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Page No 33:

Question 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

Answer:

The standard electrode potential of copper electrode is measured using standard hydrogen electrode as anode which has H₂ with a pressure of 1 bar and H⁺ ions with 1 M concentration. The concentration of Cu2+ ions is also kept 1 M. The cell satisfying all the above conditions is given in option (iii).
Hence, the correct answer is option (iii).

Page No 33:

Question 2:

Electrode potential for Mg electrode varies according to the equation EMg2+|Mg=EMg2+| Mg-0.0592 log1Mg2+.
The graph of EMg2+| Mg vs log Mg2+ is
             


                         

Answer:

The equation EMg2+|Mg=EMg2+| Mg-0.0592 log1Mg2+orEMg2+|Mg=EMg2+| Mg+0.0592 logMg2+  can be compared to the straight line equation y=mx+c.Here, y=EMg2+|Mgslope=m=0.0592 x=log [Mg2+]intercept=c=EMg2+| Mg 
Hence, the correct answer is option (ii).



Page No 34:

Question 3:

Which of the following statement is correct?
(i) ECell and ΔrG of cell reaction both are extensive properties.
(ii) ECell and ΔrG of cell reaction both are intensive properties.
(iii) ECell is an intensive property while ΔrG of cell reaction is an extensive property.
(iv) ECell is an extensive property while ΔrG of cell reaction is an intensive property.

Answer:

ECell is an example of intensive property as it is independent of mass while Δrof cell reaction is an extensive property as it depends upon mass.
Hence, the correct answer is option (iii).

Page No 34:

Question 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

Answer:

Cell electromotive force (emf) is the difference between the electrode potentials of two electrodes when no current is drawn through the cell.
Hence, the correct answer is option (ii).
 

Page No 34:

Question 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.

Answer:

An inert electrode does not take part in the cell reaction and provides surface either for oxidation or for reduction reaction. It also provides surface for conduction of electrons but not for redox reaction.
Hence, the correct answer is option (iv).

Page No 34:

Question 6:

An electro-chemical cell can behave like an electrolytic cell when ____________.
(i) Ecell = 0
(ii) Ecell > Eext
(iii) Eext > Ecell
(iv) Ecell = Eext

Answer:

When the external voltage is greater than the emf of the cell, an electrochemical cell behaves like an electrolytic cell.
Hence, the correct answer is option (iii).

Page No 34:

Question 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 viscosity of solution.
(iii) Conductivity does not depend upon solvation of ions present in solution.
(iv) Conductivity of solution increases with temperature.

Answer:

The conductivity of solutions of electrolytes depends on solvation of ions present in solution. Conductivity decreases with increase in solvation of ions.
Hence, the correct answer is option (iii).

Page No 34:

Question 8:

Using the data given below find out the strongest reducing agent.
ECr2O72-/Cr3+=1.33 V  ECl2/Cl-=1.36 VEMnO4-/Mn2+=1.51 V  ECr3+/Cr=-0.74 V
(i) Cl
(ii) Cr
(iii) Cr3+
(iv) Mn2+

Answer:

The redox couple with most negative value of electrode potential will have the least feasibility for reduction. So, Cr3+/Cr redox couple will have the least tendency for reduction among the given options. This implies that Cr will have the greatest tendency to get oxidized and will be the strongest reducing agent.
Hence, the correct answer is option (ii).



Page No 35:

Question 9:

Using the data given in Q8 and and find out which of the following is the strongest oxidising agent.
ECr2O72-/Cr3+=1.33 V  ECl2/Cl-=1.36 VEMnO4-/Mn2+=1.51 V  ECr3+/Cr=-0.74 V
(i) Cl
(ii) Mn2+
(iii) MnO4-
(iv) Cr3+

Answer:

The redox couple with the most positive value of electrode potential will most easily undergo reduction. So, MnO4-/Mn2+ redox couple will have the greatest tendency for reduction among the given options. This implies that MnO4- will have the greatest tendency to get reduced and will be the strongest oxidizing agent.
Hence, the correct answer is option (iii).

Page No 35:

Question 10:

Using the data given below and find out in which option the order of reducing power is correct.
ECr2O72-/Cr3+=1.33 V  ECl2/Cl-=1.36 VEMnO4-/Mn2+=1.51 V  ECr3+/Cr=-0.74 V

(i) Cr3+ < Cl< Mn2+ < Cr

(ii) Mn2+ < Cl< Cr3+ < Cr

(iii) Cr3+<Cl-<Cr2O72-<MnO4-

(iv) Mn2+ < Cr3+ < Cl< Cr

Answer:

More negative the electrode potential value of a redox couple, faster it will undergo oxidation and more will be the reducing power.
Hence, the correct answer is option (ii).

Page No 35:

Question 11:

Using the data given below and find out the most stable ion in its reduced form.
ECr2O72-/Cr3+=1.33 V  ECl2/Cl-=1.36 VEMnO4-/Mn2+=1.51 V  ECr3+/Cr=-0.74 V
(i) Cl
(ii) Cr3+
(iii) Cr
(iv) Mn2+

Answer:

The electrode potential value of MnO4-/Mn2+ redox couple is most positive. This implies that it will undergo reduction most easily and Mn2+ will be the most stable ion in reduced form.
Hence, the correct answer is option (iv).

Page No 35:

Question 12:

Using the data given below and find out the most stable oxidised species.
ECr2O72-/Cr3+=1.33 V  ECl2/Cl-=1.36 VEMnO4-/Mn2+=1.51 V  ECr3+/Cr=-0.74 V

(i) Cr3+

(ii) MnO4-

(iii) Cr2O72-

(iv) Mn2+

Answer:

The electrode potential value of Cr3+/Cr redox couple is most negative. This implies that it has the least tendency to undergo reduction and therefore, Cr3+ will be the most stable oxidized species.
Hence, the correct answer is option (i).

Page No 35:

Question 13:

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

Answer:

The oxidation state of aluminium is +3 in Al2O3.
Al to the power of plus 3 end exponent space plus space 3 straight e to the power of minus rightwards arrow Al
Therefore, 3 moles of electrons are required to reduce 1 mole of Al3+ to 1 mole of Al and the quantity of charge required is 3F.
1F = charge on 1 mole of electrons = charge on 1 electron x N= 1.6021 x 10-19 C x 6.023 x 1023 = 96500 C
Hence, the correct answer is option (iii).
 

Page No 35:

Question 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.

Answer:

The cell constant of a particular conductivity cell is constant for a cell.
Hence, the correct answer is option (iv).

Page No 35:

Question 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.

Answer:

The reactions taking place at anode and cathode in a lead storage battery get reversed on charging and PbSO4 is reduced to Pb on anode and oxidized to PbO2 on cathode.
Hence, the correct answer is option (i).



Page No 36:

Question 16:

m NH4OH0is equal to ______________.

(i) m NH4OH0+m NH4Cl0-HCl0

(ii) m NH4Cl0+m NaOH0-NaCl0

(iii) m NH4Cl0+m NaCl0-NaOH0

(iv) m NaOH0+m NaCl0-NH4Cl0

Answer:

According to Kohlrausch law, the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the cation and anion of the electrolyte. This law can be used to determine limiting molar conductivity of any electrolyte from individual ions.
logical and subscript straight m space open parentheses NH subscript 4 Cl close parentheses end subscript superscript 0 plus logical and subscript straight m space open parentheses NaOH close parentheses end subscript superscript 0 minus logical and subscript open parentheses NaCl close parentheses end subscript superscript 0
equals space logical and subscript straight m space left parenthesis NH subscript 4 to the power of plus right parenthesis end subscript superscript 0 space plus space logical and subscript straight m space left parenthesis Cl to the power of minus right parenthesis end subscript superscript 0 space plus space logical and subscript straight m space left parenthesis Na to the power of plus right parenthesis end subscript superscript 0 space plus space logical and subscript straight m space left parenthesis OH to the power of minus right parenthesis end subscript superscript 0 space minus space logical and subscript straight m space left parenthesis Na to the power of plus right parenthesis end subscript superscript 0 space minus space logical and subscript straight m space left parenthesis Cl to the power of minus right parenthesis end subscript superscript 0
equals logical and subscript straight m space left parenthesis NH subscript 4 OH right parenthesis end subscript superscript 0
Hence, the correct answer is option (ii).

Page No 36:

Question 17:

In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?

(i) Na+aq+e- Nas; ECell=-2.71 V

(ii) 2H2Ol  O2g+4H+aq+4e-; ECell=1.23 V

(iii) H+aq+e- 12H2g; ECell=0.00 V

(iv) Cl-aq  12Cl2g+e- ; ECell=1.36 V

Answer:

There are two reactions possible at anode during the electrolysis of aqueous sodium chloride solution.
(a) 2 straight H subscript 2 straight O open parentheses straight l close parentheses space rightwards arrow space straight O subscript 2 open parentheses straight g close parentheses plus 4 straight H to the power of plus open parentheses aq close parentheses plus 4 straight e to the power of minus semicolon space straight E subscript Cell superscript circled dash equals 1.23 space straight V
(b) Cl to the power of minus open parentheses aq close parentheses space rightwards arrow space 1 half Cl subscript 2 open parentheses straight g close parentheses plus straight e to the power of minus space semicolon space straight E subscript Cell superscript circled dash equals 1.36 space straight V
The reaction with lower Ecell value i.e. reaction (a) should take place at anode but owing to overpotential of oxygen, Cl- is oxidized to Cl2. So, reaction (b) takes place at anode.
Hence, the correct answer is option (iv).
 

Page No 36:

Question 18:

In the given question two or more than two options may be correct.

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.

Answer:

The standard electrode potential for H+/H2 redox couple is zero and any redox couple with positive value of standard electrode potential, such as Cu2+/Cu will more easily undergo reduction i.e. will be a stronger oxidizing agent. Also, it can not displace Hfrom acid.
Hence, the correct answers are options (ii) and (iv).

Page No 36:

Question 19:

In the given question two or more than two options may be correct.

ECell for some half cell reactions are given below. On the basis of these mark the correct answer.

 (a) H+aq+e-  12H2g ;                 ECell=0.00 V(b) 2H2Ol  O2g+4H+aq+4e- ;   ECell=1.23 V(c) 2SO42-aq  S2O82-aq+2e- ;     ECell=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.

Answer:

During the electrolysis of dilute sulphuric acid solution, hydrogen will be reduced at cathode while water will be oxidised at anode. This is due to excess of water as compared to SO42- in dilute sulphuric acid solution.
Hence, the correct answers are options (i) and (iii).

Page No 36:

Question 20:

In the given question two or more than two options may be correct.
ECell = 1.1for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?

(i) 1.1 = Kc

(ii) 2.303RT2FlogKc=1.1

(iii) log Kc=2.20.059

(iv) log Kc = 1.1

Answer:

At equilibrium, Ecell = 0.

So comma space 0 equals space straight E to the power of circled dash subscript cell minus fraction numerator 2.303 RT over denominator 2 straight F end fraction logK subscript straight c
space straight E to the power of circled dash subscript cell equals fraction numerator 2.303 RT over denominator 2 straight F end fraction logK subscript straight c equals 1.1
logK subscript straight c equals fraction numerator 2.2 over denominator 0.059 end fraction

Hence, the correct answers are options (ii) and (iii).



Page No 37:

Question 21:

In the given question two or more than two options may be correct.

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.

Answer:

Conductivity of an electrolytic solution depends on the following factors:
a) nature of electrolyte
b) size of the ions produced and their solvation
c) nature of the solvent and its viscosity
d) concentration of the electrolyte
e) temperature
Hence, the correct answers are options (i) and (ii).

Page No 37:

Question 22:

In the given question two or more than two options may be correct.

m0(H2O) is equal to _______________.

(i) mHCl0+mNaOH0-mNaCl0

(ii) mHNO30+mNaNO30-mNaOH0

(iii) HNO30+mNaOH0-mNaNO30

(iv) mNH4OH0+mHCl0-mNH4Cl0

Answer:

Kohlrausch law can be used to here to determine m0(H2O) from the limiting molar conductivities of given strong electrolytes. 
m0(H2O)mHCl0+mNaOH0-mNaCl0
m0(H2O)HNO30+mNaOH0-mNaNO30

Hence, the correct answers are options (i) and (iii).
 

Page No 37:

Question 23:

In the given question two or more than two options may be correct.

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.

Answer:

During the electrolysis of  aqueous solution of CuSOby using platinum electrodes, the following reactions take place:
At space cathode colon
Cu to the power of 2 plus end exponent plus space 2 straight e to the power of minus rightwards arrow Cu
At space anode colon
2 straight H subscript 2 straight O rightwards arrow straight O subscript 2 plus 4 straight H to the power of plus plus straight e to the power of minus
This is due to standard electrode potential values.
Hence, the correct answers are options (i) and (iii).



Page No 38:

Question 24:

In the given question two or more than two options may be correct.

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.
 

Answer:

During the electrolysis of aqueous solution of CuSOin the presence of Cu electrodes, the following reactions take place:
At space cathode colon
Cu to the power of 2 plus end exponent left parenthesis a q right parenthesis plus 2 straight e to the power of minus rightwards arrow Cu left parenthesis s right parenthesis
At space anode colon
Cu left parenthesis s right parenthesis rightwards arrow Cu to the power of 2 plus end exponent left parenthesis a q right parenthesis plus space 2 straight e to the power of minus

Hence, the correct answers are options (i) and (ii).

Page No 38:

Question 25:

In the given question two or more than two options may be correct.

Conductivity κ, is equal to ____________.
(i) 1RlA

(ii) G*R

(iii) m

(iv) lA

Answer:

Conductivity space kappa equals fraction numerator Cell space constant over denominator Resistance end fraction equals straight G to the power of asterisk times over R equals fraction numerator begin display style bevelled l over A end style over denominator R end fraction equals 1 over R. l over A

Hence, the correct answers are options (i) and (ii).

Page No 38:

Question 26:

In the given question two or more than two options may be correct.

Molar conductivity of ionic solution depends on ___________.
(i) temperature.
(ii) distance between electrodes.
(iii) concentration of electrolytes in solution.
(iv) surface area of electrodes.

Answer:

The molar conductivity of ionic solution varies with temperature and concentration of electrolytes in solution.
Hence, the correct answers are options (i) and (iii). 
 

Page No 38:

Question 27:

In the given question two or more than two options may be correct.

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

Answer:

In cell representation, the oxidation half cell is written on left side and the reduction half cell is written on right side.
The cell reaction for given cell is Mg + Cu2+ → Mg2+ + Cu.
So, Mg is getting oxidized to Mg2+ at anode while Cu is getting deposited at cathode via reduction of Cu2+.
Hence, the correct answers are options (ii) and (iii).
 

Page No 38:

Question 28:

Can absolute electrode potential of an electrode be measured?

Answer:

No, we can not measure the absolute electrode potential of an electrode. One of the electrodes is taken as reference, such as, SHE(Standard Hydrogen Electrode) and then, the potential of other electrode is determined by calculating the difference between the the two electrode potentials. Thus, electrode potential of an electrode can be measured only relatively.

Page No 38:

Question 29:

Can ECell or rG for cell reaction ever be equal to zero?

Answer:

No, ECell or rG for a cell reaction can never be zero. Since, SHE(Standard Hydrogen Electrode) has been assigned ECell =0  as reference, all other cells have non zero ECell or rG
rG=-nFECell

Page No 38:

Question 30:

Under what condition is ECell = 0 or ΔrG = 0?

Answer:

When the concentration of ions do not change, a state of equilibrium is attained. At this stage, ECell = 0 or Δr= 0.

Page No 38:

Question 31:

What does the negative sign in the expression EZn2+/Zn=-0.76 V mean?

Answer:

The negative sign denotes that zinc is more reactive than hydrogen and can displace it from acid solution. Also, the reduction of Zn2+ to Zn is non-spontaneous process.



Page No 39:

Question 32:

Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.

Answer:

According to Faraday's second law of electrolysis, the amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.

fraction numerator Mass space of space copper space deposited over denominator Mass space of space silver space deposited end fraction equals fraction numerator Equivalent space weight space of space copper over denominator Equivalent space weight space of space silver end fraction equals fraction numerator Molar space mass space of space copper cross times number space of space electrons space required space to space reduce space Ag to the power of plus over denominator Molar space mass space of space silver cross times number space of space electrons space required space to space reduce space Cu to the power of 2 plus end exponent end fraction
W subscript Cu over W subscript Ag equals fraction numerator straight M subscript Cu cross times n subscript Ag over denominator straight M subscript Ag cross times straight n subscript Cu end fraction
W subscript Cu over W subscript Ag equals fraction numerator 63.5 cross times 1 over denominator 108 cross times 2 end fraction equals fraction numerator 63.5 over denominator 216 end fraction

Since, equivalent weights of copper and silver are different, the masses of copper and silver deposited on the cathode will also be different.

Page No 39:

Question 33:

Depict the galvanic cell in which the cell reaction is Cu + 2Ag+ → 2Ag + Cu2+

Answer:

The galvanic cell for the given cell reaction can be depicted as follows:
Cu(s) | Cu2+(aq)  ||  Ag+(aq) | Ag(s).

Page No 39:

Question 34:

Value of standard electrode potential for the oxidation of Clions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cloxidised at anode instead of water?

Answer:

In the electrolysis of aqueous sodium chloride, Cl–  is oxidized at anode instead of water owing to overpotential of oxygen which makes the oxidation of water non-feasible.

Page No 39:

Question 35:

What is electrode potential?

Answer:

Electrode potential is the potential difference that is developed between the electrode and the electrolyte. It is known as standard electrode potential when the concentrations of all the species involved in a half-cell is unity.

Page No 39:

Question 36:

Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell?

Answer:

As zinc has lower reduction potential than copper, so, it will undergo oxidation and release electrons which will make electrode 'A' negative. The electrode 'B' will be positively charged as copper ions from the solutions will take up the electrons to deposit copper.

Page No 39:

Question 37:

Why is alternating current used for measuring resistance of an electrolytic solution?

Answer:

The use of direct current to measure the resistance of an electrolytic solution alters the concentration of ions in the solution. Therefore, alternating current is used as it maintains a constant concentration of ions in the solution. 

Page No 39:

Question 38:

A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

Answer:

When the opposing potential of 1.1V is applied to this cell, the cell reaction stops completely and no current flows through the cell.

Page No 39:

Question 39:

How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed?

Answer:

During the electrolysis of brine (aq. NaCl solution), NaOH is formed as one of the products. As it is a strong base, the pH becomes more than 7.

Page No 39:

Question 40:

Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?

Answer:

Mercury cell has a constant cell potential throughout its useful life because no ions are involved in the overall reaction of the cell. Therefore, concentration of ions also do not change throughout its lifetime.

Page No 39:

Question 41:

Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The m of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.

Answer:

Strong electrolytes already have high value of molar conductivities as they are completely dissociated into ions in solution. On diluting their solutions, only a small increase in their molar conductivities is observed as number of ions remain the same and only interionic distance increases. Therefore, electrolyte 'B' will be a strong electrolyte.



Page No 40:

Question 42:

When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.

Answer:

The following reactions take place during electrolysis of acidulated water (dil.H2SOsolution):
At cathode:4H++4e-2H2At anode:2H2OO2+4H++4e-
Since the concentration of H+ ions remains constant, the pH of the solution also remains constant and unaffected.

 

Page No 40:

Question 43:

In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

Answer:

The specific conductivity of electrolytes in an aqueous solution decreases with addition of water as number of ions per unit volume decreases on dilution.

Page No 40:

Question 44:

Which reference electrode is used to measure the electrode potential of other electrodes?

Answer:

SHE(Standard Hydrogen Electrode) is used as a reference electrode. The potential of other electrodes is determined by calculating the difference between the the two electrode potentials. The electrode potential of SHE is set to be zero.

Page No 40:

Question 45:

Consider a cell given below
Cu | Cu2+ || Cl¯ | Cl2, Pt
Write the reactions that occur at anode and cathode

Answer:

The following reactions take place:
At anode:
Cu → Cu2+ + 2e-
At cathode:
Cl2 + 2e- → 2Cl-
 

Page No 40:

Question 46:

Write the Nernst equation for the cell reaction in the Daniel cell. How will the ECell be affected when concentration of Zn2+ ions is increased?

Answer:

Cell reaction in Daniel Cell:
Zn(s) + Cu2+(aq) ⭢ Zn2+(aq) + Cu(s)

Nernst Equation:
straight E subscript cell equals straight E subscript cell superscript blank to the power of circled dash end superscript minus fraction numerator 0.059 over denominator 2 end fraction log fraction numerator left square bracket Zn to the power of 2 plus end exponent right square bracket over denominator left square bracket Cu to the power of 2 plus end exponent right square bracket end fraction

Ecell will decrease when the concentration of Zn2+ ions is increased. 
 

Page No 40:

Question 47:

What advantage do the fuel cells have over primary and secondary batteries?

Answer:

Fuel cells are advantageous in a way that they run continuously as long as the reactants are supplied to the electrodes and products are removed from the electrolytic compartment. Primary cells, on the other hand, have a limited lifetime and become dead once fully discharged while secondary batteries are rechargeable but take a long time to recharge.

Page No 40:

Question 48:

Write the cell reaction of a lead storage battery when it is discharged. How does the density of the electrolyte change when the battery is discharged?

Answer:

Cell reaction of a lead storage battery when it is discharged:

Pb(s) + PbO2(s) + 2H2SO4 → 2PbSO4(s) + 2H2O(l)

The density of the electrolyte decreases when the battery is discharged because sulphuric acid is consumed and water is obtained as product in the reaction.

Page No 40:

Question 49:

Why on dilution the Λm of CH3COOH increases drastically, while that of CH3COONa increases gradually?

Answer:

The value of Λof CH3COOH increases drastically on dilution because it is a weak electrolyte and the degree of dissociation of weak electrolytes increases on dilution and as a result, number of ions in total volume of solution also increases.
On the other hand, the value of Λof CH3COONa increases gradually as it is a strong electrolyte and strong electrolytes already have high value of molar conductivities as they are completely dissociated into ions in solution. On diluting their solutions, only a small increase in their molar conductivities is observed as number of ions remain the same and only interionic distance increases.

Page No 40:

Question 50:

Match the terms given in Column I with the units given in Column II.
 

Column I Column II
(i) Λm (a) S cm–1
(ii) ECell (b) m–1
(iii) κ (c) S cm2 mol–1
(iv) G* (d) V

Answer:

Column I Column II
(i) Λm (c) S cmmol–1
(ii) ECell (d) V
(iii) κ (a) S cm–1
(iv) G* (b) m–1

Page No 40:

Question 51:

Match the terms given in Column I with the items given in Column II.
 

Column I Column II    
(i) Λm (a) intensive property
(ii) ECell (b) depends on number of ions/volume
(iii) κ (c) extensive property
(iv) ΔrGCell (d) increases with dilution

Answer:

Column I Column II    
(i) Λm (d) increases with dilution
(ii) ECell (a) intensive property
(iii) κ (b) depends on number of ions/volume
(iv) ΔrGCell (c) extensive property

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Question 52:

Match the items of Column I and Column II.
 

Column I Column II
(i) Lead storage battery (a) maximum efficiency
(ii) Mercury cell (b) prevented by galvanisation
(iii) Fuel cell (c) gives steady potential
(iv) Rusting (d) Pb is anode, PbO2 is cathode

Answer:

Column I Column II
(i) Lead storage battery (d) Pb is anode, PbOis cathode
(ii) Mercury cell (c) gives steady potential
(iii) Fuel cell (a) maximum efficiency
(iv) Rusting (b) prevented by galvanisation



Page No 41:

Question 53:

Match the items of Column I and Column II.
 

Column I Column II
(i) κ (a) I × t
(ii) Λm (b) Λm/Λm0
(iii) α (c) κC
(iv) Q (d) G*R

Answer:

Column I Column II
(i) κ (d) G*R
(ii) Λm (c) κC
(iii) α (b) Λm/Λm0
(iv) Q (a) I × t

Page No 41:

Question 54:

Match the items of Column I and Column II.

Column I Column II
(i) Lechlanche cell (a) cell reaction 2H2 + O2 → 2H2O
(ii) Ni – Cd cell (b) does not involve any ion in solution and is used in hearing aids.
(iii) Fuel cell (c) rechargeable
(iv) Mercury cell (d) reaction at anode, Zn → Zn2+ + 2e
  (e) converts energy of combustion into electrical energy

Answer:

Column I Column II
(i) Lechlanche cell (d) reaction at anode, Zn → Zn2+ + 2e
(ii) Ni – Cd cell (c) rechargeable
(iii) Fuel cell (a) cell reaction 2H+ O→ 2H2O
(e) converts energy of combustion into electrical energy
(iv) Mercury cell (b) does not involve any ion in solution and is used in hearing aids.

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Question 55:

Match the items of Column I and Column II on the basis of data given below:

EF2/F-=2.87 V, ELi+/Li=-3.5 V, EAu3+/Au=1.4 V, EBr2/Br-=1.09 V
Column I Column II            
(i) F2 (a) metal is the strongest reducing agent
(ii) Li (b) metal ion which is the weakest oxidising agent
(iii) Au3+ (c) non metal which is the best oxidising agent
(iv) Br (d) unreactive metal
(v) Au (e) anion that can be oxidised by Au3+
(vi) Li+ (f) anion which is the weakest reducing agent
(vii) F (g) metal ion which is an oxidising agent

Answer:

Column I Column II            
(i) F2 (c) non metal which is the best oxidising agent
(ii) Li (a) metal is the strongest reducing agent
(iii) Au3+ (g) metal ion which is an oxidising agent
(iv) Br (e) anion that can be oxidised by Au3+
(v) Au (d) unreactive metal
(vi) Li+ (b) metal ion which is the weakest oxidising agent
(vii) F (f) anion which is the weakest reducing agent



Page No 42:

Question 56:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Cu is less reactive than hydrogen.
Reason : ECu2+/Cu is negative.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

Copper is less reactive than hydrogen because ECu2+/Cu is positive and it can not displace hydrogen from acid.
Hence, the correct answer is option (iii).

Page No 42:

Question 57:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : ECell should have a positive value for the cell to function.
Reason : Ecathode < Eanode

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

ECellEcathode − Eanode
The condition for a cell to function is that ECell should have a positive value and this is possible only when Ecathode > Eanode.
Hence, the correct answer is option (iii).

Page No 42:

Question 58:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Conductivity of all electrolytes decreases on dilution.
Reason : On dilution number of ions per unit volume decreases.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

The conductivity of all electrolytes in an aqueous solution decreases with addition of water as number of ions per unit volume decreases on dilution.
Hence, the correct answer is option (i).

Page No 42:

Question 59:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Λm for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.
Reason : For weak electrolytes degree of dissociation increases with dilution of solution.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

The value of Λfor weak electrolytes shows a sharp increase when the electrolytic solution is diluted because the degree of dissociation of weak electrolytes increases on dilution and as a result, number of ions in total volume of solution also increases.
Hence, the correct answer is option (i).

Page No 42:

Question 60:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Mercury cell does not give steady potential.
Reason : In the cell reaction, ions are not involved in solution.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

Mercury cell has a constant cell potential throughout its useful life because no ions are involved in the overall reaction of the cell. Therefore, concentration of ions also do not change throughout its lifetime.
Hence, the correct answer is option (v).

Page No 42:

Question 61:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Electrolysis of NaCl solution gives chlorine at anode instead of O2.
Reason : Formation of oxygen at anode requires over voltage.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

In the electrolysis of NaCl solution, Cl–  is oxidized at anode instead of water owing to overpotential of oxygen which makes the oxidation of water non-feasible. Therefore, chlorine is obtained at anode instead of oxygen.
Hence, the correct answer is option (i).

Page No 42:

Question 62:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : For measuring resistance of an ionic solution an AC source is used.
Reason : Concentration of ionic solution will change if DC source is used.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

The use of DC(Direct Current) to measure the resistance of an ionic solution alters the concentration of ions in the solution. Therefore, AC(Alternating Current) is used as it maintains a constant concentration of ions in the solution. 
Hence, the correct answer is option (i).

Page No 42:

Question 63:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Current stops flowing when ECell = 0.
Reason : Equilibrium of the cell reaction is attained.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

When there is no change in concentration of ions, a state of equilibrium is attained and current stops flowing through the cell. Thus,  Ecell = 0.
Hence, the correct answer is option (i).

Page No 42:

Question 64:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : EAg+/Ag increases with increase in concentration of Ag+ ions.
Reason : EAg+/Ag has a positive value.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

 EAg+/Ag increases with increase in concentration of Agions. This is explained by the Nernst equation.
E subscript Ag to the power of plus divided by Ag end subscript equals E subscript Ag to the power of plus divided by Ag end subscript superscript circled dash minus 0.059 space log fraction numerator 1 over denominator left square bracket Ag to the power of plus right square bracket end fraction
Also, EAg+/Ag has a positive value but this is not an explanation for the above statement.
Hence, the correct answer is option(ii).

Page No 42:

Question 65:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Copper sulphate can be stored in zinc vessel.
Reason : Zinc is less reactive than copper.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.

Answer:

Copper sulphate can not be stored in zinc vessel because zinc is more reactive than copper and will displace it from copper sulphate solution. Thus, zinc will get dissolved to form zinc sulphate solution.
Hence, the correct answer is option (iv).



Page No 43:

Question 66:

Consider the Fig. 3.2 and answer the following questions.

(i) Cell ‘A’ has ECell = 2V and Cell ‘B’ has ECell = 1.1V which of the two cells ‘A’ or ‘B’ will act as an electrolytic cell. Which electrode reactions will occur in this cell?
(ii) If cell ‘A’ has ECell = 0.5V and cell ‘B’ has ECell = 1.1V then what will be the reactions at anode and cathode?

Answer:

(i) Cell 'B' will act as electrolytic cell because it has lower Ecell than cell 'A'. The electrode reactions which will occur in this cell are:
At anode:
Cu → Cu2+ + 2e-
At cathode:
Zn2+ + 2e- → Zn

(ii) Now, cell 'A' will act as electrolytic cell because it has lower Ecell than cell 'B' and cell 'B' will act as galvanic cell. The electrode reactions which will occur in this cell are:
At anode:
Zn → Zn2+ + 2e-
At cathode:
Cu2+ + 2e- → Cu

Page No 43:

Question 67:

Consider Fig. 3.2 and answer the questions (i) to (vi) given below.

(i) Redraw the diagram to show the direction of electron flow.
(ii) Is silver plate the anode or cathode?
(iii) What will happen if salt bridge is removed?
(iv) When will the cell stop functioning?
(v) How will concentration of Zn2+ ions and Ag+ ions be affected when the cell functions?
(vi) How will the concentration of Zn2+ ions and Ag+ ions be affected after the cell becomes ‘dead’?

Answer:

(i)
 

(ii) Silver plate is cathode because it has higher E value than zinc.

(iii) The cell potential will drop to zero and it will stop functioning if salt bridge is removed.

(iv) The cell will stop functioning when Ecell becomes equal to zero i.e. when equilibrium is attained.

(v) As the cell functions, the concentration of Zn2+ ions increases because zinc is oxidized and the concentration of Agions decreases because Agions are reduced to Ag.

(vi) The concentration of Zn2+ ions and Agions will remain constant and unaffected after the cell becomes 'dead' or discharged as Ecell becomes equal to zero.

Page No 43:

Question 68:

What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?
 

Answer:

The relationship between Gibbs free energy of the cell reaction (∆rG) in a galvanic cell and the emf of the cell (Ecell) is as follows:
rG = −nFEcell
where nF is the amount of charge passed through the cell

Maximum work will be obtained from a galvanic cell when the charge is passed through it reversibly.
Maximum work = ∆rG = − nFEcell
 



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