NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev

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JEE : NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev

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MULTIPLE CHOICE QUESTIONS - I

Q.1. Which cell will measure standard electrode potential of copper electrode? 
(i) Pt (s) | H2 (g,0.1 bar) | H+ (aq.,1 M) ║ Cu2+(aq.,1M) | Cu
(ii) Pt (s) | H2 (g, 1 bar) | H+ (aq.,1 M) ║ Cu2+ (aq.,2 M) | Cu
(iii) Pt (s) | H2 (g, 1 bar) | H+ (aq.,1 M) ║  Cu2+ (aq.,1 M) | Cu
(iv) Pt(s) | H2 (g, 1 bar) | H+ (aq.,0.1 M)  ║ Cu2+ (aq.,1 M) | Cu
Ans. (iii)
Solution.
Standard electrode potential of copper electrode can be calculated by constructing a concentration cell composed of two half cell reactions in which concentration of species on left hand and right hand side are unity. In such case cell potential is equal to standard, electrode potential.
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev

Q.2. For the given cell, Mg|Mg2+|| Cu2+|Cu
(i) Mg is cathode
(ii) Cu is cathode
(iii) The cell reaction is Mg + Cu2+→ Mg2+ + Cu
(iv) Cu is the oxidising agent
Ans. (ii,iii)
Solution.

Left side of cell reaction represents oxidation half cell i.e., oxidation of Mg and right side of cell represents reduction half cell reactions i.e., reduction of copper.

  • Cu is reduced and reduction occurs at cathode.
  • Mg is oxidised and oxidation occurs at anode

Whole cell reaction can be written as
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRevHence, options (ii) and (iii) both are correct choices.


Q.3. Which of the following statement is correct?
(i) ECell and ∆rG of cell reaction both are extensive properties.
(ii) ECell and ∆rG of cell reaction both are intensive properties.
(iii) ECell is an intensive property while ∆rG of cell reaction is an extensive property.
(iv) ECell is an extensive property while ∆rG of cell reaction is an intensive property.
Ans. (iii)
Solution.
Ecell is an intensive property as it does not depend upon mass of species (number of particles) but ∆rG of the cell reaction is an extensive property because this depends upon mass of species (number of particles).

Q.4. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________.
(i) Cell potential
(ii) Cell emf
(iii) Potential difference
(iv) Cell voltage
Ans. (ii)
Solution.
Cell emf: The difference between the electrode potential of two electrodes when no current is drawn through the cell is called cell emf.

Q.5. Which of the following statement is not correct about an inert electrode in a cell?
(i) It does not participate in the cell reaction.
(ii) It provides surface either for oxidation or for reduction reaction.
(iii) It provides surface for conduction of electrons.
(iv) It provides surface for redox reaction

Ans. (iv)
Solution.
An inert electrode in a cell provides surface for either oxidation or for reduction reaction by conduction of electrons through its surface but does not participate in the cell reaction. It does not provide surface for redox reaction.

Q.6. An electrochemical cell can behave like an electrolytic cell when ____________.
(i) Ecell = 0
(ii) Ecell > Eext
(iii) Eext > Ecell
(iv) Ecell = Eext
Ans. (iii)
Solution.
If an external opposite potential is applied on the galvanic cell and increased reaction continues to take place till the opposing voltage reaches the value 1.1V.
At this stage no current flow through the cell and if there is any further increase in the external potential then reaction starts functioning in opposite direction. Hence, this works as an electrolytic cell.

Q.7. Which of the statements about solutions of electrolytes is not correct?
(i) Conductivity of solution depends upon size of ions.
(ii) Conductivity depends upon viscosiy of solution.
(iii) Conductivity does not depend upon solvation of ions present in solution.
(iv) Conductivity of solution increases with temperature.
Ans. (iii)
Solution.
Conductivity does not depend upon solvation of ions present in solution Solution consists of electrolytes is known as electrolytic solution and conductivity of electrolytic solution depends upon the following factors
(i) Size of ions As ion size increases, ion mobility decreases and conductivity decreases.
(ii) Viscosity of solution Greater the viscosity of the solvent lesser will be the conductivity of the solution.
(iii) Solvation of ions Greater the solvation of ions of an electrolyte lesser will be the electrical conductivity of the solution.
(iv) Temperature of medium Conductivity of solution increases with increase in temperature.

Q.8. Using the data given below find out the strongest reducing agent.
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(i) Cl
(ii) Cr
(iii) Cr3+
(iv) Mn2+
Ans. (ii)
Solution.
Here we use the concept of electrochemical series and standard reduction potential of the  metal.
Higher the negative value of standard reduction potential, strongest will be the reducing agent.
Here, out of given four options standard reduction potential of chromium has highest negative value hence most powerful reducing agent is chromium.

Q.9. Use the data given in Q.8 and find out which of the following is the strongest oxidising agent.
(i) Cl
(ii) Mn2+
(iii) MnO4
(iv) Cr3+
Ans. (iii)
Solution.
Higher the positive value of standard reduction potential of metal ion higher will be its oxidising capacity.
Since, E°MnO4-/Mn2+ has value equal to 1.51 V hence it is the strongest oxidising agent.

Q.10. Using the data given in Q.8 find out in which option the order of reducing power is correct.
(i) Cr3+ < Cl < Mn2+ < Cr
(ii) Mn2+ < Cl < Cr3+ < Cr
(iii) Cr3+ < Cl < Cr2O72– < MnO4
(iv) Mn2+ < Cr3+ < Cl < Cr
Ans. (ii)
Solution.
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev

On moving from top to bottom, standard reduction potential (SRP) value decreases from positive to negative which causes an increase in the reducing capacity of elements. So, the correct option is (ii).

Q.11. Use the data given in Q.8 and find out the most stable ion in its reduced form.
(i) Cl
(ii) Cr3+
(iii) Cr
(iv) Mn2+
Ans. (iv)
Solution.
MnO4-/Mn2+ has + ve value equal to 1.51 V which is highest among given four choices. So, Mn2+ is the most stable ion in its reduced form.

Q.12. Use the data of Q.8 and find out the most stable oxidised species.
(i) Cr3+
(ii) MnO4
(iii) Cr2O72–
(iv) Mn2+
Ans. (i)
Solution.
Cr3+ /Cr has most -ve value equal to -0.74 among given four choices. So, Cr3+ is the most stable oxidised species.

Q.13. The quantity of charge required to obtain one mole of aluminium from Al2O3 is ___________.
(i) 1F
(ii) 6F
(iii) 3F
(iv) 2F

Ans. (iii)
Solution.
The quantity of charge required to obtain one mole of aluminium from Al2O3 is equal to number of electrons required to convert Al2O3 to Al.
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
Hence, total 3F charge is required.

Q.14. The cell constant of a conductivity cell _________.
(i) Changes with change of electrolyte.
(ii) Changes with change of concentration of electrolyte.
(iii) Changes with temperature of electrolyte.
(iv) Remains constant for a cell.
Ans. (iv)
Solution.
Remains constant for a cell
Cell constant is defined as the ratio of length of object and area of cross section.
G = l/A
Since, l and A remain constant for any particular object hence value of cell constant always remains constant.

Q.15. While charging the lead storage battery _________.
(i) PbSO4 anode is reduced to Pb.
(ii) PbSO4 cathode is reduced to Pb.
(iii) PbSO4 cathode is oxidised to Pb.
(iv) PbSO4 anode is oxidised to PbO2.
Ans. (i)
Solution.
While charging the lead storage battery the reaction occurring on cell is reversed and PbSO4 (s) on anode and cathode is converted into Pb and PbO2 respectively The electrode reactions are as follows:
At cathode PbSO4(s) + 2e → Pb(s) + SO42− (aq) (Reduction)
At anode PbSO4(s) + 2H2O → PbO2(s) + SO42− + 4H+ + 2e (Oxidation)
Overall reaction 2PbSO4(s) + 2H2O → Pb(s) + PbO2(s) + 4H+ (aq.) + 2SO42− (aq)

Q.16. NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev is equal to ______________.
(i)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(ii)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(iii)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(iv)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
Ans. (ii)
Solution.

According to Kohlrausch law limiting molar conductivity of any salt is equal to sum of limiting molar conductivity of individual molar conductivity of cations and anions of electrolyte.
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
Hence, option (ii) is correct choice.

Q.17. In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?
(i) Na+ (aq) + e → Na (s); E°Cell = –2.71V
(ii) 2H2O (l) → O2 (g) + 4H+ (aq) + 4e ;  E°Cell = 1.23V
(iii) H+ (aq) + e → 1/2 H2 (g);  E°Cell = 0.00 V

(iv) Cl (aq) → 1/2 Cl2 (g) + e ;  E°Cell = 1.36 V
Ans. (iv)
Solution.

In case of electrolysis of aqueous NaCI oxidation reaction occurs at anode as follows
Cl (aq) → 1/2 Cl2(g) + e   E° = 1.36V
2H2O (I) → O2(g)+ 4H+(aq) + 4e   E°cell = 1.23 V
But due to lower E°cell value water should get oxidised in preference of Cl (aq).
However, the actual reaction taking place in the concentrated solution of NaCI is (iv) and not (ii) i.e., Cl2 is produced and not O2.
This unexpected result is explained on the basis of the concept of 'overvoltage', i.e., water needs greater voltage for oxidation to O2 (as it is kinetically slow process) than that needed for oxidation of Cr ions to CI2. Thus, the correct option is (iv) not (ii).


MULTIPLE CHOICE QUESTIONS - II

Note : In the following questions two or more than two options may be correct.
Q.18. The positive value of the standard electrode potential of Cu2+/Cu  indicates that ____________.
(i) This redox couple is a stronger reducing agent than the H+/H2 couple.
(ii) This redox couple is a stronger oxidising agent than H+/H2.
(iii) Cu can displace H2 from acid.
(iv) Cu cannot displace H2 from acid.
Ans. (ii,iv)
Solution.

'Lesser the E ° value of redox couple higher the reducing power"
Cu2+ + 2e → Cu    E° = 0.34V
2H + 2e → H2   E° = 0.00V
Since, 2H+/H2 has leaser SRP than Cu2+/Cu redox couple, Therefore,
(i) This redox couple is a stronger oxidising agent than H+/H2
(ii) Cu can't displace H2 from acid.
Hence, (ii) and (iv) are correct.

Q.19. E°Cell for some half cell reactions are given below. On the basis of these mark the correct answer.
(i) H+ (aq) + e → 1/2 H2 (g);  E°Cell = 0.00 V
(ii) 2H2O (l) → O2 (g) + 4H+ (aq) + 4e ;  E°Cell = 1.23V
(c) 2SO24 (aq) → S2O2–8 (aq) + 2e ; E°Cell = 1.96 V
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
(iv) In dilute sulphuric acid solution, SO42– ion will be oxidised to tetrathionate ion at anode.

Ans. (i, iii)
Solution.

During the electrolysis of dilute sulphuric acid above given three reaction occurs each of which represents particular reaction either oxidation half cell reaction or reduction half cell reaction.
Oxidation half cell reactions occur at anode are as follows
2SO42− (aq) → S2O82− + 2e−   cell = 1.96V
2H2O+ (l) → O2 (g) + 4H+ (aq) + 4e;    E°cell 1.23V
Reaction having lower value of E°cell will undergo faster oxidation.
Hence, oxidation of water occur preferentially reduction half ceil reaction occurs at cathode
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
Hence, options (i) and (ii) are correct.

Q.20. E°Cell = 1.1V  for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?
(i) 1.1 = Kc
(ii)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(iii)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(iv) log Kc = 1.1
Ans. (ii,iii)
Solution.

At state of equilibrium
ΔG = -RT log K
−nFE° = −RT2.303 log KC
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
[n = 2 for Daniel cell]
 ∵ At equilibrium E° =1.1
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev 
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
[on solving]
Hence, options (ii) and (iii) are the correct choices.

Q.21. Conductivity of an electrolytic solution depends on ________.
(i) Nature of electrolyte.
(ii) Concentration of electrolyte.
(iii) Power of AC source.
(iv) Distance between the electrodes.
Ans. (i,ii)
Solution.

Conductivity of electrolytic solution is due to presence of mobile ions in the solution. This type of conductance is known as ionic conductance. Conductivity of these type of solutions depends upon
(i) the nature of electrolyte added
(ii) size of the ion produced and their solvatian
(iii) concentration of electrolyte (i) nature of solvent and its viscosity (ii) temperature
While power of source or distance between electrodes has no effect on conductivity of electrolyte solution. Hence, options (i) and (ii) are the correct choices.

Q.22. NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRevis equal to _______________.
(i)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(ii)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(iii)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(iv)NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
Ans. (i,iii)
Solution.

This problem is based on the concept of Kohlrausch lawof independent migration of ions: is the sum of limiting molar conductivities of cation (NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev) and anion (NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev) .
i.e., 
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(i)
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(ii)
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(b) is incorrect
(iii)
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
Hence, Options (a) and (c) are the correct choices.
(iv)
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
This type of decomposition is not possible due to weak basic strength of NH4OH. This line will be placed above.

Q.23. What will happen during the electrolysis of aqueous solution of CuSO4 by using platinum electrodes?
(i) Copper will deposit at cathode.
(ii) Copper will deposit at anode.
(iii) Oxygen will be released at anode.
(iv) Copper will dissolve at anode.
Ans. (i,iii)
Solution.

For electrolysis of aqueous solution of CuSO4:
CuSO4 (aq) → CuH2++ SO42-
HO → 2H+ O2+
At anode 2O2- → O2 +2e-
At cathode Cu2+ + 2e → Cu(s)

Q.24. What will happen during the electrolysis of aqueous solution of  CuSO4 in the presence of Cu electrodes?
(i) Copper will deposit at cathode.
(ii) Copper will dissolve at anode.
(iii) Oxygen will be released at anode.
(iv) Copper will deposit at anode.
Ans. (i,ii)
Solution.

Electrolysis of CuSO4 can be represented by two half cell reactions these occurring at cathode and anode respectively as
At cathode Cu2++ 2e- → Cu(s)
At anode  Cu(s) → Cu2++ 2e
Here, Cu will deposit at cathode while copper will dissolved at anode.
Hence, options (a) and (b) are the correct choices.

Q.25. Conductivity κ , is equal to ____________.
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
(ii) G*/R
(iii) ∧m
(iv) l/A
Ans. (i,ii)
Solution.

As we know that, conductance is reciprocal of resistance and conductivity is reciprocal of resistivity.
k = 1/ρ
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
NCERT Exemplar (Part - 1) - Electrochemistry Notes | EduRev
Where, G* = Cell constant
Hence, options (i) and (ii) are the correct choices.

Q.26. Molar conductivity of ionic solution depends on ___________.
(i) Temperature.
(ii) Distance between electrodes.
(iii) Concentration of electrolytes in solution.
(iv) Surface area of electrodes.
Ans. (i,iii)
Solution.

Molar conductivity is the conductivity due to ions furnished by one mole of electrolyte in solution. Molar conductivity of ionic solution depends on
(i) Temperature: Molar conductivity of electrolyte solution increases with increase in temperature,
(ii) Concentration of electrolytes in solution: As concentration of electrolyte increases, molar conductivity decreases.

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