Class 12: Chemistry Chapter 3 solutions. Complete Class 12 Chemistry Chapter 3 Notes.
Contents
NCERT Solutions for 12th Class Chemistry: Chapter 3-Electrochemistry
NCERT 12th Chemistry Chapter 3, class 12 Chemistry chapter 3 solutions
3.1. Arrange the following metals in the order in which they displace each other from their salts.
Al, Cu, Fe, Mg and Zn
Sol: Mg, Al, Zn, Fe, Cu.
3.2. Given the standard electrode potentials, K+/K=-2. 93 V, Ag+/Ag = 0.80 V, Hg2+/Hg =0.79V, Mg2+/Mg=-2.37V, Cr3+/Cr=0.74V.
Arrange these metals in their increasing order of reducing power.
Sol: Higher the oxidation potential more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be Ag<Hg<Cr<Mg<K.
3.3. Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) —-> 7M2+(aq) + 2Ag (s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Sol. The set-up will be similar to as shown below,
(i) Anode, i. e, zinc electrode will be negatively charged.
(ii) The current will flow from silver to copper in the external circuit.
(iii) At anode: Zn(s) ——–> Zn2+(aq) + 2e–
At cathode: 2Ag+(aq) + 2e– ——–> 2Ag(s)
3.4. Calculate the standard cell potentials of the galvanic cells in which the following reactions take place.
Also calculate ∆G° and equilibrium constant for the reaction. (C.B.S.E. Outside Delhi 2008)
Sol:
3.5. Write the Nernst equation and emf of the following cells at 298 K:
Sol:
NCERT 12th Chemistry Chapter 3, class 12 Chemistry chapter 3 solutions
3.6. In the button cells widely used in watches and other devices the following reaction takes place:
Sol:
3.7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Sol: The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by K (kappa). Thus, if K is the specific conductance and G is the conductance of the solution, then
Now, if I = 1 cm and A = 1 sq.cm, then K = G.
Hence, conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte.
Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one 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 ∆m.
Variation of conductivity and molar conductivity with concentration: Conductivity always decreases with decrease in concentration, for both weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.
Molar conductivity increases with decrease in concentration. This is because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume.
3.8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.
Sol:
3.9. The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω What is the cell constant if conductivity of 0.001 M KCI solution at 298 K is 0.146 x 10-3 S cm-1?
Sol:
3.10. The conductivity of NaCl at 298 K has been determined at different concentrations and the results are given below:
Sol:
3.11. Conductivity of 0.00241 M acetic acid is 7.896 x 10-5 S cm-1. Calculate its molar conductivity. If Λm0, for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?
Sol:
3.12. How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al?
(ii) 1 mol of Cu2+ to Cu ?
(iii) 1 mol of Mn04- to Mn2+?
Sol: (i) The electrode reaction is Al3+ + 3e ——> Al
∴ Quantity of charge required for reduction of 1 mol of Al3+=3F=3 x 96500C=289500C.
(ii) The electrode reaction is Cu2+ + 2e– ——–> Cu
∴ Quantity of charge required for reduction of 1 mol of Cu2+=2F=2 x 96500=193000 C.
(iii) The electrode reaction is Mn04- ———-> Mn2+.
i.e., Mn7+ + 5e–——-> Mn2+.
∴ Quantity of charge required = 5F
=5 x 96500 C=4825000.
3.13. How much electricity in terms of Faraday is required to produce :
(i) 20·0 g of Ca from molten CaCl2
(ii) 40·0 g of Al from molten Al2O3 ?
Sol:
3.14. How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to 02 (ii) 1 mol of FeO to Fe203
Sol:
3.15. A solution of Ni(N03)2 is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Sol:
3.16. Three electrolytic cells A, B, C containing solutions of ZnS04, AgNO3 and CuS04, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 45 g of silver deposited at the cathode of call B. How long did the current flow? What mass of copper and zinc were deposited?
Sol:
3.17. Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible.
(a) Fe3+(aq) and I–(aq)
(b) Ag+(aq) and Cu(s)
(c) Fe3+(aq) and Br–(aq)
(d) Ag(s) and Fe3+(aq)
(e) Br2(aq) and Fe2+(aq).
Sol:
A particular reaction can be feasible if e.m.f. of the cell based on the E° values is positive. Keeping this in mind, let us predict the feasibility of the reactions.
3.18. Predict the products of electrolysis in each of the following.
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2S04 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Sol:
NCERT Solutions for 12th Class Chemistry: Chapter 3: Download PDF
NCERT Solutions for 12th Class Chemistry: Chapter 3-Electrochemistry
Download PDF: NCERT Solutions for 12th Class Chemistry: Chapter 3-Electrochemistry PDF
Chapterwise NCERT Solutions for Class 12 Chemistry:
- Chapter 1 : The Solid State
- Chapter 2 : Solutions
- Chapter 3 Electrochemistry
- Chapter 4 : Chemical Kinetics
- Chapter 5 : Surface chemistry
- Chapter 6 : General Principles and Processes of Isolation of Elements
- Chapter 7 : The p Block Elements
- Chapter 8 : The d and f Block Elements
- Chapter 9 : Coordination Compounds
- Chapter 10 : Haloalkanes and Haloarenes
- Chapter 11 : Alcohols Phenols and Ether
- Chapter 12 : Aldehydes Ketones and Carboxylic Acids
- Chapter 13 : Amines
- Chapter 14 : Biomolecules
- Chapter 15 : Polymers
- Chapter 16 : Chemistry in Everyday Life
About NCERT
The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website to learn more.