Question Set B

cbse 12th Standard Chemistry Electrochemistry Model Questions Papers Study Material - QB365

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Questions:

Generic 2

7 x 2 = 14

State
(a) Kohlrausch law of independent migration of ions.
(b) Faraday's first law of electrolysis.

What happens when anisole is heated with HI?

How do you convert phenol to anisole?

Explain, why oxidation states of transition elements first increases from Sc to Mn and then decreases?

The standard electrode potential of daniel is 1.1V. Calculate the standard Gibbs energy for the reaction:
\(Zn\left( s \right) +{ Cu }^{ 2+ }\left( aq \right) \rightarrow { Zn }^{ 2+ }\left( aq \right) +Cu\left( s \right) \)

Why does the conductivity of a solution decrese with dilution?

Why is Pt(IV) more stable than nickel (IV) stage?

Generic 3

7 x 3 = 21

What happens when
(i) Anisole is treated with CH₃CI/anhydrous AICl₃?
(ii) Phenol is oxidised with Na₂Cr₂O₇/H⁺?
(iii) (CH₃)₃C-OH is heated with Cu/573 K?
Write chemical equation in support of your answer.

Give the structure of the major product expected from the following reactions:
(a) Reaction of propanal with methyl magnesium bromide followed by hydrolysis.
(b) Reaction of phenol with Br₂in CS₂.
(c) Reaction of propene with diborane followed by oxidation.

10.

Compare the chemistry of the actinoids with that of lanthanoids with reference to the following :
(i) Electronic configuration
(ii) Oxidation states
(iii) Chemical reactivity

11.

From the given data of E⁰ values, answer the following questions:

\(E_{(M^{2+}/Mn)}^0\)	-0.9 1	-1.18	-0.44	-0.28	-0.25	+0.34
\(E_{(M^{2+}/Mn)}^0\)	Cr	Mn	Fe	Co	Ni	Cu

(i) Why is \(E_{(Cu^{2+}/Cu)}^0\) value exceptionally positive?
(ii) Why is \(E_{(M^{2+}/Mn)}^0\) value highly negative as compared to other elements?
(iii) Which is a stronger reducing agent Cr²+ or Fe²⁺? Give reason.

12.

Write the equations involved in the following reactions:
(i) Reimer-Tiemann reaction
(ii) Williamson synthesis

13.

Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is \(100 \ Ω\). If the resistance of the same cell, when filled with 0.02 mol L^-1 solution, is, calculate \( 520 \ Ω \) the conductivity and molar conductivity of 0.02 mol L^-1solution. (The conductivity of 0.1 mol L^-1 KCl solution is 1.29 S/m.

14.

An aqueous solution of copper sulphate, CuSO₄ was electrolysed between platinum electrodes using a current of 0.1287 ampere for 50 minutes. (Atomic mass of Cu=63.5 g mol^-1].
(a) Write the cathodic reaction.
(b) Calculate:
(i) Electric charge passed during electrolysis.
(ii) Mass of copper deposited at the cathode
[Given : 1F = 96,500 C mol^-1]

Multiple Choice Question

16 x 1 = 16

15.

Williamson's synthesis of preparing dimethyl ether is a/an

(a)

S_N1 reaction

(b)

elimination reaction

(c)

S_N2 reaction

(d)

nucleophilic addition reaction

16.

Salicylic acid reacts with Zinc dust on heating to give

(a)

Benzene

(b)

Benzoic acid

(c)

Phenol

(d)

None of these

17.

Phenol can be distinguished from ethanol by the reactions with ___________

(a)

Br₂/water

(b)

(c)

Neutral FeCl₃

(d)

All the above

18.

\(\text { Phenol } \stackrel{\mathrm{Zn} \text { , dust }}{\longrightarrow} \cdot \mathrm{X}^{\prime} \frac{\mathrm{CH}_{3} \mathrm{Cl}}{\text { Anhy. } \mathrm{AICl}_{3}} \cdot \mathrm{Y}, \frac{\text { Alkaline }}{\mathrm{KMnO}_{4}} \text { ' } \mathrm{Z} \text { ' }\)
The product 'Z' is

(a)

Benzaldehyde

(b)

Benzoic acid

(c)

Benzene

(d)

Toluene

19.

Which of the following method is used for the preparation of symmetrical and unsymmetrical ether?

(a)

Williamson's synthesis

(b)

Riemer-Tiemann reaction

(c)

Kolbe's reaction

(d)

None of the above

20.

IUPAC name of m- cresol is___________.

(a)

3-methylphenol

(b)

3-chlorophenol

(c)

3-methoxyphenol

(d)

benzene-1,3-diol

21.

Transition elements have greater tendency to form complexes because

(a)

They have vacant d orbitals

(b)

They have large size

(c)

They have large charge/ size ratio

(d)

They have two electrons in their outermost shells

22.

The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because

(a)

the 5f orbitals extend farther from the nucleus than the 4f orbitals

(b)

the 5f orbitals are more buried than the 4f orbitals

(c)

there is a similarity between 4f and 5f orbitals in their angular part of the wave function

(d)

the actinoids are more reactive than the lanthanoids

23.

Magnetic moment of 2.84 B.M is given by which of the following ion?

(a)

Ti³⁺

(b)

Ni²⁺

(c)

Cr³⁺

(d)

Mn²⁺

24.

Which of the following statements about the interstitial compounds is incorrect ?

(a)

They retain metallic conductivity

(b)

They are chemically reactive

(c)

They are much harder than the pure metal

(d)

They have higher melting points than the pure metal

25.

Which of the following reagents may be used to distinguish between phenol and benzoic acid?

(a)

Neutral FeCl₃

(b)

Aqueous NaOH

(c)

Tollen's reagent

(d)

Molisch reagent

26.

What current is to be passed for 0.25 s for deposition of a certain weight of metal which is equal to its electrochemical equivalent?

(a)

4 A

(b)

100 A

(c)

200 A

(d)

2 A

27.

\({ \wedge }_{ m }^{ 0 }\left( { NH }_{ 4 }OH \right) \) is equal to_________________.

(a)

\({ \wedge }_{ m }^{ 0 }\left( { NH }_{ 4 }OH \right) +{ \wedge }_{ m }^{ 0 }\left( { NH }_{ 4 }Cl \right) -{ \wedge }^{ 0 }(HCl)\)

(b)

\({ \wedge }_{ m }^{ 0 }\left( { NH }_{ 4 }Cl \right) +{ \wedge }_{ m }^{ 0 }\left( NaOH \right) -{ \wedge }^{ 0 }(NaCl)\)

(c)

\({ \wedge }_{ m }^{ 0 }\left( { NH }_{ 4 }Cl \right) +{ \wedge }_{ m }^{ 0 }\left( NaCl \right) -{ \wedge }^{ 0 }(NaOH)\)

(d)

\({ \wedge }_{ m }^{ 0 }\left( NaOH \right) +{ \wedge }_{ m }^{ 0 }\left( NaCl \right) -{ \wedge }^{ 0 }({ NH }_{ 4 }Cl)\)

28.

Which of the statements about solutions of electrolytes is not correct ?

(a)

Conductivity of solution depends upon size of ions

(b)

Conductivity depends upon viscosity of solution

(c)

Conductivity does not depend upon solvation of ions present in solution

(d)

Conductivity of solution increases with temperature.

29.

Which has the highest oxidizing power ?

(a)

I₂

(b)

Br₂

(c)

F₂

(d)

Cl₂

30.

Electrode potential of any electrode depends on :

(a)

nature of the metal

(b)

temperature of the solution

(c)

molarity of the solution

(d)

all of these

Generic 5

4 x 5 = 20

31.

(i) What happens when
(a) phenol reacts with bromine water?
(b) ethanol reacts with CH₃COCl/pyridine?
(c) anisole reacts with HI?
Write the chemical equations involved in the above reactions.
(ii) Distinguish between
(a) Ethanol and phenol
(b) Propan-2-ol and 2-methylpropan-2-o1

32.

(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday's is required for the reduction of 1 mole of Cu²⁺ to Cu.
(b) Calculate emf of the following cell at 298 K.
Mg(s) IMg²⁺(0.1 M) II Cu²⁺(0.01) I Cu(s)
Given, E⁰_cell = +2.71 V,
1F = 96500 C mol^-1

33.

(a) Give the preparation of potassium dichromate from chromate ore:
(b) Explain the following:
(i) Transition metals have good tendency to form complexes.
(ii) Transition metals exhibit variable oxidation states.
(c) Write the general electronic configuration of lanthanoids.

34.

EMF of Daniell cell was found using different concentrations of Zn²⁺ ion and Cu² ion. A graph was then plotted between E_cell and \(\log { \frac { \left[ { Zn }^{ 2+ } \right] }{ \left[ { Cu }^{ 2+ } \right] } } \) . The plot was found to be linear with intercept on E_cell axis equal to 1.10 V. Calculate E_cell for Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu.

Case Study Questions

2 x 4 = 8

35.

An organic compound (A) having molecular formula C6H6O gives a characteristic color with an aqueous FeCl₃ solution. (A) on treatment with CO₂ and NaOH at 400 K under pressure gives (B), which on acidification gives a compound (C). The compound (C) reacts with acetyl chloride to give (D) which is a popular pain killer.
(i) Compound (A) is
(a) 2-hexanol
(b) dimethyl ether
(c) phenol
(d) 2-methyl pentanol.
ii) Number of carbon atoms in compound (D) is
(a) 7
(b) 6
(c) 8
(d) 9
iii) The conversion of compound (A) to (C) is known as
(a) Reimer- Tiemann reaction
(b) Kolbe’s reaction
(c) Schmidt reaction
(d) Swarts reaction
(iv) Compound (A) on heating with compound (C) in presence of POCl₃ gives a compound (D) which is used
(a) in perfumery as a flavoring agent
(b) as an antipyretic
(c) as an analgesic
(d) as an intestinal antiseptic.

36.

Read the passage given below and answer the following questions :
All chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules are present in a few gram of any chemical compound varying with their atomic/molecular masses. To handle such large number conveniently, the mole concept was introduced. All electrochemical cell reactions are also based on mole concept. For example, a 4.0 molar aqueous solution of NaCI is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept.
The following questions are multiple choice questions. Choose the most appropriate answer :
(i) The total number of moles of chlorine gas evolved is

(a) 0.5

(b) 1.0

(c) 1.5

(d) 1.9

(ii) If cathode is a Hg electrode, then the maximum weight of amalgam formed from this solution is

(a) 300 g

(b) 446 g

(c) 396 g

(d) 296 g

(iii) In the electrolysis, the number of moles of electrons involved are

(a) 2

(b) 1

(c) 3

(d) 4

(iv) In electrolysis of aqueous NaCl solution when Pt electrode is taken, then which gas is liberated at cathode?

(a) H₂gas

(b) C₂ gas

(c) O₂ gas

(d) None of these

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Answers Key:

Generic 2

7 x 2 = 14

(a) The Kohlrausch law of independent migration of ions state that "limiting molar conductivity of an electrode is the sum of the individual contribution of the cation and the anion of the electrolyte."
(b) The Faraday's first law states that "the amount of chemical reaction occuring at any electrode by passing current is directly proportional to the quantity of electricity passed through the electrolyte.

methyl iodide and phenol

The use of 3d electron for formation of bond increases from Sc to Mn, causing the increase in oxidation state up to +7. The reason for Mn having highest oxidation state of +7 is due to the presence of 7 unpaired electrons in its atom. As the number of unpaired electrons decreases from Fe to Cu so there is the decrease in oxidation state.

\({ \Delta G }^{ \circ }=-n{ F{ E }^{ \circ } }_{ cell }\)
\(=-2\times 1.1\times 96500CV{ mol }^{ -1 }\)
\(=-21227J \ { mol }^{ -1 }\)
\(or \ =-21227 \ kJ \ { mol }^{ -1 }\)

Conductivity of an electrolyte solution decreases with dilution because the number of ions per unit volume furnished by an electrolyte decreases with dilution.

Pt(IV) is more stable than Ni(IV) because energy required to remove four electrons is less in case of Pt than in Ni.

Generic 3

7 x 3 = 21

(i) When anisole is treated with CH₃Cl in presence of anhydrous AlCl₃, a mixture of ortho and para methyl anisole is produced.
Anisole undergoes Friedel-Crafts alkylation reaction with an alkyl halide in presence of Lewis acid AlCl₃ as a catalyst.

(ii) When phenol is oxidized with Na₂Cr₂O₇ in presence of H⁺(acid), a conjugated diketone (benzoquinone) is produced.
In the presence of air, phenols are slowly oxidized to dark-colored mixtures containing quinones.

(iii)

10.

(i) Electronic configuration
Lanthanoids = [Xe] 4f^0-14 5d^0-1 6s²
Actinoids = [Rn] 5f^0-14 6d^0-1 7s²
(ii) Oxidation states In lanthanoids, +3 oxidation state is most common along with + 2 and + 4. While in actinoids, there is a greater range of oxidation states because 5f, 6d and 7s levels are of comparable energies.They show + 2, + 3, + 4, + 5, + 6 and + 7 oxidation states. Common oxidation state in actinoids is + 3.
(iii) Chemical reactivity Lanthanoids are less reactive than actinoids. Actually, earlier members of lanthanoids are quite reactive similar to calcium but with increasing atomic number, they behave more like aluminium. Lanthanoids react with dilute acids to liberate H₂ gas while actinoids react with boiling water and gives a mixture of oxide and hydride.

11.

(i) Sum of enthalphy of atomisation and ionisation is greater than enthalphy of hydration. Thus, \(E_{(M^{2+}/Mn)}^0\) is positive for copper.
(ii) Ionisation enthalpy is lower than hydration enthalphy due to stable 3d⁵ configuration. Thus, \(E_{(M^{2+}/Mn)}^0\) is more negative.
(iii) Cr²⁺ is a stronger reducing agent because it can lose electron to form Cr³⁺which has stable 3d³configuration (half-filled t_2g³). Further is more negative than Fe²⁺

12.

(i) Reimer-Tiemann reaction

(ii) Williamson syntheses:
\(\underset { Alkyl halide\ Sodium\ alkoxide }{ R-X+\overset { + }{ Na } -\overset { - }{ O } -{ R }' } \longrightarrow R-O-{ R }'+NaX\)
\(\underset { Alkyl\ halide\ Sodium\ alkoxide }{ { CH }_{ 3 }-{ CH }_{ 2 }-Br } +\underset { sodium\ ethoxide }{ \overset { + }{ Na } -\overset { - }{ O } -{ CH }_{ 2 }-{ CH }_{ 3 } } \longrightarrow \underset { Diethyl\ ether }{ { CH }_{ 3 }-{ CH }_{ 2 }-O-{ CH }_{ 2 }-{ CH }_{ 3 } } +NaBr\)

13.

The cell constant is given by the equation:
Cell constant = G* = conductivity × resistance
= 1.29 S/m × 100 \(\Omega\) = 129 m^–1 = 1.29 cm^–1
= \(\frac{G^{*}}{R}=\frac{129 \mathrm{~m}^{-1}}{520 \Omega}\) = 0.248 S m^–1
Concentration = 0.02 mol L^–1
= 1000 × 0.02 mol m–3 = 20 mol m^–3
Molar conductivity = \(A_{m}=\frac{\kappa}{c}\)
= \(\frac{248 \times 10^{-3} \mathrm{Sm}^{-1}}{20 \mathrm{~mol} \mathrm{~m}^{-3}}\) = 124 × 10^–4 S m²mol^–1
Alternatively, \(\kappa=\frac{1.29 \mathrm{~cm}^{-1}}{520 \Omega}\) = 0.248 × 10^–2 S cm^–1
and Λ _m = κ × 1000 cm3 L^–1 molarity^–1
= \(\frac{0.248 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1} \times 1000 \mathrm{~cm}^{3} \mathrm{~L}^{-1}}{0.02 \mathrm{~mol} \mathrm{~L}^{-1}}\)
= 124 S cm² mol^–1

14.

\(t=50\times 60=3000s\)
\(I=0.1287A\)
\((a) \ { Cu }^{ 2+ }+{ 2e }^{ - }\longrightarrow Cu(s) \ At \ cathode\)
\((b)(i) Q=I\times t=0.1287\times 3000=386.1C\)
\((ii) \ m=Z\times I\times t=\frac { 63.5 }{ 2\times 96500 } \times 386.1=0.127g\)

Multiple Choice Question

16 x 1 = 16

15.

16.

(b)
Benzoic acid

17.

(a)
Br₂/water

18.

(b)
Benzoic acid

19.

(b)
Riemer-Tiemann reaction

20.

(a)
3-methylphenol

21.

(a)
They have vacant d orbitals

22.

(a)
the 5f orbitals extend farther from the nucleus than the 4f orbitals

23.

(b)
Ni²⁺

24.

(b)
They are chemically reactive

25.

(a)
Neutral FeCl₃

26.

(a)
4 A

27.

(b)
\({ \wedge }_{ m }^{ 0 }\left( { NH }_{ 4 }Cl \right) +{ \wedge }_{ m }^{ 0 }\left( NaOH \right) -{ \wedge }^{ 0 }(NaCl)\)

28.

29.

30.

(d)
all of these

Generic 5

4 x 5 = 20

31.

(ii) (a) Phenol gives violet colouration with FeCl₃, solution but ethanol does not.

(b) Propan-2-ol when warmed with l₂, in NaOHg ives yellow precipitate of iodoform, while 2-methylpropan-2-ol does not respond to this test.

32.

(a) Faraday's first law of electrolysis It states that the amount of chemical reaction occurring at an electrode by passing current is proportional to the quantity of electricity passed through the electrolyte. Charge required for the reduction of 1 mole of Cu²⁺ to Cu = 2F
(b) \(Cu^{ 2+ }Mg\rightarrow Mg^{ 2+ }+Cu\)
Given, E^_O _Cell= +271 V
By using Nernst equation,
\(E_{ cell }=E^{ 0 }_{ cell }-\frac { 0.059 }{ n } log\frac { Mg^{ 2+ } }{ Cu^{ 2+ } } \)
Here, n = 2 and E⁰ _cell = + 2.71 V
\(\therefore E_{ cell }=2.71-\frac { 0.059 }{ 2 } log\frac { 0.1 }{ 0.01 } \)
\(=2.71-\frac { 0.059 }{ 2 } log10\)
= 2.71- 0.0295
\([\therefore \) log₁₀ = 1]
= 2.68 V

33.

(i) The transition elements exhibit variable oxidation states. The variable oxidation states of transition metals are due to the participation of ns and (n - 1) d-electrons. This is because of the very small difference between the energies of (n - 1) d and ns orbitals. For the first five elements, the minimum oxidation state is equal to the number of electrons ---in the 4s orbitals and the other oxidation states are equal to the sum of 4s and some of the 3d-electrons. The highest oxidation state is equal to the sum of 4s and 3d electrons. For the remaining elements, the minimum oxidation state is equal to electrons in 4s-orbitals and the maximum oxidation state is not equal to the sum of 4s and 3d electrons. In general, the oxidation state increases up to the middle and then decreases.

34.

For Daniell cell, Zn + Cu²⁺⇾ Zn²⁺+ Cu
\(E_{cell}=E^0_{cell}-{0.0591\over 2}log{[Zn^{2+}]\over [Cu^{2+}]}\)
It is the equation of straight line (y = C + mx).
Intercept = E^O_cell= 1·10 V (Given)
\(E_{cell}\equiv 1.10-{0.0591\over 2}log{0.1\over 0.01}=1.10-0.0295=1.0705V\)

Case Study Questions

2 x 4 = 8

35.

i. (c) phenol
ii. (d) 9
iii.(b) Kolbe’s reaction
iv.(d) as an intestinal antiseptic.

36.

(i) (b) : \(n_{\mathrm{NaCl}}=\frac{4 \times 500}{1000}=2 \mathrm{~mol}\)
\(\therefore\) \(n_{\mathrm{Cl}_{2}}=1 \mathrm{~mol}\)
(ii) (b) : n_Nadeposited = 2 mol
\(\therefore\) n_Na_ _Hgformed = 2 mol
\(\therefore\) Mass of amalgam formed = 2 x 223 = 446 g
(iii) (a)
(iv) (a)