Search

CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from Chapter 3 - Electrochemistry

The students appearing for CBSE Class 12th Chemistry Examination 2020 can go through the important questions and answers for Chapter 3 - Electrochemistry to prepare themselves for the upcoming examination. 

Mar 2, 2020 19:08 IST
facebook IconTwitter IconWhatsapp Icon
Electrochemistry
Electrochemistry

CBSE Class 12th Chemistry Examination will be held on March 7, 2020. Therefore, for the students appearing in the CBSE Class 12th Examination 2020,  we have listed several important questions and answers as per the latest syllabus by the CBSE Board.

Key Points to be mentioned while writing the answers to the below mentioned important questions: 

Question 1- Two half-reactions of an electrochemical cell are given below :

MnO 4 (aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I), E° = 1.51 V

Sn2+ (aq) → Sn4+ (aq) + 2e, E° = + 0.15 V.

Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured.

Answer: Redox Reaction Equation: 2MnO 4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O

cell= + 1.36 V (Product Favourd)

(Hint- E°cell = E°cathode – E°anode

Question 2- Determine the values of equilibrium constant (Kc) and ΔG° for the following reaction :

Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s),

E° = 1.05 V

(1F = 96500 C mol-1)

Answer: ΔG° = -202.65 KJ mol-1 and Kc = 0.35 x 107

(Hint- Use formulas: ΔG° = -nFE° and log Kc = ΔG°/ 2.303 RT )

Question 3- Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens when charging the battery? 

Answer: At Anode: Pb + SO4-2 → PbSO4 + 2e

                   At Cathode : PbO2 + SO4-2+ 4H+ + 2e → PbSO4 + 2H2O

When the battery is charged, the reaction is reversed. PbSO4 on anode and cathode is converted into Pb and PbO2 respectively.

Question 4- The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity.

Answer: Resistivity, R = 87.135 Ω cm  

              Conductivity, K = 0.01148 S cm-1 

              Molar Conductivity, Λm = 229.6 S cm2 mol-1

(Hint- Use formulas: R = ρl / A ; K= 1 / ρ ; Λm = K x 1000 / C)

Question 5-State Kohlrausch law of independent migration of ions. 

Answer: Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. 

 λ°m (AxBy) = xλ°+ + yλ°

Thus, if λ°Na+ and λ°Cl are limiting molar conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation:  λ°m (NaCl) = λ°Na+ + λ°Cl

Question 6- Define: 

1- Rate Constant (k)

2- Activation energy (Ea)

3- Conductivity (K)

4- Molar Conductivity (Λm)

5- Corrosion 

Answer: 1- Rate Constant (k): Rate constant is equal to the rate of reaction when the molar concentration of each of the reactants is unity. It is a proportionality constant. It is denoted by k. 

2- Activation energy (Ea): The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex. It is denoted by Ea

3- Conductivity (K): Conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. It is represented by K. 

4- Molar Conductivity (Λm): It is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are 1 cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by Λm

5- Corrosion: It is an electrochemical phenomenon where the deterioration of a substance occurs because of its reaction to the environment. 

Question 7-Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of CuSO4.

(Molar mass of Cu = 63.5 g mol-1,1 F = 96500 C mol-1)

Answer: Time, t = 1930 seconds

Question 8- Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) 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: α = 0.1

Question 9- Following reactions may occur at the cathode during the electrolysis of aqueous CuCl2 solution using Pt electrodes:

Cu2+ (aq) +  2e- → Cu (s) ; E° = +0.34 V

H+ (aq) + e- → 1/2 H2 (g) ; E° = 0.00 V

On the basis of their standard electrode potential values, which reaction is feasible at the cathode and why?

Answer:   Cu2+ (aq) + 2e → Cu will be feasible at Cathode as Cu2+ aq has higher reduction potential. 

Question 10- A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential is measured 0.422 V. Determine the concentration of the silver ion in the cell.

Given : E°Ag+/Ag = + 0.80 V, E° Cu2+/Cu = + 0.34 V. 

Answer: [ Ag+] = 13.93 M

(Hint- Use Nernst equation)

Question 11- A voltaic cell is set up at 25°C with the following half cells :

Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M)

Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.

(Log 8 × 10-6 = -0.54)

Answer: Equation for the reaction that occurs when the cell generates an electric current  is : 2Al + 3Ni+2 → 2Al+3 + 3Ni

              E° cell = 1.46 volts

Question 12- A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Furthermore 18 g of water is added to this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate

(i) the molecular mass of solute and

(ii) vapour pressure of water at 298 K. 

Answer: i) Molecular Mass, MB = 34 g/mol

ii) Vapour pressure, Po = 3.4 kPa

Question 13- When a certain conductance cell was filled with 0.1 M KCl, it has a resistance of 85 ohms at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 ohms. Calculate the molar conductance of the electrolyte at this concentration.

[Specific conductance of 0.1 M KCl = 1.29 × 10-2 ohm-1 cm-1]

Answer: Λm = 219.23 Ω-1 cm2 mol-1

Question 14- Calculate emf of the following cell at 25°C :

Fe | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1 bar) | Pt(s)

(Fe2+ | Fe) = -0.44 V,  E°(H+ | H2) = 0.00V

Answer: E cell = 0.4105 Volts

Question 15- One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO3)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver nitrate solution used.

(E° Zn2+/ Zn = -0.76 V;  E° Ag+/ Ag= + 0.80 V)

Answer: Cell Reaction is  Zn (s) + 2Ag+ (aq) - Zn2+ (aq) + 2Ag(s)

concentration of silver nitrate solution used: [Ag+] = 4.406 x 10-2 M

Question 16- Calculate ΔrG° and e.m.f. (E) that can be obtained from the following cell under the standard conditions at 25°C :

Zn (s) | Zn2+ (aq) || Sn2+ (aq) | Sn (s) 

Answer: ΔrG° = -119660 J mol-1

                E°cell = 0.62 

Question 17- Define: 

1- Fuel Cells

2-  Faraday’s first law of electrolysis

3- Secondary batteries

4- Degree of dissociation

5- Electrochemical cell 

6- Primary batteries

Answer:

1- Fuel Cells: They convert the energy produced during combustion of fuels like H2, CH4, etc. directly into electrical energy.

2-  Faraday’s first law of electrolysis: The amount of chemical reaction and hence the mass of any substance deposited/liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.

ω ∝ Q (∵ Q = I × t)

ω = = ZIt

3- Secondary batteries: The cells which can be recharged on passing an electric current through them in the opposite direction and can be used again. e.g. Lead-acid storage cell.

4- Degree of dissociation: It is the measure of the extent to which an electrolyte gets dissociated into its constituent ions. Higher the degree of dissociation, higher will be its molar conductance. 

α = λ /  λ°m

5- Electrochemical cell:  A device which converts chemical energy into electrical energy. It is produced as a result of redox reaction taking place in the electrolyte.

6- Primary batteries: These batteries can not be recharged or reused. 

The above-mentioned questions are important for the CBSE Class 12th Chemistry Examination 2020. The students appearing for the CBSE Class 12th Chemistry Examination 2020 will find these questions helpful while preparing for the upcoming examination. Below we have mentioned links for the important questions that you might be interested in looking into: 

CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from Chapter 1 - The Solid State

CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from Chapter 2 - Solutions

Related Stories