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Unit 3 Electrochemistry Chemistry Worksheet for Class 12
Class 12 Chemistry students should refer to the following printable worksheet in Pdf in Class 12. This test paper with questions and solutions for Class 12 Chemistry will be very useful for tests and exams and help you to score better marks
Class 12 Chemistry Unit 3 Electrochemistry Worksheet Pdf
Question. The standard electrode potential (E°) values of Al3+/Al, Ag+/Ag, K+/K and Cr3+/Cr are –1.66 V,0.80 V, –2.93 V and –0.74 V, respectively. The correct decreasing order of reducing power of the metal is
(a) Ag > Cr > Al > K
(b) K > Al > Cr > Ag
(c) K > Al > Ag > Cr
(d) Al > K > Ag > Cr
Answer. B
Question. A button cell used in watches function as following :
Zn(s) + Ag2O(s) + H2O(l) ⇔ 2Ag(s) + Zn2+(aq) + 2OH–(aq)
If half cell potentials areZn2+(aq) + 2e– → Zn(s); E° = – 0.76 V
Ag2O(s) + H2O(l) + 2e– → 2Ag(s) + 2OH–(aq); E° = 0.34 V
The cell potential will be
(a) 0.84 V
(b) 1.34 V
(c) 1.10 V
(d) 0.42 V
Answer. C
Question. Standard reduction potentials of the half reactions are given below :
F2(g) + 2e– → 2F–(aq) ; E° = + 2.85 V
Cl2(g) + 2e– → 2Cl–(aq) ; E° = + 1.36 V
Br2(l) + 2e– → 2Br–(aq) ; E° = + 1.06 V
I2(s) + 2e– → 2I–(aq) ; E° = + 0.53 V
The strongest oxidising and reducing agents respectively are
(a) F2 and I–
(b) Br2 and Cl–
(c) Cl2 and Br–
(d) Cl2 and I2
Answer. A
Question. Standard electrode potentials of three metals X, Y and Z are –1.2 V, + 0.5 V and – 3.0 V respectively.
The reducing power of these metals will be
(a) Y > Z > X
(b) Y > X > Z
(c) Z > X > Y
(d) X > Y > Z
Answer. C
Question. Standard electrode potential for Sn4+/Sn2+ couple is +0.15 V and that for the Cr3+/Cr couple is –0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be
(a) + 1.19 V
(b) + 0.89 V
(c) + 0.18 V
(d) + 1.83 V
Answer. B
Question. A solution contains Fe2+, Fe3+ and I– ions. This solution was treated with iodine at 35°C. E° for Fe3+/Fe2+ is + 0.77 V and E° for I2/2I– = 0.536 V. The favourable redox reaction is
(a) I2 will be reduced to I–
(b) there will be no redox reaction
(c) I– will be oxidised to I2
(d) Fe2+ will be oxidised to Fe3+.
Answer. C
Question. Consider the following relations for emf of an electrochemical cell
(i) EMF of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
(ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)
Which of the above relations are correct?
(a) (iii) and (i)
(b) (i) and (ii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Answer. D
Question. On the basis of the following E° values, the strongest oxidizing agent is
[Fe(CN)6]4– → [Fe(CN)6]3– + e– ; E° = –0.35 V
Fe2+ → Fe3+ + e– ; E° = –0.77 V
(a) Fe3+
(b) [Fe(CN)6]3–
(c) [Fe(CN)6]4–
(d) Fe2+
Answer. A
Question. A hypothetical electrochemical cell is shown below :
A/ A+ (x M)||B+(y M)|B
The emf measured is + 0.20 V. The cell reaction is
(a) A + B+ → A+ + B
(b) A+ + B → A + B+
(c) A+ + e– → A; B+ + e– → B
(d) the cell reaction cannot be predicted.
Answer. A
Question. E°Fe2+/Fe = – 0.441 V and E°Fe3+/Fe2+ = 0.771 V, the standard EMF of the reaction Fe + 2Fe3+ → 3Fe2+ will be
(a) 0.111 V
(b) 0.330 V
(c) 1.653 V
(d) 1.212 V
Answer. D
Question. Standard electrode potentials are Fe2+/Fe;
E° = –0.44 and Fe3+/Fe2+; E° = 0.77. Fe2+, Fe3+ and Fe blocks are kept together, then
(a) Fe3+ increases
(b) Fe3+ decreases
(c) Fe2+/Fe3+ remains unchanged
(d) Fe2+ decreases.
Answer. B
Question. Electrode potential for the following half-cell reactions are
Zn → Zn2+ + 2e–; E° = + 0.76 V;
Fe → Fe2+ + 2e–; E° = + 0.44 V.
The EMF for the cell reaction
Fe2+ + Zn → Zn2+ + Fe will be
(a) – 0.32 V
(b) + 1.20 V
(c) – 1.20 V
(d) + 0.32 V
Answer. D
Question. An electrochemical cell is set up as :
Pt; H2 (1 atm)|HCl(0.1 M) || CH3COOH (0.1 M) |H2 (1 atm); Pt. The e.m.f. of this cell will not be zero, because
(a) acids used in two compartments are different
(b) e.m.f. depends on molarities of acids used
(c) the temperature is constant
(d) pH of 0.1 M HCl and 0.1 M CH3COOH is not same.
Answer. D
Question. Standard reduction potentials at 25°C of Li+|Li, Ba2+|Ba, Na+|Na and Mg2+|Mg are –3.05, –2.90, –2.71 and –2.37 volt respectively. Which one of the following is the strongest oxidising agent?
(a) Ba2+
(b) Mg2+
(c) Na+
(d) Li+
Answer. B
Question. A solution of potassium bromide is treated with each of the following. Which one would liberate bromine?
(a) Hydrogen iodide
(b) Sulphur dioxide
(c) Chlorine
(d) Iodine
Answer. C
Question. For the cell reaction :
2Fe3+(aq) + 2I–(aq) → 2Fe2+ (aq) + I2(aq)
E°cell = 0.24 V at 298 K. The standard Gibbs’ energy (ΔrG°)of the cell reaction is
[Given that Faraday constant, F = 96500 C mol–1]
(a) 23.16 kJ mol–1
(b) –46.32 kJ mol–1
(c) –23.16 kJ mol–1
(d) 46.32 kJ mol–1
Answer. B
Question. In the electrochemical cell :
Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu, the emf of this Daniell cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E2.
From the followings, which one is the relationship between E1 and E2? (Given, RT/F = 0.059)
(a) E1 < E2
(b) E1 > E2
(c) E2 = 0 ≠ E1
(d) E1 = E2
Answer. B
Question. If the E°cell for a given reaction has a negative value,which of the following gives the correct relationships for the values of ΔG° and Keq ?
(a) ΔG° > 0; Keq < 1
(b) ΔG° > 0; Keq > 1
(c) ΔG° < 0; Keq > 1
(d) ΔG° < 0; Keq < 1
Answer. A
Question. The pressure of H2 required to make the potential of H2 electrode zero in pure water at 298 K is
(a) 10–10 atm
(b) 10–4 atm
(c) 10–14 atm
(d) 10–12 atm.
Answer. C
Question. A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be
(a) 0.118 V
(b) 1.18 V
(c) 0.059 V
(d) 0.59 V
Answer. D
Question. Consider the half-cell reduction reaction
Mn2+ + 2e– → Mn, E° = – 1.18 V
Mn2+ → Mn3+ + e–, E° = – 1.51 V
The E° for the reaction,
3Mn2+ → Mn0 + 2Mn3+,
and possibility of the forward reaction are respectively
(a) – 4.18 V and yes
(b) + 0.33 V and yes
(c) + 2.69 V and no
(d) – 2.69 V and no.
Answer. D
Question. The electrode potentials for, Cu2+(aq) + e– → Cu+(aq) and Cu+ (aq) + e– → Cu(s) are + 0.15 V and + 0.50 V respectively. The value of E°Cu2+/Cu will be
(a) 0.500 V
(b) 0.325 V
(c) 0.650 V
(d) 0.150 V
Answer. B
Question. For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25 °C. The value of standard Gibbs energy, DG° will be (F = 96500 C mol–1)
(a) – 89.0 kJ
(b) – 89.0 J
(c) – 44.5 kJ
(d) – 98.0 kJ
Answer. A
Question. Given :
(i) Cu2+ + 2e– → Cu, E° = 0.337 V
(ii) Cu2+ + e– → Cu+, E° = 0.153 V
Electrode potential, E° for the reaction,
Cu+ + e– → Cu, will be
(a) 0.90 V
(b) 0.30 V
(c) 0.38 V
(d) 0.52 V
Answer. D
Question. Standard free energies of formation (in kJ/mol) at 298 K are –237.2, –394.4 and –8.2 for H2O(l), CO2(g) and pentane(g) respectively. The value of E°cell for the pentane-oxygen fuel cell is
(a) 1.0968 V
(b) 0.0968 V
(c) 1.968 V
(d) 2.0968 V
Answer. A
Question. The equilibrium constant of the reaction :
Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s);
E° = 0.46 V at 298 K is
(a) 2.0 × 1010
(b) 4.0 × 1010
(c) 4.0 × 1015
(d) 2.4 × 1010
Answer. C
Question. The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C.The equilibrium constant of the reaction would be
(a) 2.0 × 1011
(b) 4.0 × 1012
(c) 1.0 × 102
(d) 1.0 × 1010
(Given F = 96500 C mol–1, R = 8.314 J K–1 mol–1)
Answer. D
Question. On the basis of the information available from the reaction, 4/3Al + O2 → 2/3Al2O3, DG = –827 kJ mol–1 of O2, the minimum e.m.f. required to carry out an electrolysis of Al2O3 is (F = 96500 C mol–1)
(a) 2.14 V
(b) 4.28 V
(c) 6.42 V
(d) 8.56 V
Answer. A
Question. For the disproportionation of copper 2Cu+ → Cu2+ + Cu, E° is (Given : E° for Cu2+/Cu is 0.34 V and E° for Cu2+/Cu+ is 0.15 V)
(a) 0.49 V
(b) –0.19 V
(c) 0.38 V
(d) –0.38 V
Answer. C
Question. E° for the cell, Zn | Zn2+(aq) ||Cu2+(aq) | Cu is 1.10 V at 25°C, the equilibrium constant for the reaction Zn + Cu2+ (aq) → Cu + Zn2+ (aq) is of the order
(a) 10+18
(b) 10+17
(c) 10–28
(d) 10+37
Answer. D
Question. The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76 × 10–3 S cm–1 at 298 K is
(a) 2.88 S cm2/mol
(b) 11.52 S cm2/mol
(c) 0.086 S cm2/mol
(d) 28.8 S cm2/mol
Answer. B
Question. At 25 °C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm–1 cm2 mol–1 and at infinite dilution its molar conductance is 238 ohm–1 cm2 mol–1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is
(a) 4.008%
(b) 40.800%
(c) 2.080%
(d) 20.800%
Answer. A
Question. Limiting molar conductivity of NH4OH [i.e., L°m(NH4OH)] is equal to
(a) Λ°m(NH4Cl) + Λ°m(NaCl) – Λ°m(NaOH)
(b) Λ°m(NaOH) + Λ°m(NaCl) – Λ°m(NH4Cl)
(c) Λ°m(NH4OH) + Λ°m(NH4Cl) – Λ°m(HCl)
(d) Λ°m(NH4Cl) + Λ°m(NaOH) – Λ°m(NaCl)
Answer. D
Question. Molar conductivities (Λ°m) at infinite dilution of NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. (Λ°m) for CH3COOH will be
(a) 425.5 S cm2 mol–1
(b) 180.5 S cm2 mol–1
(c) 290.8 S cm2 mol–1
(d) 390.5 S cm2 mol–1
Answer. D
Question. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
(a) increase in ionic mobility of ions
(b) 100% ionisation of electrolyte at normal dilution
(c) increase in both i.e., number of ions and ionic mobility of ions
(d) increase in number of ions.
Answer. A
Question. The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mho cm2 and at infinite dilution is 400 mho cm2. The dissociation constant of this acid is
(a) 1.25 × 10–6
(b) 6.25 × 10–4
(c) 1.25 × 10–4
(d) 1.25 × 10–5
Answer. D
Question. Kohlrausch’s law states that at
(a) infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
(b) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
(c) finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
(d) infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.
Answer. A
Question. Equivalent conductances of Ba2+ and Cl– ions are 127 and 76 ohm–1 cm–1 eq–1 respectively. Equivalent conductance of BaCl2 at infinite dilution is
(a) 139.5
(b) 101.5
(c) 203
(d) 279
Answer. A
Question. The specific conductance of a 0.1 N KCl solution at 23°C is 0.012 ohm–1 cm–1. The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be
(a) 0.918 cm–1
(b) 0.66 cm–1
(c) 1.142 cm–1
(d) 1.12 cm–1
Answer. B
Question. On heating one end of a piece of a metal, the other end becomes hot because of
(a) energised electrons moving to the other end
(b) minor perturbation in the energy of atoms
(c) resistance of the metal
(d) mobility of atoms in the metal.
Answer. A
Question. On electrolysis of dil. sulphuric acid using platinum (Pt) electrode, the product obtained at anode will be
(a) hydrogen gas
(b) oxygen gas
(c) H2S gas
(d) SO2 gas.
Answer. B
Question. The number of Faradays (F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca = 40 g mol–1) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer. A
Question. During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is
(a) 55 minutes
(b) 110 minutes
(c) 220 minutes
(d) 330 minutes.
Answer. B
Question. The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 × 10–19 C)
(a) 6 × 1023
(b) 6 × 1020
(c) 3.75 × 1020
(d) 7.48 × 1023
Answer. C
Question. When 0.1 mol MnO42– is oxidised, the quantity of electricity required to completely oxidise MnO42– to MnO4– is
(a) 96500 C
(b) 2 × 96500 C
(c) 9650 C
(d) 96.50 C
Answer. C
Question. The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be
(a) 5.4 g
(b) 10.8 g
(c) 54.0 g
(d) 108.0 g
Answer. D
Question. How many grams of cobalt metal will be deposited when a solution of cobalt(II) chloride is electrolyzed with a current of 10 amperes for 109 minutes?
(1 Faraday = 96,500 C; Atomic mass of Co = 59 u)
(a) 4.0
(b) 20.0
(c) 40.0
(d) 0.66
Answer. B
Question. Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, at. mass of Al = 27 g mol–1)
(a) 8.1 × 104 g
(b) 2.4 × 105 g
(c) 1.3 × 104 g
(d) 9.0 × 103 g
Answer. A
Question. 4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H+ ions in solution by the same quantity of electric charge will be
(a) 44.8 L
(b) 22.4 L
(c) 11.2 L
(d) 5.6 L
Answer. D
Question. In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam. The reason for this is
(a) Hg is more inert than Pt
(b) more voltage is required to reduce H+ at Hg than at Pt
(c) Na is dissolved in Hg while it does not dissolve in Pt
(d) conc. of H+ ions is larger when Pt electrode is taken.
Answer. B
Question. A 5 ampere current is passed through a solution of zinc sulphate for 40 minutes. The amount of zinc deposited at the cathode is
(a) 0.4065 g
(b) 65.04 g
(c) 40.65 g
(d) 4.065 g
Answer. D
Question. Sodium is made by the electrolysis of a molten mixture of about 40% NaCl and 60% CaCl2 because
(a) Ca++ can reduce NaCl to Na
(b) Ca++ can displace Na from NaCl
(c) CaCl2 helps in conduction of electricity
(d) this mixture has a lower melting point than NaCl.
Answer. D
Question. When CuSO4 is electrolysed using platinum electrodes,
(a) copper is liberated at cathode, sulphur at anode
(b) copper is liberated at cathode, oxygen at anode
(c) sulphur is liberated at cathode, oxygen at anode
(d) oxygen is liberated at cathode, copper at anode.
Answer. B
Question. On electrolysis of dilute sulphuric acid using platinum electrodes, the product obtained at the anode will be
(a) hydrogen
(b) oxygen
(c) hydrogen sulphide
(d) sulphur dioxide.
Answer. B
Question. A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as
(a) dynamo
(b) Ni-Cd cell
(c) fuel cell
(d) electrolytic cell.
Answer. C
Question. The efficiency of a fuel cell is given by
(a) DG/DS
(b) DG/DH
(c) DS/DG
(d) DH/DG
Answer. B
Question. Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
(a) zinc is lighter than iron
(b) zinc has lower melting point than iron
(c) zinc has lower negative electrode potential than iron
(d) zinc has higher negative electrode potential than iron.
Answer. D
Question. The most convenient method to protect the bottom of ship made of iron is
(a) coating it with red lead oxide
(b) white tin plating
(c) connecting it with Mg block
(d) connecting it with Pb block.
Answer. B
Question. To protect iron against corrosion, the most durable metal plating on it, is
(a) copper plating
(b) zinc plating
(c) nickel plating
(d) tin plating.
Answer. B
Please click on below link to download CBSE Class 12 Chemistry Electrochemistry Question Bank Set A
CBSE Class 12 Chemistry Unit 3 Electrochemistry Worksheet
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