohm–1 cm2 eq–1] (6) At 298K, the molar conductivities at infinite dilution of sodium propionate

(CH3CH2COONa), HCl and NaCl are 85.9, 426.1 and 126 ohm–1 cm2 mol–1 respectively. Calculate the molar conductivity of propionic acid at infinite dilution.

[Ans: 386.0 ohm–1 cm2 mol–1] (7) The molar conductivity of a weak monobasic acid at infinite dilution is 387 ohm–1

cm2 mol–1. The 0.02M solution of the same acid has a specific conductivity of 3.3 x 10–4 ohm–1 cm2 mol–1. Calculate the degree of dissociation of the acid at the given concentration.

[Ans: 0.0426] (8) The molar conductivity of CH3COOH at infinite dilution is found to be 387 ohm–1

cm2 mol–1 but at a dilution of 1g mol in 1000 litres it is 55 ohm–1 cm2 mol–1. What is percent dissociation of the acid at the given dilution?

[Ans: 14.21%] (9) Calculate the molar conductivity at infinite dilution for CH3COOH if the molar

conductivities at infinite dilution for NaCl, HCl and CH3COONa are 126.45, 426.16 and 91.0 ohm–1 cm2 mol–1 respectively.

[Ans: 390.71 ohm–1 cm2 mol–1] (10) A potential difference of 0.25 volts is observed when an electrode system M/M2+

(1 mole L–1) is connected to a standard hydrogen electrode. If the direction of the flow of current is from hydrogen electrode to the metal electrode, calculate the standard potential of the electrode.

[Ans: E°M2+/M = - 0.25 volts]

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(11) The standard reduction potential of the electrode X/X+ (1mol L–1) is +0.80 volts. Predict the direction of the flow of electrons when it is connected to a standard hydrogen electrode.

[Ans: From hydrogen electrode to X electrode] (12) Using the standard electrode potentials predict the reaction, if any, that occurs

between the following: (a) Fe3+ (aq) and I– (aq) (d) Ag (s) and Fe3+ (aq) (b) Ag+ (aq) and Cu (s) (e) Br2 (aq) and Fe2+ (aq) (c) Fe3+ (aq) and Br – (aq)

(13) Calculate the standard electrode potential of the Ni2/Ni electrode, if the cell potential of the cell Ni | Ni2+ (0.01M) ∥ Cu2+ (0.1M) | Cu is 0.59V. Given E°

Cu2+/Cu = + 0.34V.

[Ans: - 0.2205V] (14) Consider a cell composed of two half cells :

(i) Cu (s) | Cu2+ (aq) and (ii) Ag (s) | Ag+ (aq) Calculate : (a) the standard cell potential, and (b) the cell potential when [Cu2+] is 2M and [Ag+] is 0.05M. Given, E°

Cu2+/Cu = +0.33V, E°Ag+/Ag = + 0.80V, R = 8.31 JK–1 mol–1, F=96500C mol–1.

[Ans: 0.46 V, 0.374v] (15) Calculate the cell potential at 298 K:

Zn (s) | Zn2+ (0.1 M) ∥ Sn2+ (0.001 M) | Sn (s) Given: E°

Zn2+/Zn = -0.76V, E°Sn2+/Sn = -0.14V

(Gas constant, R = 8.314JK–1mol–1, Faraday constant , F = 96500 C mol–1) [Ans: 0.5609 V]

(16) The standard potentials are given as: E°

Cu2+/Cu = -0.34V and E°Ag+/Ag = - 0.80V

Calculate the cell potential (E) for the cell containing 0.100 M Ag+ and 4.00 M Cu2+ at 298 K.

[Ans:0.3832 V] (17) Calculate the quantity of electricity that would be required to reduce 12.3g of

nitrobenzene to aniline, if the current efficiency is 50%. [Ans: 115800 C]

(18) How many molecules of chlorine should be deposited from sodium chloride in 1 minute by a current of 300 mA?

[Ans: 5361 x 1019] (19) A constant current was passed through a solution of AuCl–4 between gold

electrodes. After a period of 14 minutes, the cathode recorded an increase in mass of 1.808g. Calculate the quantity of current passed in solution. What was the magnitude of current? (Atomic mass of gold = 197.0)

[Ans : 2656.9C, 3.163 amp] (20) A current of 4 amperes was passed for 2 hours through a solution of copper

sulphate when 5.0g of copper was deposited. Calculate the current efficiency. [Ans: 52.77%]

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(21) When a conductivity cell was filled with 0.02 M KCl, it had a resistance of 82.4 ohm at 25°C and when filled with 0.005 N K2SO4 it had a resistance of 326 ohm. Calculate: (i) cell constant (ii) specific conductivity, and (iii) equivalent conductivity of 0.005N K2SO4 solution. The specific conductance of 0.02 M KCl is 0.002768 ohm–1 cm–1 .

[Ans: (i) 0.2281 cm–1 (ii) 6.997 x 10–4 ohm–1 cm–1 (iii) 139.94 ohm–1 cm2 eq–1] (22) The resistance of a 0.5M solution of an electrolyte in a cell was found to be

45 ohm. Calculate the molar conductivity of the solution if the electrodes in the cell are 2.2 cm apart and have an area of 3.8 cm2.

[Ans: 25.73 ohm–1 cm–2 mol–1] (23) During electrolysis, 4828.4mg of iodine requires 3671.3 coulombs of electricity for

its deposition at the anode. Calculate the value of Faraday’s constant. [Ans: 96489.25 C]

(24) Calculate the volume of Cl2 at STP produced during electrolysis of MgCl2 which produces 6.50g of Mg. (At.wt. of Mg = 24.3)

[Ans: 5.99 L] (25) A current of 1.70 amp is passed through 500 mL of 0.16 M solution of ZnSO4 for

230 seconds with a current efficiency of 90%. Find the molarity of Zn2+ after the deposition of zinc. Assume that the volume of the solution remains constant during electrolysis.

[Ans: 0.1564 M] (26) How many grams of silver could be plated out on a serving tray by electrolysis of a

solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? Density of silver is 10.5 g cm–3.

[Ans: 272.88 g, 1.02 x 104 cm2] (27) Calculate the quantity of electricity required to reduce 6.15 g of nitrobenzene to

aniline if the current efficiency is 68 per cent. If potential drop across the cell is 7.0 volts, calculate the energy consumed in the process.

[Ans: 42573.5 C, 298.014 kJ] (28) For the cell

Zn | Zn2+ (a = 1) ∥ Cu2+ (a = 1) | Cu Given EZn/Zn2+ = 0.761 volt; ECu2+/Cu = +0.339 volt, volt (i) Write the cell reaction. (ii) Calculate the emf and free energy change at 298 K involved in the cell. [Faraday’s constant, F=96500 coulomb equivalent–1]

(29) (a) A current of 10 A is passed for 80 min. and 27 seconds through a cell containing dilute sulphuric acid. (i) How many moles of oxygen gas will be liberated at the anode? (ii) Calculate the amount of zinc deposited at the cathode when another cell

containing ZnSO4 solution is connected in series (Zn = 65). (b) Calculate:

Ecell at 25°C for the reaction Zn + Cu2+ (0.20M) → Zn2+ (0.50 M) + Cu Given: E° (Zn2+/Zn) = - 0.76 volt, E° (Cu2+/Cu) = 0.34 volt.

(c) Give a reason for the following: Specific conductance decreases with dilution whereas equivalent conductance increases with dilution.

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(30) (a) Mention any two factors affecting the electrode potential of a metal. (b) For the following cell, calculate the emf: Al/Al3+ (0.01 M)//Fe2+ (0.02 M)/Fe Given: E°Al+3/Al = -1.66V, E° Fe+2/Fe = -0.44V.

(31) Consider the reduction 2Ag+ + Cd → 2Ag + Cd2+. The reduction potentials of Ag+/Ag and Cd2+/Cd are + 0.80 volt and -0.40 volt respectively. (i) Give the cell representation. (ii) What is the standard cell emf, E°? (iii) What will be the emf of the cell if concentration of Cd2+ is 0.1M and Ag+ is

0.2M? (iv) Will the cell work spontaneously for the condition given in (iii) above?