Question Set D

cbse 12th Standard Chemistry Electrochemistry Important Questions And Answers Study Material - QB365

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Questions:

Case Study Questions

15 x 4 = 60

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.

A compound (X) containing C, H and O is unreactive towards sodium. It also does not react with Schiff’s reagent. On refluxing with an excess of hydroiodic acid, (X) yields only one organic product (Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorous and iodine. The compound (Z) on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60.
The following questions are multiple choice question. Choose the most appropriate answer:
(i) The compound (X) is an
(a) acid
(b) aldehyde
(c) alcohol
(d) ether
(ii) The IUPAC name of the acid formed is
(a) methanoic acid
(b) ethanoic acid
(c) propanoic acid
(d) butanoic acid
iii) Compound (Y) is
(a) ethyl iodide
(b) methyl iodide
(c) propyl iodide
(d) mixture of (a) and (b)
or
Compound (Z) is
(a) methanol
(b) ethanol
(c) propanol
(d) butanol
(iv) Compound (X) on treatment with excess of Cl₂ in presence of light gives
(a) α-chlorodiethyl ether
(b) α, α’-dichlorodiethyl ether
(c) perchlorodiethyl ether
(d) none of these

Table given below has compounds and their pKa values. Study the table and answer the questions based on table and related studied concepts.

Compound	pKa
Ethanol	15.9
Phenol	9.98
o-cresol	10.28
p-cresol	10.14
m-cresol	10.08
o-nitro phenol	7.23
p-nitro phenol	7.15
m-nitro phenol	8.40
2, 4-dinitro phenol	4.0
Picric acid	0.71
m-methoxy phenol	9.65
o-methoxy phenol	9.96
p-methoxy phenol	10.21
m-amino phenol	9.87

(a) Which phenolic compound is most acidic?
(b) What is relationship between pKa and acidic character?
(c) Why are cresols weaker acids than phenol?
(d) Which has more electron withdrawing effect (-I)-I effect -OCH₃ or -NH₂? Why?
(e) Why is o-fluoro phenol weakest acid than pand m-fluoro phenol?

Observe the following table showing boiling points of alcohol, molar mass. Study the table and answer the questions based on table and related studied concept.

Alcohol	Boiling Point	Molar Mass
CH₃OH	64°C	32 g mol^-1
C₂H₅OH	78°C	46 g mol^-1
C₃H₇OH (n-propyl alcohol)	97°C	60 g mol^-1
Isopropyl alcohol	82.5^oC	60 g mol^-1
n-butanol	118°C	74 g mol^-1
Isobutyl alcohol	108°C	74 g mol^-1
Butan-2-ol	100°C	74 g mol^-1
Tert. butyl alcohol	83°C	74 g mol^-1

(a) Why do alcohols have higher boiling points than haloalkanes, ethers, aldehydes and ketones?
(b) Why does tertiary butyl alcohol have lower boiling point than n-butyl alcohol?
(c) How does boiling point vary with increase in carbon chain?
(d) How is solubility of alcohol vary with increase in molar mass?
(e) Which alcohol is most acidic and why?

Alcohols play very important role in our daily life. Ordinary sprit used as an antiseptic contains methanol. Ethanol is present in cough syrups, tonics, wine, beer and whisky, Sugar, starch, cellulose are carbohydrates which also contain large number -OH groups. Phenol is also an antiseptic in low concentration (0.2%) where as 2% solution of phenol is used as disinfectant. The fragrance of rose is due to citronellol (unsaturated alcohol). Phenol is used for preparation of many useful compounds like aspirin, methyl salicylate (Iodex) and phenyl salicylate (salol) used as intestinal antiseptic.
(a) How is phenol prepared from cumene? What is advantage of this method?
(b) How is phenol converted into salicylic acid?
(c) Convert phenol to picric acid
(d) Distinguish between phenol and benzyl alcohol?
(e) Why does phenol turn pink after long standing?

Observe the table given below belonging to 3d series, their first, second, third ionisation enthalpy and \(\mathbf{E}_{\mathbf{M}^{2+} / \mathbf{M}}^{\circ}\) and \(\mathbf{E}_{\mathbf{M}^{3+} / \mathbf{M}^{2+}}^{\circ}\) and answer the questions that follow based on table and related concepts.

Element	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
1^stionisation enthalpy	631	656	650	652	717	762	758	736	745	905 kJ ml ^-1
II^nd ionisation enthalpy	1245	1320	1376	1635	1513	1564	1648	1757	1962	1736
III^rd ionisation enthalpy	2451	2721	2874	2995	3258	2964	3238	3401	3561	3839
\(\mathbf{E}_{\mathbf{M}^{2+} / \mathbf{M}}^{\circ}\) in volts	-	-1.63V	-1.18V	-0.91V	-1.18V	-0.44V	-0.28V	-0.25V	+0.34V	-0.76V
\(\mathbf{E}_{\mathbf{M}^{3+} / \mathbf{M}^{2+}}^{\circ}\)in volts	-	-0.37V	-0.26V	-0.41V	+1.57V	+0.77V	+1.97V	-	-	-

(a) Why does zinc have highest first ionisation enthalpy?
(b) Why is 3 rd ionisation enthalpy of Mn high?
(c) Why is Cr^{3 +} more stable than Cr²⁺?
(d) Why is \(\mathbf{E}_{\mathbf{M n}^{2+} / \mathbf{M n}}^{\circ}-\mathbf{1 . 1 8 V} ?\)
(e) Why is \(\mathbf{E}_{\mathbf{C n}^{2+} / \mathrm{Cn}}^{\circ}+\mathbf{0 . 3 4 V} ?\)
(f) Why is Fe³⁺ more stable than Fe²⁺?
(g) Why is Mn ³ ⁺good oxidising agent and \(\mathbf{E}_{\mathbf{M n}^{3+} / \mathrm{Mn}^{2+}}^{\circ}=\mathbf{1 . 5 7 V} ?\)

Observe the graph shown in figure between Am (molar conductivity) Vs \(\sqrt{\mathrm{C}}\) (Molar concentration) and answer the questions based on graph.

(a) The curve 'V' is for KCI or CH₃COOH?
(b) What is intercept on \(\Lambda\)_m axis for 'X' equal to?
(c) Give mathematical equation representing straight line.
(d) What is slope equal to?
(e) What happens to molar conductivity on dilution in case of weak electrolyte and why?

Read the passage given below and answer the following questions:
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M_(s) | M⁺(aq.; 0.05 molar) || M⁺(aq; 1 molar) |M_(s).
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) For the above cell,

(a) \(E_{\text {cell }}<0 ; \Delta G>0\)

(b) \(E_{\text {cell }}>0 ; \Delta G<0\)

(c) \(E_{\text {cell }}<0 ; \Delta G^{\circ}>0\)

(d) \(E_{\text {cell }}>0 ; \Delta G^{\circ}<0\)

(ii) The value of equilibrium constant for a feasible cell reaction is

(a) < 1

(b) = 1

(c) > 1

(d) zero

(iii) What is the emf ofthe cell when the cell reaction attains equilibrium?

(a) 1

(b) 0

(c) > 1

(d) < 1

(iv) The potential of an electrode change with change in

(a) concentration of ions in solution	(b) position of electrodes
(c) voltage of the cell	(d) all of these

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

10.

Read the passage given below and answer the following questions:
Standard electrode potentials are used for various processes:
(i) It is used to measure relative strengths of various oxidants and reductants.
(ii) It is used to calculate standard cell potential.
(iii) It is used to predict possible reactions.
A set of half-reactions (in acidic medium) along with their standard reduction potential, E^o (in volt) values are given below
\(\mathrm{I}_{2}+2 e^{-} \rightarrow 2 \mathrm{I}^{-} ; \quad E^{\circ}=0.54 \mathrm{~V}\)
\(\mathrm{Cl}_{2}+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-} ; \quad E^{\circ}=1.36 \mathrm{~V}\)
\(\mathrm{Mn}^{3+}+e^{-} \rightarrow \mathrm{Mn}^{2+} ; \quad E^{\circ}=1.50 \mathrm{~V}\)
\(\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} ; \quad E^{\circ}=0.77 \mathrm{~V}\)
\(\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} ; E^{\circ}=1.23 \mathrm{~V}\)
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Which of the following statements is correct?

(a) CI^- is oxidised by O₂	(b) Fe²⁺is oxidised by iodine
(c) I^- is oxidised by chlorine.	(d) Mn²⁺ is oxidised by chlorine

(ii) Mn³⁺ is not stable in acidic medium, while Fe³⁺is stable because

(a) O₂ oxidises Mn²⁺ to Mn³⁺

(b) O₂ oxidises both Mn²⁺ to Mn³⁺ and Fe²⁺to Fe³⁺

(c) Fe^3-oxidises H₂O to O₂

(d) Mn³⁺ oxidises H₂O to O₂

(iii) The strongest reducing agent in the aqueous solution is

(a) I^-

(b) Cl^-

(c) Mn2⁺

(d) Fe²⁺

(iv) The emf for the following reaction is
\(\mathrm{I}_{2}+\mathrm{KCl} \rightleftharpoons 2 \mathrm{KI}+\mathrm{Cl}_{2}\)

(a) -0.82 V

(b) +0.82 V

(c) -0.73 V

(d) +0.73 V

11.

Read the passage given below and answer the following questions:
Transition metal oxides are compounds formed by the reaction of metals with oxygen at high temperature. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic V₂O₃ to less basic V₂O₄ and to amphoteric V₂O_5・V₂O₄ dissolves in acids to give VO²⁺ salts. Transition metal oxides are commonly utilized for their catalytic activity and semiconductive properties. Transition metal oxides are also frequently used as pigments in paints and plastic. Most notably titatnium dioxide. One of the earliest application of transition metal oxides to chemical industry involved the use of vanadium oxide for catalytic oxidation of sulfur dioxide to sulphuric acid. Since then, many other applications have emerged, which include benzene oxidation to maleic anhydride on vandium oxides; cyclohexane oxidation to adipic acid on cobalt oxides. An important property of the catalyst material used in these processes is the ability of transition metals to change their oxidation state under a given chemical potential of reductants and oxidants.
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Which oxide of vanadium is most likely to be basic and ionic ?

(a) VO

(b) V₂O₃

(c) VO₂

(d) V₂O₅

(ii) Vanadyl ion is

(a) VO²⁺

(b) VO₂⁺

(c) V₂O⁺

(d) VO₄^3-

(iii) The oxidation state of vanadium in V₂O₅ is

(a) +5/2

(b) +7

(c) +5

(d) +6

(iv) Identify the oxidising agent in the following reaction.

(a) V₂O₅

(b)Ca

(c) V

(d) None of these

12.

Read the passage given below and answer the following questions:
Molar conductivity of ions are given as product of charge on ions to their ionic mobilities and Faraday's constant.
\(\lambda_{A^{n+}}=n \mu_{A^{n+}} F\) (here \(\mu\) is the ionic mobility of Aⁿ⁺).
For electrolytes say A_XB_y, molar conductivity is given by
\(\lambda_{m\left(A_{x} B_{y}\right)}=x_{n} \mu_{A^{n+}} F+y_{m} \lambda_{A^{m}-F}\)

Ions	Ionic mobility
K⁺	7.616 x 10^{- 4}
Ca²⁺	12.33 x 10^-4
Br^-	8.09 x 10^{- 4}
\(\mathrm{SO}_{4}^{2-}\)	16.58 x 10^{- 4}

The following questions are multiple choice questions. Choose the most appropriate answer
(i) At infinite dilution, the equivalent conductance of CaSO₄ is

(a) 256 x 10^-4

(b) 279

(c) 23.7

(d) 2.0 x 10^{- 8}

(ii) If the degree of dissociation of CaSO₄ solution is 10% then equivalent conductance of CaSO₄ is

(a) 3.59

(b) 36.9

(c) 27.9

(d) 30.6

(iii) What is the unit of equivalent conductivity?

(a) ohm^-1 cm² eq^-1	(b) ohm cm²eq^-1
(c) ohm^-1 cm eq^-1	(d) ohm cm² eq^-1

(iv) If the molar conductance value of Ca²⁺ and Cl^-at infinite dilution are 118.88 x 10^-4m² mho mol^-1 and 77.33 x 10^-4 m²mho mol^-1respectively then the molar conductance of CaCl₂ (in m² mho mol^-1) will be

(a) 120.18 x 10^{- 4}

(b) 135 x 10^-4

(c) 273.54 x 10^-4

(d) 192.1 x 10^-4

13.

Read the passage given below and answer the following questions:
The unique behaviour of Cu, having a positive E^oaccounts for its inability to liberate H₂ from acids. Only oxidising acids (nitric and hot concentrated sulphuric acid) react with Cu, the acids being reduced. The stability of the half-filled (d⁵) subshell in Mn²⁺ and the completely filled (d¹⁰ ) configuration in Zn²⁺ are related to their E^o (M³⁺ /M²⁺) values. The low value for Sc reflects the stability of Sc³⁺ which has a noble gas configuration. The comparatively high value for Mn shows that Mn²⁺(d⁵) is particularly stable, whereas a comparatively low value for Fe shows the extra stability of Fe³⁺ (d⁵). The comparatively low value for V is related to the stability of V²⁺ (half-filled t_2g level).
The following questions are multiple choice questions.Choose the most appropriate answer :
(i) Standard reduction electrode potential of Zn²⁺ /Zn is - 0.76 V. This means

(a) ZnO cannot be reduced to Zn by H₂ under standard conditions

(b) Zn cannot liberate H₂ with concentrated acids

(c) Zn is generally the anode in an electrochemical cell

(d) Zn is generally the cathode in an electrochemical cell.

(ii) E^ovalues for the couples Cr³⁺/Cr²⁺ and Mn³⁺ /Mn²⁺ are -0.41 and +1.51 volts respectively. These values suggest that

(a) Cr²⁺ acts as a reducing agent whereas Mn3+ acts as an oxidizing agent

(b) Cr²⁺ is more stable than Cr³⁺ state

(c) Mn³⁺ is more stable than Mn²⁺

(d) Cr²⁺ acts as an oxidizing agent whereas Mn³⁺ acts as a reducing agent

(iii) The reduction potential values of M, Nand O are +2.46, -1.13 and -3.13 Y respectively. Which of the following order is correct regarding their reducing property?

(a) O>N>M

(b) O>M>N

(c) M>N>O

(d) M>O>N

(iv) Which of the following statements are true?
(i) Mn²⁺ compounds are more stable than Fe²⁺ towards oxidation to +3 state.
(ii) Titanium and copper both in the first series of transition metals exhibits +1 oxidation state most frequently.
(iii) Cu⁺ ion is stable in aqueous solutions.
(iv) The E⁰ value for the Mn³⁺ /Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺.

(a) (ii) and (iii)

(b) (i) and (iv)

(c) (i) and (iii)

(d) (ii) and (iv)

14.

Read the passage given below and answer the following questions:
The transition elements have incompletely filled d-subshells in their ground state or in any of their oxidation states. The transition elements occupy position in between s- and p-blocks in groups 3-12 of the Periodic table. Starting from fourth period, transition elements consists of four complete series : Sc to Zn, Y to Cd and La, Hf to Hg and Ac, Rf to Cn. In general, the electronic configuration of outer orbitals of these elements is (n - 1) d^1-10 ns^1-2. The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n - 1)d¹⁰ n². All the transition elements have typical metallic properties such as high tensile strength, ductility, malleability. Except mercury, which is liquid at room temperature, other transition elements have typical metallic structures. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. Transition metal also forms alloys. An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.
The following questions are multiple choice questions. Choose the most appropriate answer :
(i) Which of the following characteristics of transition metals is associated with higher catalytic activity?

(a) High enthalpy of atomisation	(b) Variable oxidation states
(c) Paramagnetic behaviour	(d) Colour of hydrated ions

(ii) Transition elements form alloys easily because they have

(a) same atomic number	(b) same electronic configuration
(c) nearly same atomic size	(d) same oxidation states.

(iii) The electronic configuration of tantalum (Ta) is

(a) \([\mathrm{Xe}] 4 f^{0} 5 d^{1} 6 s^{2}\)	(b) \([\mathrm{Xe}) 4 f^{14} 5 d^{2} 6 s^{2}\)
(c) \([\mathrm{Xe}] 4 f^{14} 5 d^{3} 6 s^{2}\)	(d) \(\left[\mathrm{Xe}\left]4 f^{14} 5 d^{4} 6 s^{2}\right.\right.\)

(iv) Which one of the following outer orbital configurations may exhibit the largest number of oxidation states?

(a) 3d⁵4s¹

(b) 3d⁵4s²

(c) 3d²4s²

(d) 3d³4s²

15.

Read the passage given below and answer the following questions:
The f-block elements are those in which the differentiating electron enters the (n -2) forbital. There are two series of f-block elements corresponding to filling of 4f and 5f-orbitals. The series of 4f- orbitals is called lanthanides. Lanthanides show different oxidation states depending upon stability of f⁰, f⁷ and f¹⁴ configurations, though the most common oxidation states is +3. There is a regular decrease in size of lanthanides ions with increase in atomic number which is known as lanthanide contraction.
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) The atomic numbers of three lanthanide elements X, Y and 2 are 65, 68 and 70 respectively, their Ln³⁺ electronic configuration is

(a) 4f⁸, 4f¹¹, 4f¹³

(b) 4f¹¹, 4f⁸ , 4f¹³

(c) 4f^o, 4f², 4f¹¹

(d) 4f³, 4f⁷, 4f⁹

(ii) Lanthanide contraction is observed in

(a) Gd

(b) At

(c) Xe

(d) Te

(iii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

(a) Cerium (Z = 58)

(b) Europium (Z = 63)

(c) Lanthanum (Z = 57)

(d) Gadolinium (Z = 64)

(iv) Identify the incorrect statement among the following.

(a) Lanthanojd contraction is the accumulation of successive shrinkages.

(b) The different radii of Zr and Hf due to consequence of the lanthanoid contraction.

(c) Shielding power of 4f electrons is quite weak.

(d) There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.

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Answers Key:

Case Study Questions

15 x 4 = 60

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

(i) d
(ii) b
(iii) a or b
(iv) c

(a) Picric acid
(b) Lower the pKa, more will be acidic character.
(c) It is because -CH₃ groups are electron releasing, destabilise phenoxide ion.
(d) -OCH₃ has more -I effect than -NH₂ because oxygen is more electronegative than nitrogen.
(e) It is because in a-fluoro phenol, there is strong intramolecular H-bonding.

(a) It is due to inter molecular H-bonding.
(b) It is because it has spherical shape, least surface area, less van der Waals' forces of attraction, hence lower boiling point than n-butyl alcohol.
(c) Boiling point increases with increase in carbon chain
(d) Solubility of alcohol decreases with increase in molar mass.
(e) CH₃OH is most acidic because \(\mathrm{CH}_{3} \mathrm{O}^{\ominus}\) is most stable among \(\mathrm{RO}^{\ominus}\) .

(a)

Acetone is obtained which is useful by product.
(b)

(c)

(d) Add neutral FeCI₃ . Phenol gives violet colour where as benzyl alcohol does not.
(e) It is due to oxidation.

(a) It is because Zn has completely filled '4s' orbital which is stable.
(b) It is because after losing 2 electrons, it becomes 3d⁵ which is more stable.
(c) It is because \(\mathrm{Cr}^{3+}\left(t_{2 g}^{3}\right)\) half filled t_2g orbitals are more stable than Cr²⁺(3d⁴).
(d) It is due to low enthalpy of atomisation, sublimation enthalpy, ionisation enthalpy.
(e) It is due to high ionisation enthalpy and low hydration enthalpy.
(f) Fe³⁺(3d⁵) is half filled which is more stable than Fe²⁺(3d⁶).
(g) It is because it can gain one electron easily to from Mn²⁺(3d⁵) which is more stable that is why\(\mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}}^{\circ}=1.57 \mathrm{~V}\)

(a) It is for CH₃COOH.
(b) It is equal \(\Lambda\)° (limiting molar conductivity).
(c) \(\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-\mathrm{A} \sqrt{\mathrm{C}}\)
(d) Slope = -A
(e) \(\Lambda\)m for weak electrolyte increases sharply on dilution because both number of ions as well as mobility of ions increases.

(i) (b) : \(\begin{array}{l} M \longrightarrow M^{+}+e^{-} \\ (1 \cdot M)(0.05 M) \end{array}\)
For concentration cell, \(E_{\text {cell }}=-\frac{0.059}{1} \log \frac{0.05}{1}\)
\(E_{\text {cell }}=-\frac{0.059}{1} \log \left(5 \times 10^{-2}\right)\)
\(E_{\text {cell }}=-\frac{0.059}{1}[(-2)+\log 5]-0.059(-2+0.698)\)
= -0.059(-1.302) = 0.0768
\(\Delta G=-n F E_{\text {cell }}\)
If E_cellis positive, \(\Delta G\) is negative.
(ii) (c) : \(K=\operatorname{antilog}\left(\frac{n E^{\circ}}{0.0591}\right)\)
For feasible cell, E^o is positive, hence from the above equation K > 1 for a feasible cell reaction.
(iii) (b)
(iv) (a)

(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)

10.

(i) (c) : The half cell having the higher reduction potential will undergo reduction process.
(ii) (d) : Electrode potential of Mn³⁺ is higher than O₂.
(iii) (a) : Due to least electrode potential value.
(iv) (a) : Half reactions :

\(\mathrm{I}_{2}+2 e^{-} \rightarrow 2 \mathrm{I}^{-}\)	Reduction E^o = 0.54 V
\(2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-}\)	Oxidation E^o = -1.36 V
	-------------------------------
	e.m.f = -0.82 V

11.

(i) (a): Oxide of V in lowest oxidation state, i.e., VO is basic and ionic in character.
(ii) (a): Vanadyl ion is VO²⁺ where V is in +4 oxidation state.
(iii) (c)
(iv) (a)

12.

(i) (b) : Equivalent conductance of CaSO₄ :
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=\lambda_{\mathrm{Ca}^{2+}}^{\infty}+\lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}\)
\(\lambda_{\mathrm{Ca}^{2+}}^{\infty}=\left(\mu_{\mathrm{Ca}^{2+}}\right) F ; \lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}=\left(\mu_{\mathrm{SO}_{4}^{2-}}\right) F\)
\(\mu_{\mathrm{Ca}^{2+}}\) and \(\mu_{\mathrm{SO}_{4}^{2-}}\) - are ionic mobilities.
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=F(12.33+16.58) \times 10^{-4}\)
= 96500 x 10^{- 4}x 28.91 = 279
(ii) (c) : \(\alpha=\frac{\Lambda_{C}}{\Lambda^{\infty}} \Rightarrow 0.1=\frac{\Lambda_{C}}{279} \Rightarrow \Lambda_{C}=27.9\)
(iii) (a)
(iv) (c) : \(\Lambda_{m\left(\mathrm{CaCl}_{2}\right)}^{\circ}=\lambda_{\mathrm{Ca}^{2+}}^{\circ}+2 \lambda_{\mathrm{Cl}^{-}}^{\circ}\)
= (118.88 x 10^{- 4} ) + 2(77.33 x 10^{- 4} )
= 273.54 x 10^-4 m² mho mol^-1

13.

(i) (a)
(ii) (a): Lesser and negative reduction potential indicates that Cr²⁺ is a reducing agent. Higher and positive reduction potential indicates that Mn³⁺ is a stronger oxidizing agent.
(iii) (a) : The electrode which has more reduction potential is a good oxidizing agent and has least reducing power.
(iv) (b): (i) It is because Mn²⁺ has 3d⁵ electronic configuration which has extra stability.
(ii) Not titanium but copper, because with + 1 oxidation state an extra stable configuration, 3d¹⁰ results.
(iii) It is not stable as it undergoes disproportionation;
2Cu⁺(aq) ➝Cu²⁺(aq) + Cu(s). The E^o value for this is favourable.
(iv) Much larger third ionisation energy ofMn (where the required change is d⁵ to d⁴) is mainly responsible for this.

14.

(i) (b): The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states to form complexes.
(ii) (c) : Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals.
(iii) (c)
(iv) (b): Greater the number of valence electrons, more will be the number of oxidation states exhibited by the element.

15.

(i) (a):Terbium (65), 4f⁸; Dysprosium (Dy), 4f⁹; Ytterbium (Yb), 4f¹³.
(ii) (a)
(iii) (a)
(iv) (b): The almost identical radii of Zr (160 pm) and Hf(159 pm), a consequence of lanthanoid contraction.