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NCERT Exemplar Class 12 Chemistry Chapter 3 Electrochemistry

NCERT Exemplar Class 12 Chemistry Chapter 3 Electrochemistry

Edited By Sumit Saini | Updated on Sep 16, 2022 04:33 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Chemistry solutions chapter 3 deals with the practical application and production of electricity from the energy released at the time of spontaneous chemical reactions and its use for conducting non-spontaneous chemical transformations. Class 12 Chemistry NCERT exemplar solutions chapter 3 shows the diagrammatic and practical formation of different types of chemicals through an electrochemical process. The electrochemical process for conversion of chemical energy into electric energy through batteries and fuel cells are used at a large scale in a variety of devices and are considered very efficient and environment friendly. The concepts in NCERT along with NCERT exemplar Class 12 Chemistry solutions chapter 3 are very useful for Board exams as well as the JEE Main and NEET entrance exams which you can make more convenient by using the NCERT exemplar Class 12 Chemistry solutions chapter 3 pdf download function.

This Story also Contains
  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: MCQ (Type 1)
  2. Question:1
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: MCQ (Type 2)
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Short Answer Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Matching Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Assertion and Reason Type
  7. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Long Answer Type
  8. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3 Electrochemistry- Main Subtopics
  9. NCERT Exemplar Class 12 Chemistry Solutions
  10. NCERT Exemplar Class 12 Chemistry Solutions Chapter 3 Electrochemistry -Important Topics

Also, check - NCERT Solutions for Class 12 Chemistry

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: MCQ (Type 1)

Question:1

Which cell will measure standard electrode potential of copper electrode?
(i) \; Pt (s)| H_{2}(g,0.1 bar)| H^+ (aq.,1 M)|| Cu^{2+}(aq.,1M) |Cu
(ii) \; Pt (s) |H_{2}(g,1 bar) |H^+ (aq.,1 M)|| Cu^{+2}(aq.,2M)| Cu
(iii) \; Pt (s) | H_{2}(g,1 bar)| H^+ (aq.,1 M) || Cu^{2+}(aq.,1M) |Cu
(iv) \; Pt (s) |H_{2}(g,1 bar) |H^+ (aq.,0.1 M) ||Cu^{2+}(aq.,1M)| Cu

Answer:

The answer is the option (iii) On connecting copper electrode to standard hydrogen electrode, it acts as cathode and its standard electrode potential can be measured.
E^{0}=E_{R}^{0}-E_L^0=E_R^0-0=E_R^0
Pt (s)\left | H_{2}(g,1\; bar) \right |H^{+}(aq., 1 M)\parallel Cu^{2+}(aq., 1\; M)\mid Cu
will measure standard electrode potential of copper electrode.
The standard electrode potential is measured for a given cell by coupling it with the standard hydrogen electrode where pressure of hydrogen gas is maintained at one bar. Concentration of the H+ ion in the solution is one molar and so is the concentration of the oxidized and reduced forms of the species.

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Question:2

Electrode potential for Mg electrode varies according to the equation
E_{Mg^{2+}/Mg}=E^{\circleddash }_{Mg^{2+}/Mg}-\frac{0.0591}{2}\; log \frac{1}{[Mg^{2+}]} The graph of E_{Mg^{2+}/Mg} Vs log [Mg^{2+}] is

Answer:

The answer is the option (ii) E_{\frac{Mg2+}{Mg}}=E_{{\frac{Mg^{2+}}{Mg}}}^o-\frac{0.059}{2}log[Mg^{2+}]
The given equation is that of a straight line with a positive slope and a non-zero intercept

Question: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.
Answer:

The answer is the option (iii) While Ecell is independent of the mass of species or the number of particles, it is an intensive property, whereas ΔrG depends on the number of particles and is an extensive property.

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:

The answer is the option (ii) when no current is drawn through the cell, the difference between the electrode potentials of two electrodes is EMF.

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 the reduction reaction.
(iii) It provides a surface for conduction of electrons.
(iv) It provides a surface for a redox reaction.
Answer:

The answer is the option (iv) Inert electrodes act only as source or sink for electrons. They do not undergo redox reaction and merely provide the surface for the reaction.

Question:6

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

Answer:

The answer is the option (iii) An electrochemical cell can behave like an electrolytic cell when there is an application of an external opposite potential on the galvanic cell and reaction is not inhibited until the opposing voltage reaches the value 1.1 V. No current flows through the cell when this happens. Reaction will function in the opposite direction on increasing the external potential any further.

Question:7

Which of the statements about solutions of electrolytes is not correct?
(i) The conductivity of the solution depends upon the size of ions.
(ii) Conductivity depends upon the viscosity of the solution.
(iii) Conductivity does not depend upon solvation of ions present in solution.
(iv) The conductivity of the solution increases with temperature.
Answer:

The answer is the option (iii) Conductivity decreases with an increasing salvation of ions.

Question:8

Using the data given below to find out the strongest reducing agent.
E^{\circleddash }_{Cr_{2}O{_{7}}^{2-}/Cr^{3+}}=1.33 \; V
E^{\circleddash }_{Cl_{2}/Cl^{-}}=1.36 \; V
E^{\circleddash }_{MnO{_{4}}^{-}/Mn^{2+}}=1.51 \; V
E^{\circleddash }_{Cr^{3+}/Cr}=-0.74\; V
(i)\; Cl^{-}
(ii)\; Cr
(iii)\; Cr^{3+}
(iv)\; Mn^{+2}
Answer:

The answer is the option (ii)\; Cr is the strongest reducing agent as it is the only one with a negative standard reduction potential

Question:9

Use the data given in Q.8 and find out which of the following is the strongest oxidising agent.
(i) \; Cl^{-}
(ii) \: Mn^{2+}
(iii) \; MnO^{-}_{4}
(iv) Cr^{3+}
Answer:

The answer is the option (iii). Oxidizing capacity increases with a positive increase in the standard reduction potential. MnO^{4-} with the highest standard reduction potential is the strongest oxidizing agent.

Question:10

Using the data given in Q.8 find out in which option the order of reducing power is correct.
(i) Cr^{3+} < Cl^{-} < Mn^{2+} < Cr
(ii) Mn^{2+} < Cl^{-}< Cr^{+3} < Cr
(iii) Cr^{3+} < Cl^{-}< Cr_{2}O_{7}^{2-} < MnO^{-}_{4}
(iv) Mn^{2+} < Cr^{3+} < Cl^{-}< Cr
Answer:

The answer is the option (ii). A decrease in reduction potential corresponds with an increase in reduction power. The order of reducing power is (ii) Mn^{2+} < Cl^{-}< Cr^{+3} < Cr

Question:11

Use the data given in Q.8 and find out the most stable ion in its reduced form.
(i) \; Cl^{-}
(ii) \; Cr^{3+}
(iii) \; Cr
(iv)\; Mn^{2+}
Answer:

The answer is the option (iv) Due to a higher E^{o} value, Mn^{2+} is most stable in its reduced form.

Question:12

Use the data of Q.8 and find out the most stable oxidised species.
(i) \; Cr^{3+}
(ii) \; MnO^{-}_{4}
(iii)\; Cr_{2}O_{7}^{2-}
(iv) \; Mn^{2+}
Answer:

The answer is the option (i). Due to the lowest E^{o}value, Cr^{3+} is most stable in its oxidized form.

Question:13

The quantity of charge required to obtain one mole of aluminium from Al_{2}O_{3} is ___________.
(i) 1F
(ii) 6F
(iii) 3F
(iv) 2F
Answer:

The answer is the option (iii) Al_{2}O_{3}\rightarrow 2Al^{3+}+ 3O^{2-}
Al^{3+}+3e^{-} \rightarrow Al (for 1 mole)
To obtain 1 mole Al from Al_{2}O_{3}, we require 3F charge

Question:14

The cell constant of a conductivity cell _____________.
(i) changes with the change of electrolyte.
(ii) changes with the change of concentration of electrolyte.
(iii) changes with the temperature of the electrolyte.
(iv) remains constant for a cell.
Answer:

The answer is the option (iv) The cell constant of a conductivity cell remains constant for a cell.

Question:15

While charging the lead storage battery ______________.
(i) PbSO_{4} anode is reduced to Pb
(ii) PbSO_{4} cathode is reduced to Pb
(iii) PbSO_{4} cathode is oxidised to Pb
(iv) PbSO_{4} anode is oxidised to PbO_{2}
Answer:

The answer is the option (i)
Reduction at Anode: PbSO_4 (s)+2e^-\rightarrow Pb(s)+SO_{4}^{2-} (aq) (Reduction)
Oxidation at Cathode: PbSO_4 (s)+2H_2 O\rightarrow PbO_2 (s)+SO_{4}^{2-}+4H^++2e^-
Overall reaction : 2PbSO_4 (s)+2H_2 O\rightarrow Pb(s)+PbO_2 (s)+4H^++2SO_{4}^{2-}

Question:16

\wedge ^{0}_{m(NH_{4}OH)} is equal to ______________.
(i) \wedge ^{0}_{m(NH_{4}OH)} + \wedge ^{0}_{m(NH_{4}Cl)}- \wedge ^{0}_{(HCl)}
(ii) \wedge ^{0}_{m(NH_{4}Cl)} + \wedge ^{0}_{m(NaOH)}- \wedge ^{0}_{(NaCl)}
(iii) \wedge ^{0}_{m(NH_{4}Cl)} + \wedge ^{0}_{m(NaCl)}- \wedge ^{0}_{(NaOH)}
(iv) \wedge ^{0}_{m(NaOH)} + \wedge ^{0}_{m(NaCl)}- \wedge ^{0}_{(NH_{4}Cl)}
Answer:

The answer is the option (ii) NH_{4}Cl\rightleftharpoons NH_4^++Cl^-
NaCl\rightleftharpoons Na^{+}+Cl^{-}
NaOH\rightleftharpoons Na^{+}+OH^{-}
NH_{4}OH\rightleftharpoons NH_{4}^{+}+OH^{-}
\wedge ^{0}_{m(NH_{4}Cl)} + \wedge ^{0}_{m(NaOH)}- \wedge ^{0}_{(NaCl)}

Question:17

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

(i) Na^{+}(aq) + e^{-} \rightarrow Na (s); E^{o}_{Cell} = -2.71V

(ii) 2H_{2}O (l) \rightarrow O_{2}(g) + 4H^+(aq) + 4e^{-} ; E^{o}_{Cell} = 1.23V

(iii) H^{+}(aq) + e^{-}\rightarrow \frac{1}{2}H_{2} (g); E^{o}_{Cell} = 0.00V

(iv) Cl^{-}(aq) \rightarrow \frac{1}{2} Cl_{2}(g) + e^{-} ; E^{o}_{Cell} = 1.36V
Answer:

The answer is the option (ii) Upon electrolysis H2O gives-

2H_2 O\rightarrow O_2+4H^++4e^-;E_{cell}^{0}=1.23V

Since, E_{cell}^{0}=1.23V. Hence this half cell reaction occurs at anode.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: MCQ (Type 2)

Question:18

The positive value of the standard electrode potential of {Cu^{2+} / Cu indicates that ____________.
(i) this redox couple is a stronger reducing agent than the {H^{+}/{H_{2}} couple.
(ii) this redox couple is a stronger oxidising agent than {H^{+}/{H_{2}}.
(iii) Cu can displace H_{2} from acid.
(iv) Cu cannot displace H_{2} from acid.
Answer:

The answer is the option (ii, iv) Decrease in E° indicates an increase in reducing power
\\Cu^{2+}+2e^-\rightarrow Cu ; \; E^0=0.34V\\ 2H^++2e^-\rightarrow H_2 ;\; E^0=0.00V
As, E° value for Cu^{2+}is higher -
(i) It is the stronger oxidizing agent
(ii) It can't displace H_{2}

Question:19

E^{\Theta }_{cell} for some half cell reactions are given below. Based on these mark the correct answer.
(i) H^{+}(aq) + e^{-} \rightarrow \frac{1}{2}H_{2}(g) ; E^{o}_{cell} = 0.00V
(ii) 2H_{2}O (l) \rightarrow O_{2} (g) + 4H^{+}(aq) + 4e^{-}; E^{o}_{cell} = 1.23V
(iii) 2SO_{4}^{2-} (aq) \rightarrow S_{2}O_{8}^{2-} (aq) + 2e^{-} ; E^{o}_{cell} = 1.96V
(i) In dilute sulphuric acid solution, hydrogen will be reduced at the cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at the anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at the anode.
(iv) In dilute sulphuric acid solution, SO_{4}^{2-} ion will be oxidised to tetrathionate ion at the anode.
Answer:

The answer is the option (ii, iii) In dilute sulphuric acid solution -
Reduction at cathode: H^++e^-\rightarrow \frac{1}{2} H_2
Oxidation at anode : 2H_2 O \rightarrow O_2+4H^++4e^-
In concentrated sulphuric acid solution, sulphate (SO{_{4}}^{2-}) ions oxidize to form tetrathionate (S_{2}O{_{8}}^{2-}) ions.

Question:20

E^{\Theta }_{Cell} = 1.1V for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?
(i) \; 1.1 = K_{c}
(ii) \;\frac{2.303RT}{2 F} logK_{c} = 1.1
(iii)\; log K_{c} = 2.2/0.059
(iv) log K_{c} = 1.1
Answer:

The answer is the option (ii,iii)
E_{cell}^{0}=\frac{2.303\; RT}{nF} logK_{c}=\frac{0.059}{2}logK_{c}
1.1=\frac{0.059}{2}logK_{c}
logK_{c}=\frac{2.2}{0.059}

Question:21

Conductivity of an electrolytic solution depends on ____________.
(i) nature of electrolyte.
(ii) the concentration of electrolyte.
(iii) power of AC source.
(iv) distance between the electrodes.
Answer:

The answer is the option (i, ii). Mobile ions are responsible for the conductivity of electrolyte solution and this phenomenon is termed ionic conductance. It depends only on the following factors–

  1. the nature of electrolyte added

  2. size of the ion produced and their solvation

  3. concentration of electrolyte

  4. nature of solvent and its viscosity

  5. temperature

Question:22

\wedge _{m(H_{2}O)}^{0} is equal to ______________.
(i) \wedge_{m(HCl)}^{0} + \wedge _{m(NaOH)}^{0} - \wedge_{m(NaCl)}^{0}
(ii) \wedge_{m(HNO_{3})}^{0} + \wedge _{m(NaNO_{3})}^{0} - \wedge_{m(NaOH)}^{0}
(iii) \wedge_{m(HNO_{3})}^{0} + \wedge _{m(NaOH)}^{0} - \wedge_{m(NaNO_{3})}^{0}
(iv) \wedge_{m(NH_{4}OH)}^{0} + \wedge _{m(HCl)}^{0} - \wedge_{m(NH_{4}Cl)}^{0}
Answer:

The answer is the option (i,iv)
\wedge _{m(H_{2}O)}^{0}=\wedge _{m(HCl)}^{0}+\wedge _{m(NaOH)}^{0}-\wedge _{m(NaCl)}^{0}
\wedge_{m(NH_{4}OH)}^{0} + \wedge _{m(HCl)}^{0} - \wedge_{m(NH_{4}Cl)}^{0}= \wedge _{m(H_{2}O)}^{0}
Molar conductivity of water is equal to the sum of the molar conductivities of constituent ions. However, NH_{4}OH doesn’t undergo complete decomposition as it is a weak electrolyte.

Question:23

What will happen during the electrolysis of an aqueous solution of CuSO_{4} by using platinum electrodes?
(i) Copper will deposit at the cathode.
(ii) Copper will deposit at the anode.
(iii) Oxygen will be released at anode.
(iv) Copper will dissolve at the anode.
Answer:

The answer is the option (i,iii)
CuSO_4\rightleftharpoons Cu^{2+}+SO_{4}^{2-}
H_{2}O\rightleftharpoons H^{+}+OH^{-}
At cathode : Cu^{2+}+2e^{-}\rightarrow Cu; E_{cell}^{0}=0.34 \; V
H^{+}+e^{-}\rightarrow\; \frac{1}{2} H_{2}; E_{cell}^{0}=0.00 \; V
At cathode, the reaction with higher E^{0} is preferred
At anode :2SO_{4}^{2-}-2e^{-}\rightarrow S_{2}O_{8}^{2-}; E_{cell}^{0}=1.96 V
2H_{2}O\rightarrow O_{2}+4H^{+}+4e^{-};E_{cell}^{0}=1.23V
At anode, the reaction with lower E^{o} is preferred.

Question:24

What will happen during the electrolysis of an aqueous solution of CuSO_{4} in the presence of Cu electrodes?
(i) Copper will deposit at the cathode.
(ii) Copper will dissolve at the anode.
(iii) Oxygen will be released at anode.
(iv) Copper will deposit at the anode.
Answer:

The answer is the option (i,ii)
At Cathode : Cu^{2+}+2e^{-}\rightarrow Cu(s)
At anode : Cu(s)\rightarrow Cu^{2+}+2e^{-}
Cathode will witness deposite of Cu and Anode will witness its dissolution

Question:25

Conductivity κ , is equal to ____________.
(i) \frac{1}{R} \frac{l}{A}
(ii) \frac{G^{*}}{R}
(iii) \wedge_{m}
(iv) \frac{l}{A}
Answer:

The answer is the option (i,ii)
(i) \frac{1}{R} \frac{l}{A}
(ii) \frac{G^{*}}{R}

Question:26

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

The answer is the option (i, iii). Temperature and concentration of electrolytes determine the molar conductivity of an ionic solution

Question:27

For the given cell, Mg\left | Mg^{2+} \right |\left | Cu^{2+} \right |Cu
(i) Mg is cathode
(ii) Cu is cathode
(iii) The cell reaction is Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu
(iv) Cu is the oxidising agent
Answer:

The answer is the option (ii, iii). At cathode, Cu is reduced and at anode, Mg is oxidized.
(ii) Cu is cathode
(iii) The cell reaction is Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Short Answer Type

Question:28

Can absolute electrode potential of an electrode be measured?

Answer:

No, absolute electrode potential of an electrode cannot be measured.

Question:29

Can E^{\Theta }_{cell} or \Delta _{r}G^{\Theta } for cell reaction ever be equal to zero?

Answer:

E^{o}and \Delta G^{o}can never be zero for a cell reaction.

Question:30

Under what condition is E_{cell} = 0 or \Delta _{r}G= 0 \;?
Answer:

Upon completely discharging or at equilibrium, E_{cell} = 0 and \Delta _{r}G= 0.

Question:31

What does the negative sign in the expression E^{\Theta }_{Zn^{2+}/Zn}=-0.76\; V mean ?
Answer:

E^{\Theta }_{Zn^{2+}/Zn}=-0.76\; V
Zinc has a higher reducing power than Hydrogen Sa it has a negative E^{0} while it is zero for Hydrogen
Zn+H_{2}SO_{4}\rightarrow ZnSO_{4}+H_{2}

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 the same or different? Explain your answer.
Answer:

According to Faraday's second law of electrolysis amount of different substance liberated by same quantity or electricity passes through electrolyte solution is directly proportional to their chemcial equivalent weight.
W1/W2=E1/E2
where, E1 and E2 have different values depending upon number of electrons required to reduce the metal ion.
Hence, massess of Cu and Ag deposited will be different.

Question:33

Depict the galvanic cell in which the cell reaction is Cu + 2Ag^+\rightarrow 2Ag + Cu^{2+}
Answer:

Cu\left | Cu^{2+} \right |\left | Ag^{+} \right |Ag

Question:34

Value of standard electrode potential for the oxidation of Cl^{-} ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl^{-} oxidised at anode instead of water?

Answer:

Under the conditions of electrolysis of aqueous sodium chloride, oxidation of water at anode requires overpotential hence Cl- is oxidised instead of water

Question:35

What is electrode potential?
Answer:

The electrical potential difference set up between the metal and its solution is called electrode potential.

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:

‘A’ has negative and ‘B’ has positive polarity.

Question:37

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

Direct current will lead to electrolysis of the solution, which will change the concentration of ions in the solution. Use of Alternating current will prevent this from happening.

Question:38

A galvanic cell has an 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:

Cell reaction ceases if the opposing potential is equal to the electrical potential.

Question:39

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

Upon electrolysis of Brine solution, the following reactions take place:
Cathode :2H^{+}+2e^{-}\rightarrow H_{2}
Anode : 2Cl^{-}\rightarrow Cl_{2}+2e^{-}
The remaining Na^{+} and OH^{-} ions are responsible for turning the solution basic and thus increase the pH.

Question:40

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

Answer:

The mercury cell has a constant cell potential throughout its useful life because the Ions are not involved in the overall cell reaction of mercury cells.

Question:41

Solutions of two electrolytes ‘A’ and ‘B are diluted. The \Lambda _{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:

The lower increase of \wedge _{m} for Electrolyte B is due to complete ionization of the same. Hence, It is the stronger electrolyte.

Question:42

When acidulated water (dil.H_{2}SO_{4} solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
Answer:

The following reaction occur :
At Anode : 2H_{2}O\rightarrow O_{2}+4H^{+}+4e^{-}
At cathode : 4H^{+}+4e^{-}\rightarrow 2H_{2}(\uparrow)
As the concentration of H+ ions is maintained, there will be no change in pH.

Question:43

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

Concentration of ions will decrease on addition of water, which in turn will reduce the electrical conductivity.

Question:44

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

The standard hydrogen electrode is used as the reference electrode. For other electrodes, we measure the electrode potential considering the electrode potential for standard hydrogen electrode to be zero.

Question:45

Consider a cell given below
Cu\left | Cu^{2+} \right |\left | Cl^{-} \right |Cl_{2},Pt
Write the reactions that occur at anode and cathode.
Answer:

The given cell is :
Cu (s)\left | Cu^{2+} \right |\left | Cl^{-} \right |Cl_{2}(pt)
Anode : Cu (s)\rightarrow Cu^{2+}+2e^{-}
Cathode : Cl_{2}(g)+2e^{-}\rightarrow 2Cl^{-}

Question:46

Write the Nernst equation for the cell reaction in the Daniel cell. How will the ECell be affected when the concentration of Zn^{2+} ions is increased?
Answer:

Daniel Cell :
Zn\; (s)\left |Zn^{2+} \right |\left | Cu^{2+} \right |Cu\; (s)
Anode :Zn(s)\rightarrow Zn^{2+}+2e^{-}
Cathode : Cu^{2+}+2e^{-}\rightarrow Cu (s)
Overall cell reaction : Zn (s)+Cu^{2+}\rightleftharpoons Zn^{2+}+Cu (s)
Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}
E_{cell}=E_{cell}^{0}-\frac{0.059}{2}log \; \frac{[Zn^{2+}]}{[Cu^{2+}]}
Therefore, with increasing concentration of Zn^{2+}, the cell potential decreases.

Question:47

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

Primary battery come with limited number of reactants and can’t be reused once they are discharged. Secondary batteries take a considerable amount of time to recharge. Fuel cell is superior to both primary and secondary batteries as they operate without any breaks as long as you keep supplying the reactants

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:

During Discharge: Pb+PbO_2+2H_2 SO_4\rightarrow 2PbSO_4+2H_2 O
As water is formed during the discharge process, the concentration of electrolyte reduces.

Question:49

Why on dilution the \wedge _{m} of CH_{3}COOH increases drastically, while that of CH_{3}COONa increases gradually?
Answer:

For weak electrolytes like CH_{3}COOH, on dilution, the concentration of ions increases due to the increase in degree of dissociation, but for strong electrolytes like CH_{3}COONa, the number of ions remain constant upon dilution

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Matching Type

Question:50

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

Column I

Column II

(i) \wedge _{m}

(a) S cm-1

(ii) E _{cell}

(b) m-1

(iii) \kappa

(c) S cm2 mol -1

(iv) G^{*}

(d) V

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Answer:

(i —> c), (ii —> d), (iii —> a), (iv —> b)

Question:51

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

Column I

Column II

(i) \wedge _{m}

(a) intensive property

(ii) E_{cell}^{\Theta }

(b) depends on number of ions/volume

(iii) \kappa

(c) extensive property

(iv) \Delta_{r} G_{cell}

(d) increases with dilution

Answer:

(i -> d), (ii -> a), (iii -> b), (iv -> c)

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 galvanization

(iii) Fuel cell

(c) gives steady potential

(iv) Rusting

(d) Pb is anode, PbO2 is cathode

Answer:

(i —> d), (ii -> c) (iii -> a), (iv -> b)

Question:53

Match the items of Column I and Column II.

Column I

Column II

(i) k

(a)I\times t

(ii)\wedge _{m}

(b)\wedge _{m}/\wedge_{m}^{0}

(iii) \; \alpha

(c)\frac{k}{c}

(iv) Q

(d)\frac{G^{*}}{R}

Answer:

(i -> d), (ii -> c), (iii -> b), (iv -> a)

Question:54

Match the items of Column I and Column II.

Column I

Column II

i) Lechlanche cell

(a) cell reaction 2H_{2}+O_{2}\rightarrow 2H_{2}O

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\rightarrow Zn^{2+}+2e^{-}


(e) converts energy of combustion into electrical energy.

Answer:

(i -> d), (ii -> c), (iii —> a, e), (iv -> b)

Question:55

Match the items of Column I and Column II on the basis of data given below:
E_{F_{2}/F^{-}}^{\circleddash}=2.87\; V, E_{Li^{+}/Li}^{\circleddash}=-3.5\; V, E_{Au^{3+}/Au}^{\circleddash}=1.4\; V, E_{Br_{2}/Br^{-}}^{\circleddash}=1.09\; V,

Column I

Column II

(i)\; F_{2}

(a) metal is the strongest reducing agent

(ii)\; Li

(b) metal ion which is the weakest oxidising agent

(iii)\; Au^{3+}

(c) non metal which is the best oxidizing agent

(iv)\; Br^{-}

(d) unreactive metal

(v)\; Au

(e) anion that can be oxidised by Au^{3+}

(vi) \; Li^{+}

(f) anion which is the weakest reducing Agent

(vii) \; F^{-}

(g) metal ion which is an oxidising agent

Answer:

(i->c),(ii->a),(iii->g),(iv->e),(v->d),(vi->b),(vii->g,f)

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Assertion and Reason Type

Question:56

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion:Cu is less reactive than hydrogen.

Reason:E_{Cu^{2+}/Cu}^{\circleddash } is negative
(i) Both Assertion and Reason are true and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (iii) As E_{Cu^{2+}/Cu}^{\circleddash } is positive, Copper is less reactive than hydrogen.

Question:57

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion:E_{Cell} should have a positive value for the cell to function.

Reason:E_{cathode}<E_{anode}
(i) Both Assertion and Reason are true and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (iii) E_{cell}=E_{cathode}-E_{anode}. To have a positive value of E_{cell}, E_{cathode}>E_{anode}

Question:58

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
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 for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (i) Upon dilution, the concentration of ions decreases and hence, conductivity also decreases.

Question:59

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion: \wedge _{m} 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 for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (i) Degree of dissociation of weak electrolytes increases on dilution, which results in a sharp increase in \wedge _{m} values.

Question:60

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
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 for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (v). Mercury cell maintains a constant cell potential because the electrolyte isn’t consumed in the cell process

Question:61

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion: Electrolysis of NaCl solution gives chlorine at anode instead of O_{2}.
Reason: Formation of oxygen at anode requires overvoltage.
(i) Both Assertion and Reason are true and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (i). Though the E^{o} value for the formation of oxygen is lower than that for the formation of chlorine, it is not formed because it requires overvoltage.

Question:62

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
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 for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (i). Direct current will lead to electrolysis of the solution, which will change the concentration of ions in the solution. Use of Alternating current will prevent this from happening

Question:63

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion: Current stops flowing when E_{cell}=0.
Reason: Equilibrium of the cell reaction is attained.
(i) Both Assertion and Reason are true and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (i) At equilibrium, E_{cell}=0 and no current flows.

Question:64

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion:E_{Ag^{+}/Ag} increases with increase in concentration of Ag^{+} ions.
Reason:E_{Ag^{+}/Ag} has a positive value.
(i) Both Assertion and Reason are true and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (ii) Both Assertion and Reason are true and the Reason is not correct explanation for Assertion

Question:65

In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
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 for Assertion.
(ii) Both Assertion and Reason are true and the Reason is not correct explanation for 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 answer is the option (iv) As Zinc is more reactive than Copper, Zinc dissolves in CuSO_{4} solution.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3: Long Answer Type

Question:66

Consider the following figure and answer the following questions.

(i) Cell ‘A’ has Ecell = 2 V and cell ‘B’ has Ecell = 1.1 V 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 = 0.5 V and cell ‘B’ has Ecell = 1.1 V, what will be the reactions at anode and cathode?
Answer:

(i) As the potential of ‘B’ is lower than the potential of ‘A’, it will act as the electrolytic cell. The reactions at electrode ‘B’ are shown below:
Cathode : Zn^{2+}+2e^{-}\rightarrow Zn(s)
Anode: Cu (s)\rightarrow Cu^{2+}+2e^{-}
(ii) At higher potential, Cell ‘B’ acts as a galvanic cell and the reactions will be:
Anode :Zn(s)\rightarrow Zn^{2+}+2e^{-}
Cathode : Cu^{2+}+2e^{-}\rightarrow Cu(s)

Question:67

Consider the figure given below and answer the questions (i) to (vi) that follow.

(i) Redraw the diagram to show the direction of electron flow.
(ii) Is silver plate anode or cathode?
(iii) What will happen if salt bridge is removed?
(iv) When will the cell stop functioning?
(v) How will concentration of Zn^{+2} ions and Ag^{+} ions be affected when the cell functions ?
(vi) How will the concentration of Zn^{+2} ions and Ag^{+} ions be affected after the cell becomes ‘dead’?

Answer:

(i) The cell reaction can be summarised as:
Zn(s)\left | Zn^{+2} \right |\left | Ag^{+} \right |Ag
Electrons move from Zn to Ag.
(ii) Due to a higher standard reduction potential, Silver will act as Cathode and in the external circuit, electrons will flow from zinc anode to silver cathode.
(iii) Removal of salt bridge will lead to a sudden drop in potential to zero.
(iv) If the potential reaches zero (or the cell is discharged), all reactions will cease and the cell will stop functioning.
(v) Nernst equation for the cell is: 0.059,
E=E^{0}-\frac{0.059}{2}log\frac{[Zn^{2+}]}{[Ag^{+}]^{2}}
With increase in concentration of [Zn^{+2}], cell potential will decrease, and with an increase in concentration of [Ag^{+}], cell potential will increase.

(vi) At equilibrium (discharged state, potential drop to zero), the concentration of [Zn^{+2}] and [Ag^{+}] will not change.

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 required relationship between Gibbs free energy and the emf in a galvanic cell is \Delta G=-nFE
The maximum work will be obtained from a galvanic cell W_{max}=nFE^{0} where,
E=cell potential
E^{0} = Standard emf of the cell

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3 Electrochemistry- Main Subtopics

  • Electrochemical Cells

  • Galvanic Cells

  • Measurement of Electrode Potential

  • Nernst Equation

  • Equilibrium Constant from Nernst Equation

  • Electrochemical Cell and Gibbs Energy of Reaction

  • The conductance of Electrolytic Solutions

  • Measurement of the Conductivity of Ionic Solutions

  • Variation of Conductivity and Molar Conductivity with Concentration

  • Electrolytic Cells and Electrolysis

  • Products of Electrolysis

  • Batteries

  • Primary Batteries

  • Secondary Batteries

  • Fuel Cells

  • Corrosion

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NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Chemistry Solutions Chapter 3 Electrochemistry -Important Topics

NCERT Exemplar Class 12 Chemistry chapter 3 solutions gives a descriptive representation of working of an electrochemical cell also known is as Daniel cell, galvanic cell and introduces various new terms such as resistivity, conductivity, molar conductivity, ionic conductivity and electronic conductivity of different ionic solutions.

  • NCERT exemplar Class 12 Chemistry chapter 3 solutions provides various formulas/methods for calculating conductivity and molar conductivity of electrolytic solutions. The lesson gives a brief about Kohlrausch law, its application and defines the Nernst equation for calculating emf and equilibrium constant of galvanic cells.

  • NCERT exemplar Class 12 Chemistry solutions chapter 3 also reflects upon detailed explanation about corrosion as an electrochemical process along with the formation of primary and secondary batteries, fuel cells, and also the process of electrolysis. The lesson covers a variety of topics relating to thermodynamics.

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NCERT Exemplar Class 12 Solutions

Check NCERT Solutions for Class 12 Chemistry

Must read NCERT Solution subject wise

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Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. Q: Can I download the solutions for this chapter?

Ans: Yes, you can download the NCERT exemplar Class 12 Chemistry chapter 3 solutions easily from the website.

2. Q: How many questions are there in this chapter?

Ans: There are a total 6 exercises with a total of 136 questions that are divided into 2 sets of MCQs, short answer, long answer, matching type and assertion & reason questions in this chapter.

3. Q: How can I ace my boards with NCERT books?

Ans: You must be well versed with the chapters in the NCERT textbooks. You must solve the questions after each chapter and more questions. Furthermore, you need to keep practice and solve new questions to ace your boards.

4. Q: Who has prepared these solutions?

Ans: The Class 12 Chemistry NCERT exemplar solutions chapter 3 are prepared by our expert team who refer various advanced books for the precise solutions.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
Since I have finished my 10th boards of CBSE, I want to get an admission to a new school. But admission procedure in many good schools have been done. What can be done now?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


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Read Complete Answer
Garima Sihag 06 May,2024
I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

Read Complete Answer
Garima Sihag 07 May,2024
I did cbse class 12th science with computers in 2023 and did not make 75% for NITS. I plan to give NIOS in october of 2024 but will that make me eligible to take admission in NITs in 2024 or do I have to wait till 2025 because results of NIOS come out after NIT session starts. Thank you for reply!
  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.
Read Complete Answer
Garima Sihag 07 May,2024
I got 96% in cbse class 12th and 86% in cbse class 12th(Science-PCM). I am B.tech Graduate.Can I get the IIM Bangalore call if I score above 99.90%ile in CAT

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

Read Complete Answer
Garima Sihag 07 May,2024
i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer
Upasana Barman 24 August,2022
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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