NCERT Solutions for Class 12 Chemistry Part 1 Chapter 3
Electrochemistry Class 12
Chapter 3 Electrochemistry Exercise Solutions
In text : Solutions of Questions on Page Number : 68
How would you determine the standard electrode potential of the systemMg2+| Mg?
(aq)
The standard electrode potential of Mg2+| Mg can be measured with
respect to the standard hydrogen electrode,
represented by Pt(s), H2(g)(1 atm) | H+ (1 M).
A cell, consisting of Mg | MgSO4(aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.
Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.
Here, for the standard hydrogen electrode is zero.
∴
Can you store copper sulphate solutions in a zinc pot?
Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution.
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.
; = - 0.77 V
This implies that the substances having higher reduction potentials than
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
For hydrogen electrode, , it is given that pH = 10
∴[H+] = 10 - 10M
Now, using Nernst equation:
=
= - 0.0591 log 1010
Calculate the emf of the cell in which the following reaction takes place:
Answer :
Applying Nernst equation we have:
= 1.05 - 0.02955 log 4 × 104
= 1.05 - 0.02955 (log 10000 + log 4)
= 1.05 - 0.02955 (4 + 0.6021)
The cell in which the following reactions occurs:
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Here, n = 2, T = 298 K
We know that:
= - 2 × 96487 × 0.236
= - 45541.864 J mol - 1
= - 45.54 kJ mol - 1
Again, - 2.303RTlog Kc
= 7.981
∴Kc= Antilog (7.981)
Why does the conductivity of a solution decrease with dilution?
Suggest a way to determine the value of water.
Applying Kohlrausch's law of independent migration of ions, the value of water can be determined as follows:
The molar conductivity of 0.025 mol L-1 methanoic acid is
Calculate its degree of dissociation and dissociation constant. Given λ °(H+)
Answer :
C = 0.025 mol L - 1
Now, degree of dissociation:
Thus, dissociation constant:
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
I = 0.5 A
t = 2 hours = 2 × 60 × 60 s = 7200 s Thus, Q = It
= 0.5 A × 7200 s
= 3600 C
We know that number of electrons. Then,
Suggest a list of metals that are extracted electrolytically.
What is the quantity of electricity in coulombs needed to reduce 1 mol of
Answer :
The given reaction is as follows:
Therefore, to reduce 1 mole of , the required quantity of electricity will be:
=6 F
= 6 × 96487 C
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO2) as the cathode, and a 38% solution of sulphuric acid (H2SO4) as an electrolyte.
When the battery is in use, the following cell reactions take place:
At anode:
At cathode:
The overall cell reaction is given by,
When a battery is charged, the reverse of all these reactions takes place.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Electrons released at the anodic spot move through the metallic object and go to another spot of the object.
There, in the presence of H+ ions, the electrons reduce oxygen. This spot behaves as the cathode. These H+ ions come either from H2CO3, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.
The reaction corresponding at the cathode is given by,
The overall reaction is:
Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide i.e., rust.
Exercise : Solutions of Questions on Page Number : 92
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
The following is the order in which the given metals displace each other from the solution of their salts. Mg, Al, Zn, Fe, Cu
Given the standard electrode potentials, K+/K = -2.93V, Ag+/Ag = 0.80V,
Mg2+/Mg = -2.37 V, Cr3+/Cr = - 0.74V
Answer :
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
Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show:
(ii) The carriers of the current in the cell.
Answer :
The galvanic cell in which the given reaction takes place is depicted as:
(i) Zn electrode (anode) is negatively charged.
(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.
(iii) The reaction taking place at the anode is given by,
The reaction taking place at the cathode is given by,
Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Answer :
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
In the given equation,
n = 6
F = 96487 C mol - 1
= +0.34 V
Then, = - 6 × 96487 C mol - 1 × 0.34 V
= - 196833.48 CV mol - 1
= - 196833.48 J mol - 1
= - 196.83 kJ mol - 1
Again,
= - RT ln K
= 34.496
K = antilog (34.496)
= 3.13 × 1034
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
Here, n = 1. Then,
= - 1 × 96487 C mol - 1 × 0.03 V
= - 2894.61 J mol - 1
= - 2.89 kJ mol - 1
Again,
= 0.5073
K = antilog (0.5073)
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)
(iii) Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) | Pt(s)
Answer :
(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:
= 0.52865 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:
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)
Answer :
= 1.104 V
We know that,
= - 2 × 96487 × 1.04
= - 213043.296 J
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
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.
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
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 conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm-1. Calculate its molar conductivity.
Given,
ĂŽÂş = 0.0248 S cm - 1
c = 0.20 M
Molar conductivity,
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 x 10-3 S cm-1.
Given,
Conductivity, ĂŽÂş = 0.146 × 10 - 3 S cm - 1
Resistance, R = 1500 Ω Cell constant = ĂŽÂş × R
= 0.146 × 10 - 3 × 1500
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
102 × ĂŽÂş/S m - 1 1.237 11.85 23.15 55.53 106.74
Answer :
Given,
ĂŽÂş = 1.237 × 10 - 2 S m - 1, c = 0.001 M
Then, ĂŽÂş = 1.237 × 10 - 4 S cm - 1, c ˝ = 0.0316 M1/2
= 123.7 S cm2 mol - 1
Given,
ĂŽÂş = 11.85 × 10 - 2 S m - 1, c = 0.010M
Then, ĂŽÂş = 11.85 × 10 - 4 S cm - 1, c ˝ = 0.1 M1/2
= 118.5 S cm2 mol - 1
Given,
ĂŽÂş = 23.15 × 10 - 2 S m - 1, c = 0.020 M
Then, ĂŽÂş = 23.15 × 10 - 4 S cm - 1, c1/2 = 0.1414 M1/2
= 115.8 S cm2 mol - 1
Given,
ĂŽÂş = 55.53 × 10 - 2 S m - 1, c = 0.050 M
Then, ĂŽÂş = 55.53 × 10 - 4 S cm - 1, c1/2 = 0.2236 M1/2
= 111.1 1 S cm2 mol - 1
Given,
ĂŽÂş = 106.74 × 10 - 2 S m - 1, c = 0.100 M
Then, ĂŽÂş = 106.74 × 10 - 4 S cm - 1, c1/2 = 0.3162 M1/2
= 106.74 S cm2 mol - 1
Conductivity of 0.00241 M acetic acid is 7.896 × 10 - 5 S cm - 1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol - 1, what is its dissociation constant?
Given, ĂŽÂş = 7.896 × 10 - 5 S m - 1
c = 0.00241 mol L - 1
Then, molar conductivity,
= 32.76S cm2 mol - 1
Again, = 390.5 S cm2 mol - 1
Now,
= 0.084
= 1.86 × 10 - 5 mol L - 1
How much charge is required for the following reductions:
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of to Mn2+.
(i)
Required charge = 3 F
= 3 × 96487 C
= 289461 C
Required charge = 2 F
= 2 × 96487 C
= 192974 C
i.e.,
Required charge = 5 F
= 5 × 96487 C
How much electricity in terms of Faraday is required to produce
(ii) 40.0 g of Al from molten Al2O3.
(i)
Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium
= 1 F
(ii)
Electricity required to produce 27 g of Al = 3 F
Therefore, electricity required to produce 40 g of Al
How much electricity is required in coulomb for the oxidation of
(ii) 1 mol of FeO to Fe2O3.
(i) According to the question,
Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F
= 2 × 96487 C
= 192974 C
(ii)
Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F
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?
Given, Current = 5A
Time = 20 × 60 = 1200 s
Charge = current × time
= 5 × 1200
= 6000 C
Nickel deposited by 2 × 96487 C = 58.71 g
Therefore, nickel deposited by 6000 C
= 1.825 g
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?
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by =
= 1295.43 C
Given,
Current = 1.5 A
Time
= 863.6 s
= 864 s
= 14.40 min Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit
= 0.439 g of Zn
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br- (aq)
(v) Br2 (aq) and Fe2+ (aq).
Since for the overall reaction is positive, the reaction between Fe3+ and I - is feasible.
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Since for the overall reaction is negative, the reaction between Fe3+ and Br - is not feasible.
(aq) (aq)
Since E for the overall reaction
is negative, the reaction between
Ag(s) and Fe3+(aq) is not feasible.
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Predict the products of electrolysis in each of the following:
(ii) An aqueous solution of AgNO3with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
(i) At cathode:
The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.
(ii) At cathode:
The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.
For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The reaction with a higher value of takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
At anode:
At the anode, the reaction with a lower value of is preferred. But due to the over-potential of oxygen, Cl - gets oxidized at the anode to produce Cl2 gas.