NCERT 12 Chemistry Electrochemistry Chapter 3 exercises 1

Q – 3.1. How would you determine the standard electrode potential of Mg2+ 1 Mg ?
Ans. We will set up a cell consisting of Mg l MgSO4 (1 M) as one electrode ( by dipping a magnesium wire in 1 M MgSO4 solution) and standard hydrogen electrode Pt, H2 (1 atm) H+ (1 M) as the second electrode and measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that electrons flow from magnesium electrode to hydrogen electrode, i.e., oxidation lakes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as :
chemistry chapter 3 ex 1 qes 1
Q – 3.2. Can you store copper sulphate solution in a zinc pot?
Ans. Zinc is more reactive than copper. Hence, it displaces copper from copper sulphate solution as follows :
Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
Thus, zinc reacts with CuSO4 solution. Hence, we cannot store copper sulphate solution in a zinc pot.
chemistry chapter 3 ex 1 qes 2
Q – 3.3. Consult the table of the standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.
Ans. Oxidation of ferrous ions means that the following reaction should occur :
Fe2+ → Fe3+ + e- ; E0 ox = – 0.77 V
Only those substance can oxidize Fe2+ to Fe3+ which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reaction is positive. This is so for elements lying below Fe3+/Fe2+ in the electrochemical series,
i.e., Br2, Cl2, and F2.
Q – 3.4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
chemistry chapter 3 ex 1 qes 4
Q – 3.5. Calculate the emf of the cell in which the following reaction takes place:
Ni (s) + 2 Ag+ (0.002 M) + 2 Ag (s)
Given that E0cell – 1.05 V
Ans. Applying Nernst equation to the given cell reaction,
chemistry chapter 3 ex 1 qes 5
Q – 3.6. The cell in which following reaction accurs:
2Fe3+ (aq) + 2 I- (aq) → 2 Fe2+ (aq) + I2 (s)
Has E0cell = 0.236 V 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Ans. 2Fe3+ + 2 e- → 2 Fe2+ or 2 I- → I2 + 2 e-
Hence, for the given cell reaction, n = 2.
chemistry chapter 3 ex 1 qes 6
Q – 3.7. Why does the conductivity of a solution decrease with dilution?
Ans. Conductivity of a solution is the conductance of ions present is a unit volume if the solution. On dilution the number of ions per unit volume decreases. Hence, the conductivity decreases.
Q – 3.8. Suggest a way to determine the ⋀0m value of water.
chemistry chapter 3 ex 1 qes 7
Q – 3.9. The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissociation constant. Given λ0 (H+) = 349.6 S cm2 mol-1 and λ0 (HCoo-) = 54.6 S cm2 mol-1.
chemistry chapter 3 ex 1 qes 9
Q – 3.10. If a current of 0-5 ampare flows through a metallic wire for 2 hours, then how many electrons flow through the wire?
Ans. Q (Coulombs) = I (ampere) × t (sec)
= (0.5 ampere) × (2×60×60 s) = 3600 C
A flow of 1 F, i.e., 96500 C is equivalent to flow of 1 mole of electrons, i.e., 6.02 × 1023 electrons 3600chemistry chapter 3 ex 1 qes 10
Q – 3.11. Suggest a list of metals that are extracted electrolytically.
Ans. Sodium, calcium, magnesium and aluminum.
Q – 3.12. Consider the reaction:
Cr2O72- + 14 H+ + 6e- → 2 Cr3+ 7 H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72- = ?
Ans. From the given reaction, 1 mol of Cr2O72- ions require
6 F = 6 × 96500 C = 579000 C of electricity for reduction to Cr3+.
Q – 3.13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Ans. A lead storage battery consists of anode of lead, cathode of a grid of lead packed with lead dioxide (PbO2) and 38% solution of sulphuric acid as electrolyte. When the battery is in use, the following reactions take place:
chemistry chapter 3 ex 1 qes 13
Overall reaction :
Pb (s) + PbO2 (s)+2 H2SO4 (aq) →2 PbSO4 (s) + 2H2O(l)
On charging the battery, the reverse reaction takes place,
i.e., PbSO4 deposited on the electrodes is converted back into Pb and PbO2 and H2SO4 is regenerated.
Reverse of reaction (i) will be reduction and hence will take place at cathode. Reverse of reaction (ii) will be oxidation and hence will take place at anode.
Q – 3.14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Ans. Methane and methanol.
Q- 3.15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Ans. The water layer present on the surface of iron (especially in the rainy season) dissolves acidic oxides of the air like CO2, So2 etc. to form acids which dissociate to give H+ ions
chemistry chapter 3 ex 1 qes 15
In the presence of H+ ions, iron starts losing electrons at some spot to form ferrous ions, i.e., its oxidation takes place. Hence, this spot acts as the anode:
Fe (s) →Fe2+ (aq) + 2 e-
The electrons thus released move through the metal to reach another spot where H+ ions and the dissolved oxygen take up these electrons and reduction reaction takes place. Hence, This spot acts as the cathode:
O2 (g) + 4 H+(aq) + 4 e- → 2 H2O (l)
The overall reaction is :
2F (s) + O2 (g) + H+ (aq) → 2 Fe2+ (aq) + 2 H2O(l)
Thus, an electrochemical cell is set up on the surface.
Ferrous ions are further oxidized by the atmospheric oxygen to ferric ions which combine with water molecules to from hydrated ferric oxide, Fe2O3. xH2O, which is rust.

Updated: November 15, 2022 — 2:26 pm

Leave a Reply

Your email address will not be published.