Chapter 2 Electrochemistry Exercises NCERT Solutions Class 12 Chemistry- PDF Download
Page No. 59
2.1. Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn
Solution
The following is the order in which the given metals displace each other from the solution of their salts. Mg, Al, Zn, Fe, Cu
2.2. Given the standard electrode potentials,
K+/K = −2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = −2.37 V, Cr3+/Cr = − 0.74V
Arrange these metals in their increasing order of reducing power.
Solution
The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag.
Hence, the reducing power of the given metals increases in the following order: Ag < Hg < Cr < Mg < K
2.3. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) ⟶ Zn2+(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.
Solution
The galvanic cell in which the reaction takes place will be depicted as:
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
(i) Anode, i.e., zinc electrode, will be negatively charged.
(ii) The current will flow from silver to zinc in the external circuit.
(iii) At anode: Zn(s) ⟶ Zn2+(aq) + 2e-
At cathode: Ag+(aq) + e- ⟶ Ag(s)
2.4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ΔrGθ and equilibrium constant of the reactions.
Solution
(i)
The galvanic cell of the given reaction is depicted as:
Cr(s) | Cr3+(aq) || Cd2+(aq) | Cd(s)
Now, the standard cell potential is
Eθcell = EθR - EθL
= -0.40-(-0.74)
= +0.34 V
△tGθ = -nFEθcell
In the given equation,
n = 6
F = 96487 C mol−1
Eθcell = +0.34 V
Then, △rGθ = -6×96487 Cmol−1×0.34V
= −196833.48 CV mol−1
= −196833.48 J mol−1
= −196.83 kJ mol−1
Again,
△rGθ = -RT ln K
⇒△rGθ = -2.303 RT ln K
= 34.496
∴ K = antilog (34.496)
= 3.13 × 1034
(ii)
The galvanic cell of the given reaction is depicted as:
Fe2+(aq) | Fe3+(aq) || Ag+(aq) | Ag(s)
Now, the standard cell potential is
Eθcell = EθR - EθL
= 0.80 - 0.77
= 0.03 V
Here, n = 1.
Then, △rGθ = -nFEθcell
= −1 × 96487 C mol−1 × 0.03 V
= −2894.61 J mol−1
= −2.89 kJ mol−1
Again, △rGθ = -2.303RT lnK
= 0.5073
∴ K = antilog (0.5073)
= 3.2 (approximately)
2.5. Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s) | Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s)
(ii) Fe(s) | Fe2+(0.001M) || H+(1M)|H2(g)(1bar) | Pt(s)
(iii) Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) | Pt(s)
(iv) Pt(s) | Br2(l) | Br−(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s).
Solution
(i) For the given reaction, the Nernst equation can be given as:
= 2.7 − 0.02955
= 2.67 V (approximately)
(ii) For the given reaction, the Nernst equation can be given as:
= 0.44 - 0.02955(-3)
= 0.52865 V
= 0.53 V (approximately)
(iii) For the given reaction, the Nernst equation can be given as:
= 0.14 − 0.0295 × log125
= 0.14 − 0.062
= 0.078 V
= 0.08 V (approximately)
(iv) For the given reaction, the Nernst equation can be given as:
= -1.09 - 0.02955×log(1.11×107)
= -1.09 - 0.2955(0.0453+7)
= -1.09 - 0.208
= -1.298 V
2.6. In the button cells widely used in watches and other devices the following reaction takes place:
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH−(aq)
Determine △rGθ and E° for the reaction.
Solution
∴ E° = 1.104V
We know that,
△rG° = -nFE°
= −2 × 96487 × 1.04
= −213043.296 J
= −213.04 kJ
2.7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Solution
Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol κ. If ρ is resistivity, then we can write:
The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length i.e.
(Since, a = 1, l = 1)
Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
∴ Λm = kV
Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.
The variation of Λm with c for strong and weak electrolytes is shown in the following plot:
2.8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm−1. Calculate its molar conductivity.
Solution
Given,
κ = 0.0248 S cm−1
c = 0.20 M
∴ Molar conductivity,
= 124 Scm2mol−1
2.9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146×10−3 S cm−1.
Solution
Given,
Conductivity, κ = 0.146 × 10−3 S cm−1
Resistance, R = 1500 Ω
∴ Cell constant = κ × R
= 0.146 × 10−3 × 1500
= 0.219 cm−1
2.10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
|
Concentration/M |
0.001 |
0.010 |
0.020 |
0.050 |
0.100 |
|
102×κ/S m–1 |
1.237 |
11.85 |
23.15 |
55.53 |
106.74 |
Calculate Λm for all concentrations and draw a plot between Λm and c12 Find the value of Λ°m.
Solution
Given,
κ = 1.237 × 10−2 S m−1, c = 0.001 M
Then, κ = 1.237 × 10−4 S cm−1, c½ = 0.0316 M1/2
= 123.7 S cm2 mol−1
Given,
κ = 11.85 × 10−2 S m−1, c = 0.010M
Then, κ = 11.85 × 10−4 S cm−1, c½ = 0.1 M1/2
= 118.5 S cm2 mol−1
Given,
κ = 23.15 × 10−2 S m−1, c = 0.020 M
Then, κ = 23.15 × 10−4 S cm−1, c1/2 = 0.1414 M1/2
= 115.8 S cm2 mol−1
Given,
κ = 55.53 × 10−2 S m−1, c = 0.050 M
Then, κ = 55.53 × 10−4 S cm−1, c1/2 = 0.2236 M1/2
= 111.1 1 S cm2 mol−1
Given,
κ = 106.74 × 10−2 S m−1, c = 0.100 M
Then, κ = 106.74 × 10−4 S cm−1, c1/2 = 0.3162 M1/2
= 106.74 S cm2 mol−1
Now, we have the following data:
|
C1/2/M1/2 |
0.0316 |
0.1 |
0.1414 |
0.2236 |
0.3162 |
|
Λm(Scm2mol-1) |
123.7 |
118.5 |
115.8 |
111.1 |
106.74 |
Since the line interrupts Λm at 124.0 S cm2 mol−1, Λ°m = 124.0 S cm2 mol−1.
2.11. Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if Λ0m for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?
Solution
Given, κ = 7.896 × 10−5 S m−1
c = 0.00241 mol L−1
Then, molar conductivity,
= 32.76S cm2 mol−1
Again, Λ0m = 390.5 S cm2 mol−1
= 0.084
= 1.86 × 10−5 mol L−1
2.12. 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 MnO4- to Mn2+.
Solution
(i) Al3+ + 3e- → Al
∴ Required charge = 3 F
= 3 × 96487 C
= 289461 C
(ii) Cu2+ + 2e- → Cu
∴ Required charge = 2 F
= 2 × 96487 C
= 192974 C
(iii) MnO4- → Mn2+
i.e. Mn7+ + 5e- → Mn2+
∴ Required charge = 5 F
= 5 × 96487 C
= 482435 C
2.13. How much electricity in terms of Faraday is required:
(i) 20.0 g of Ca from molten CaCl2
(ii) 40.0 g of Al from molten Al2O3
Answer
(i) According to the question,
Ca2+ + 2e- → Ca
Electricity required to produce 40g of Calcium = 2F
Therefore, electricity required to produce 20g of calcium
= 1F
(ii) According to the question,
Al3+ + 3e- → Al
Electricity required to produce 27g of Al = 3F
Therefore, electricity required to produce 40g of Al
= 4.44F
2.14. How much electricity is required in coulomb for the oxidation of:
(i) 1 mol of H2O to O2.
(ii) 1 mol of FeO to Fe2O3.
Solution
According to the question,
Now, we can write:
Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F
= 2 × 96487 C
= 192974 C
(ii) According to the question,
Fe2+ → Fe3+ + e-1
Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F
= 96487 C
2.15. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Solution
Given,
Current = 5A
Time = 20 × 60 = 1200 s
∴ Charge = current × time
= 5 × 1200
= 6000 C
According to the reaction,
Nickel deposited by 2 × 96487 C = 58.71 g
Therefore,
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
2.16. Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Solution
According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore,
= 1295.43 C
Given,
Current = 1.5 A
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore,
= 0.426 g of Cu
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore,
= 0.439 g of Zn
2.17. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+(aq) and I−(aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br− (aq)
(iv) Ag(s) and Fe3+ (aq)
(v) Br2 (aq) and Fe2+ (aq).
Solution
(i)
Since E° for the overall reaction is positive, the reaction between Fe3+(aq) and I-(aq) is feasible.
(ii)
Since E° for the overall reaction is positive, the reaction between Ag+ (aq) and Cu(s) is feasible.
(iii)
Since E° for the overall reaction is negative, the reaction between Fe3+(aq) and Br−(aq) is not feasible.
(iv)
Since E° for the overall reaction is negative, the reaction between Ag (s) and Fe3+(aq) is not feasible.
(v)
Since E∘ for the overall reaction is positive, the reaction between Br2(aq) and Fe2+(aq) is feasible.
2.18. 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 H2SO4 with platinum electrodes.
(iv) (iii) A dilute solution of CuCl2 with platinum electrodes.
Solution
(i) At cathode:
The following reduction reactions compete to take place at the cathode.
Ag+(aq) + e- → Ag(s); E° = 0.80V
H+(aq) + e- → ½ H2(g); E° = 0.00V
The reaction with a higher value of E° takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by NO3- ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.
(ii) At cathode:
The following reduction reactions compete to take place at the cathode.
Ag+(aq) + e- → Ag(s); E° = 0.80V
H+(aq) + e- → ½ H2(g); E° = 0.00V
The reaction with a higher value of E° takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
Since Pt electrodes are inert, the anode is not attacked by NO3- ions. Therefore, OH- or NO3- ions can be oxidized at the anode. But OH- ions having a lower discharge potential and get preference and decompose to liberate O2.
OH- → OH + e-
4OH- → 2H2O + O2
(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.
H+(aq) + e- → ½ H2(g)
At the anode, the following processes are possible
2H2O(l) → O2(g) + 4H+(aq) + 4e-; E° = +1.23V ...(i)
2SO42-(aq) → S2O62-(aq) + 2e-; E° = +1.96V ...(ii)
For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The following reduction reactions compete to take place at the cathode.
Cu2+(aq) + 2e- → Cu(s); E° = 0.34V
H+(aq) + e- → ½ H2(g); E° = 0.00V
The reaction with a higher value of E° takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
At anode:
The following oxidation reactions are possible at the anode.
The following reduction reactions compete to take place at the cathode.
Cl-(aq) → ½ Cl2 + e-; E° = 1.36V
H2O(g) → O2(g) + 4H+(aq) + 4e-; E° = +1.23V
At the anode, the reaction with a lower value of E° is preferred. But due to over potential of oxygen, Cl- gets oxidised at the anode to produce Cl2 gas.