NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry are part of NCERT Solutions for Class 12 Chemistry. Here we have given NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry.
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NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry
NCERT IN-TEXT QUESTIONS
How would you determine the standard electrode potential of the system ; Mg2+/Mg ?
n order to determine E° value of Mg2+/Mg electrode, an electrochemical cell is set up in which a Mg electrode dipped in 1 M MgSO4 solution acts as one half cell (oxidation half cell) while the standard hydrogen electrode acts as the other half cell (reduction half cell). The deflection of voltmeter placed in the cell circuit is towards the Mg– electrode indicating the flow of current. The cell may be represented as :
Mg/Mg2+ (1 M) || H+(l M)/H2(1 atm), Pt
The reading as given by voltmeter gives
The expected value of standard electrode potential (E°) = -2-36 V.
Can you store copper sulphate solution in a zinc pot ?
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn2+(aq) + 2e– → Zn(s) ; E° = – 0·76 V
Cu2+(aq) + 2e– → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu2+ ions and redox reaction will immediately set in.
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.
Consult the table on standard electrode potentials and suggest three substances that can oxidise Fe2+ ions under suitable conditions.
The oxidation of Fe2+ ions to Fe3+ ions proceeds as follows :
Fe2+ → 3+ + e– ; = – 0·77 V
Only those substances can oxidise Fe2+ ions to Fe3+ ions which can accept electrons released during oxidation or are placed above iron in electrochemical series. These are : Cl2(g), Br2(g) and Cr2 ions (in the acidic medium).
Calculate the potential of hydrogen electrode in contact with a solution with pH equal to 10.
For hydrogen electrode, H+ + e– → 1/2H2
Applying Nernst equation,
Calculate e.m.f. of the cell in which the following reaction takes place
Ni(s) + 2Ag+(0·002M) → Ni2+(0·160M) + 2Ag(s) Given that = 1.05 V. (C.B.S.E. Outside Delhi 2015)
The cell in which the following reaction occurs
2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(s) has = 0-236 V at 298 K.
Calculate standard Gibbs energy and equilibrium constant for the reaction.
The two half reactions are :
2Fe3+ + 2e– → 2Fe2+ and 2I– → I2 + 2e–
For the above reaction, n = 2
Why does the conductivity of a solution decrease with dilution ?
The conductivity of a solution is linked with the number of ions present per unit volume. With dilution, these decrease and the corresponding conductivity or specific conductance of the solution decreases.
Suggest a way to determine the for water.
Water (H2O) is a weak electrolyte. Its molar conductance at infinite dilution i.e., can be determined in terms of for strong electrolytes. This is in accordance with Kohlrausch’s Law.
The molar conductance of 0·025 mol L-1of methanoic acid is 46·15 cm2 mol-1. Calculate its degree of dissociation and dissociation constant. Given λ°(H+) = 349·6 S cm2 mol-1 and λ°(HCOO–) = 54·6 S cm2 mol-1.
If a current of 0·5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire ?
Quantity of charge (Q) passed = Current in amperes x Time in seconds = (0·5 A) X (2 x 60 x 60 s)
= 3600 As = 3600 C
No. of electrons flowing through the wire by passing a charge of one faraday (96500 C) = 6·022 x 1023
No. of electrons flowing through the wire by passing a charge of 3600 C
Suggest a list of metals which can be extracted electrolytically.
The highly reactive metals having large -ve E° values, which can themselves act as powerful reducing agents can be extracted electrolytically. The process is known as electrolytic reduction. For details, consult Unit-6. For example, sodium, potassium, calcium, magnesium etc.
Consider the reaction :
Cr2 + 14H+ + 6e– → 2Cr3+ + 7H2O.
What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr2 ions ?
The quantity of electricity in coulombs is 6 F or 6 x 96500 C = 5·76 x 105 C.
Write the chemistry of recharging the lead storage battery highlighting all the materials that are involved during recharging.
Chemical reactions while recharging :
Chemical reactions while recharging :
2PbSO4 + 2H2O → PbO2 + Pb + 2H2SO4
Electricity is passed through the electrolyte PbSO4 which is converted into PbO2 and Pb.
Recharging is possible in this case because the PbSO4 formed during discharging is a solid and sticks to the electrodes. Therefore, it can either take up or give electrons during recharging.
Suggest two materials other than hydrogen that can be used as fuels in the fuel cells.
Methane (CH4) and methanol (CH3OH) can also be used as fuels in place of hydrogen in the fuel cells.
Explain how rusting of iron can be envisaged as setting up of an electrochemical cell.
Iron (Fe) is involved in the redox-reaction that is carried in the electrochemical cell which is set up. As a result, it slowly dissolves and the metal surface gets rusted or corroded.
The redox-reaction may be described as follows :
At anode : Fe (s) undergoes oxidation to release electrons
F2(s) → Fe2+(aq) + 2e– ….(oxidation)
At cathode : The electrons which are released participate in the reduction reaction and combine with H+ ions released fromcarbonic acid (H2CO3) formed by the combination of CO2 and H2O present.
Arrange the following metals in the order in which they displace each other from their salts.
Al, Cu, Fe, Mg and Zn
Mg, Al, Zn, Fe, Cu.
Given the standard electrode potentials
K+/K = – 2·93 V, Ag+/Ag = 0·80 V
Hg2+/Hg = 0·79 V ; Mg2+/Mg = – 2·37V, Cr3+/Cr = – 0·74 V
Arrange these metals in increasing order of their reducing power.
It may be noted that lesser the E° value for an electrode, more will be reducing power. The increasing order of reducing power is :
Ag+/Ag < Hg2+/Hg < Cr3+/Cr < Mg2+/Mg < K+/K
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show :
(i) which electrode is negatively charged ?
(ii) the carriers of the current in the cell.
(iii) individual reaction at each electrode. (C.B.S.E. Delhi 2008)
The galvanic cell for the reaction is
Zn(s) | Zn2+(aq) || 2Ag+(aq) | 2Ag(s)
(i) Zinc electrode (anode) is negatively charged.
(ii) From silver to zinc in external circuit
(iii) Anode : Zn (s) → Zn2+(aq) + 2e–
Cathode : 2Ag+(aq) + 2e– → 2Ag(s)
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)
Write the Nernst Equation and calculate e.m.f. of the following cells at 298 K :
(i) Mg(s) | Mg2+ (0·001 M) || Cu2+ (0·0001 M) | Cu(s) (C.B.S.E. Delhi 2008, 2013)
(ii) Fe(s) | Fe2+ (0·001 M) || H+ (1 M) | H2(g) (1 bar) | Pt(s)
(iii) Sn(s) | Sn2+ (0·050 M) || H+ (0·02 M) | H2(g) (1 bar) | Pt(s) (C.B.S.E. Outside Delhi 2013, 2015)
(iv) Pt(s) | Br2(l) | Br– (0·010 M) || H+ (0·030 M) | H2(g) (1 bar) | Pt(s)
In the button cell widely used in watches and in other devices, the following reaction takes place :
Zn (s) + Ag2O (s) + H2O (l) → Zn2+ (aq) + 2Ag (s) + 2OH– (aq)
Determine E° and ∆G° for the reaction. (C.B.S.B. Delhi 2005, Outside Delhi 2006 Supp., 2008, C.B.S.E. Sample Paper 2010)
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
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 = lsq.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.
The conductivity of 0·20 M solution of KCl at 298 K is 0·0248 S cm-1.Calculate its molar conductivity.
(C.B.S.E. Delhi 2008, 2013)
The resistance of a conductivity cell containing 0·001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0·001M KCl solution at 298 K is 0·146 x 10-3 S cm-2 ? (C.B.S.E. Outside Delhi 2007, 2008, 2013)
The conductivity of sodium chloride solution at 298 K has been determined at different concentrations and results are given below :
Calculate molar conductivity for all the concentrations and draw a plot between and . Find the value from the graph.
(unit conversion factor)
can be obtained on extrapolation to zero concentration along Y-axis. It is 124·0Scm2mol-1.
Conductivity of 0·00241 M acetic acid is 7·896 x 10-5 S cm-1. Calculate the molar conductivity. If A° for acetic acid is 390·5 S cm2 mol-1, what is its dissociation constant ? (C.B.S.E. Delhi 2008, C.B.S.E. Outside Delhi 2016)
How much charge is required for the reduction of :
(i) 1 mol of Al3+ to Al
(ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of Mn to Mn2+.
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 ?
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.
For the oxidation of two moles of FeO, charge required = 2 F
For the oxidation of one mole of FeO, charge required = 1 F = 96500 C.
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5·0 ampere for 20 minutes. What weight of Ni will be produced at the cathode ? (Atomic mass of Ni = 58·7). (Jharkhand Board 2009)
Three electrolytic cells A, B and C containing electrolytes zinc sulphate, silver nitrate and copper sulphate respectively, were connected in series. A steady current of 1·50 ampere was passed through them until 1·45 g of silver were deposited at the cathode of cell B. How long did the current flow ? What weight of copper and of zinc were deposited ? (Atomic mass of Cu = 63·5 ; Zn = 65·3; Ag = 108) (C.B.S.E. Outside Delhi 2008, Jharkhand Board 2010)
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).
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.
Predict the products of electrolysis of each of the following :
(i) An aqueous solution of AgNO3 using silver electrodes.
(ii) An aqueous solution of AgNO3 using platinum electrodes.
(iii) A dilute solution of H2SO4 using platinum electrodes. (C.B.S.E. Outside Delhi 2007)
(iv) An aqueous solution of CuCl2 using platinum electrodes. (C. B. S. E. Sample Paper 2010)
(i) An aqueous solution of AgNO3 using silver electrodes :
Both AgNO3 and water will ionise in aqueous solution
At cathode : Ag+ ions with less discharge potential are reduced in preference to H+ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag+ (aq) + e– → Ag (deposited)
At anode : An equivalent amount of silver will be oxidised to Ag+ ions by releasing electrons.
(ii) An aqueous solution of AgNO3 using platinum electrodes :
In this case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode: Ag+ ions will be reduced to Ag which will get deposited at the cathode.
At anode: Both and OH– ions will migrate. But OH– ions with less discharge potential will be oxidised in preference to ions which will remain in solution.
Thus, as a result of electrolysis, silver is deposited on the cathode while O2 is evolved at the anode. The solution will be acidic due to the presence of HNO3.
(iii) A dilute solution of H2SO4 using platinum electrodes :
On passing current, both acid and water will ionise as follows :
At cathode: H+ (aq) ions will migrate to the cathode and will be reduced to H2.
Thus, H2 (g) will be evolved at cathode.
At anode: OH– ions will be released in preference to ions because their discharge potential is less. They will be oxidised as follows :
Thus, O2 (g) will be evolved at anode. The solution will be acidic and will contain H2SO4.
(iv) An aqueous solution of CuCl2 using platinum electrodes :
The electrolysis proceeds in the same manner as discussed in the case of AgNO3 solution. Both CuCl2 and H2O will ionise as follows :
At cathode: Cu2+ ions will be reduced in preference to H+ ions and copper will be deposited at cathode
Cu2+ (aq) + 2e– → Cu (deposited)
At anode: C– ions will be discharged in preference to OH– ions which will remain in solution.
Cl– → Cl– + e–; Cl + Cl → Cl2 (g)
Thus, Cl2 will be evolved at anode.
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