CBSE Class 12 Chemistry Electrochemistry Worksheet Set 03

Question. Why on dilution the Lm of CH₃COOH increases drastically, while that of CH₃COONa increases gradually?
Answer. In case of CH₃COOH which is a weak electrolyte, the number of ions increases on dilution due to an increase in degree of dissociation resulting in drastic increase in Lm.
CH₃COOH+H₂O CH₃COO⁻ +H₃O ⁺
In the case of CH3COONa which is a strong electrolyte, the number of ions remains the same but the inter-ionic attraction decreases resulting in gradual increase in Lm.

Question. From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell.
Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life?
Answer. (i) Mercury cell
(ii) Fuel cell 
(iii) Lead storage cell
(iv) Dry cell 

Question. Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell.
Answer. Dry cell/Leclanche cell
Anode: Zn(s) → Zn²⁺ + 2e^– 
Cathode: MnO₂ + NH₄⁺ + e^– → MnO(OH) + NH₃

Question. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases 1.5 times while that of ‘A’ increases 25 times. Which of the two is a strong electrolyte ? Justify your answer.
Answer. ‘B’ is a strong electrolyte. B is a strong electrolyte which is completely dissociated into ions, but on dilution interionic forces overcome and ions are free to move. So there is slight increase in molar conductivity on dilution. 

Question. In a galvanic cell, the following cell reaction occurs:
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
E^o_cell = +1.56 V
(i) Is the direction of flow of electrons from zinc to silver or silver to zinc?
(ii) How will concentration of Zn²⁺ ions and Ag⁺ ions be affected when the cell functions?
Answer. (i) Zinc to silver 
(ii) Concentration of Zn²⁺ ions will increase and Ag⁺ ions will decrease.

Question. Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Λ_m) is
39.05 S cm² mol^-1. Given Λ˚_(H⁺) = 349.6 S cm² mol-1 and Λ°_(CH3COO^–) = 40.9 S cm² mol^-1.
Answer. Λ°_CH3COOH = Λ°_CH3COO^– + Λ°_H⁺ 
= 40.9 + 349.6 = 390.5 S cm²/mol 
Now, α = Λ_m/Λ°_m
= 39.05/390.5 = 0.1 

Question. Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell.
Answer. Mercury cell 
Anode: Zn(Hg) + 2OH^– → ZnO(s) + H₂O + 2e^–
Cathode: HgO + H₂O + 2e^– → Hg(l) + 2OH^– 

Question. Iron displaces copper from copper sulphate solution but Pt does not why?
Answer. Electrode potential of Fe is more than electrode potential of Cu. So, Fe displaces Cu from copper sulphate while electrode potential of Pt is less than Cu. Due to this reason, Pt cannot displace Cu from copper sulphate.

Short Answer Type Questions-II

Question. (a) The cell in which the following reaction occurs:
2 Fe³⁺ (aq) + 2 I^– (aq) → 2 Fe²⁺ (aq) + I₂ (s)
has E°_cell = 0.236 V at 298 K. Calculate the standard Gibb’s energy of the cell reaction.
(Given: 1 F = 96,500 C mol^–1)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours ?
(Given: 1 F = 96,500 C mol^–1) 
Answer. (a) ΔG° = – nFE°_cell 
n = 2
ΔG° = – 2 × 96500 C /mol × 0.236 V
= – 45548 J/mol
= – 45.548 kJ/mol
(b) Q = I t = 0.5 × 2 × 60 × 60
= 3600 C
96500 C = 6.023 × 10²³ electrons
3600 C = 2.25 × 10²² electrons

Question. The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55 × 10³ ohm. Calculate its molar conductivity. 
Answer. A = pr²
= 3.14 × 0.5 × 0.5 cm²
= 0.785 cm² 
l = 45.5 cm
G* = l/A = 45.5 cm/0.785 cm²
= 57.96 cm^–1
k = G*/R
= 57.96 cm^–1/4.55 × 103 Ω
= 1.27 × 10^–2 S cm^–1
∧_m = k × 1000/C
= [1.27 × 10^–2 S cm^–1] × 1000/0.05 mol/cm³
= 254.77 S cm² mol^–1

Question. Consider the following reaction:
Cu(s) + 2Ag⁺(aq) → 2Ag(s) + Cu²⁺(aq)
(i) Depict the galvanic cell in which the given reaction takes place.
(ii) Give the direction of flow of current.
(iii) Write the half-cell reactions taking place at cathode and anode.
Answer. (i) Cu(s) | Cu²⁺(aq) || Ag⁺ (aq) | Ag(s)
(ii) Current will flow from silver to copper electrode in the external circuit. 
(iii) Cathode: 2Ag⁺(aq) + 2e^– → 2Ag(s)
Anode: Cu(s) → Cu²⁺ (aq) + 2e^– 

Question. (a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO₃ for 15 minutes.
(Given: Molar mass of Ag = 108 g mol^–1,1 F = 96500 C mol^–1)
(b) Define fuel cell. 
Answer.
(a) m = ZI
= (108× 2× 15× 60)/1× 96500 
= 2.01 g (or any other correct method)
(b) Cells that convert the energy of combustion of fuels directly into electrical energy. 

Question. (a) For the reaction
2AgCl (s) + H₂ (g) (1 atm) → 2Ag(s)+2H⁺ (0.1 M)+2Cl^–(0.1 M),
ΔG°= – 43600 J at 25°C.
Calculate the e.m.f. of the cell.
[log 10^–n = –n]
(b) Define fuel cell and write its two advantages.
Answer.(a) ΔGo = – nFE^o 
–43600 = – 2 × 96500 ×E^o
Eo = 0.226 V
E = E^o – 0.059/2 log ([H⁺]² [Cl^–]² / [H₂]) 
= 0.226 – 0.059/2 log[ (0.1)² × (0.1)² ] / 1 
= 0.226 – 0.059 /2 log 10^-4
= 0.226 + 0.118 = 0.344 V
(Deduct half mark if unit is wrong or not written)
(b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol etc.) directly into electrical energy are called fuel cells. 
Advantages: High efficiency, non polluting (or any other suitable advantage)

Please click on below link to download CBSE Class 12 Chemistry Electrochemistry Worksheet Set A