Given: l = I amp, t = 16 minute, V =?, 1F = 96500 c mol^-1
m = Z x I x t
mO₂ = \(\frac{16}{2 \times 96500} \times 1 \times 16 \times 60=\frac{15360}{193000}\) 
= 0.08 g 
Volume at STP = \(\frac{\text { Mass of } \mathrm{O}_{2}}{\text { Molar macs }} \times 22400\) 
= \(\frac{0.08}{32} \times 22400=\frac{1792}{32}\) = 56 mL.

Transition metals form a large variety of alloys. This is due to the fact that transition metals have almost similar sizes and therefore, the atoms of one metal can easily substitute the atoms of other metals in their lattices. When we cool the mixture solution of two or more transition metals, solid alloys are generally formed. For example, Mn dissolves in molten iron to form manganese steel. The alloys formed are hard, having high melting points. These are also resistant to corrosion in comparison to the parent metals. Tungsten is also used for preparing stainless steel.

1° > 2° > 3° is order reactivity towards Na and K because 1° alcohols are most acidic; whereas 3° alcohols are least acidic.

A primary cell is the cell which the redox reaction occurs only once and the cell becomes dead after some time. It cannot used again.

Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d5) configuration which has extra stability.

(i) The potential of a mercury cell is approximately 1.35 V. It remains constant as in the overall cell reaction no ions are involved in solution whose concentration can change during the life span of the cell. Thus, it lasts longer.
(ii) When DC current is used, it cause the polarization of electrodes and electrolyte which leads to error in measurement. In AC there is no polarization due to non-continuous flow. Therefore, direct current is not used in determination of electrolytic conductance.

(i) CH₃CH₂CH₂OCH₃ + HBr \(\rightarrow\) CH₃Br + CH₃CH₂CH₂OH

(iii) (CH₃)₃C - OC₂H₅ + HI \(\rightarrow\) (CH₃)₃C - I + C₂H₅OH

(i) It is because -OH group is electron releasing, activating, therefore, it undergoes electrophilic substitution more readily than benzene.
(ii) It is due to repulsion between lone pair of electrons with bonded pair of electrons.
(iii) It is because tert. carbocation is more stable which combines with 1- to form Tert. butyl iodide and methanol.

(a) Lanthanoids, mostly show +3 oxidation state but some of them show +2 and +4 oxidation states also due to the stability of electronic configuration (4f⁰, 4f ⁷ and 4f¹⁴ ).
(b) (i) They show variable oxidation states.
(ii) They form coloured ions.
(c) Zn may not be regarded as transition metal because neither Zn nor Zn²⁺ have incompletely filled d-orbital.

(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 3d³ configuration (half-filled t_2g³). Further is more negative than Fe²⁺

\(Ag\left( s \right) |{ Ag }^{ + }({ 10 }^{ -3 }M)||{ Cu }^{ 2+ }({ 10 }^{ -1 }M)|Cu(s)\)
The cell reaction is:
 \(Cu(s)+{ 2Ag }^{ + }(aq)\longrightarrow { Cu }^{ 2+ }(aq)+2Ag(s)\)
\(\ e={ { E }^{ ° } }_{ cell }-\frac { 0.059 }{ 2 } log\frac { \left[ { Cu }^{ 2+ } \right] }{ { \left[ Ag \right] }^{ 2 } } \)
\({ E }^{ ° }=0.46V\)
\(E=0.46-\frac { 0.059 }{ 2 } log\frac { \left( { 10 }^{ -1 } \right) }{ { \left( { 10 }^{ -3 } \right) }^{ 2 } } \)
\(=0.46-\frac { 0.059 }{ 2 } log{ 10 }^{ 5 }\)
\(=0.46-\frac { 0.059 }{ 2 } \times 5\)
\( =0.46-0.1475\)
\(=0.3125V\)

(i) Propane-1, 2, 3-triol
(ii) 2-Methylphenol
(iii) 4-Methylphenol.

\(Zn(s)+{ Cu }^{ 2+ }(aq)\rightleftharpoons { Zn }^{ 2+ }(aq)+Cu(s)\)
\(At \ anode \ Zn\longrightarrow { Zn }^{ 2+ }+{ 2e }^{ - }\)
\({ At \ cathode \ Cu }^{ 2+ }+{ 2e }^{ - }\longrightarrow Cu(s)\)
\( { { E }^{ ° } }_{ cell }=\left( { { E }^{ ° } }_{ { Cu }^{ 2+ }/Cu }-{ { E }^{ ° } }_{ { Zn }^{ 2+ }/Zn } \right) =+0.34V-(\_ 0.76V)=1.10V\)
\( { { E }^{ ° } }_{ cell }=\frac { 0.0591 }{ 2 } \ logK\)
\( logK=\frac { 2\times { { E }^{ ° } }_{ cell } }{ 0.0591 } =\frac { 2\times 1.10V }{ 0.0591 } =37.2250\)
\( K=Antilog \ 37.2250\)
\(K=1.679\times { 10 }^{ 37 }\)

In an electrolytic cell, electrons enter the cathode and leave at the anode. Thus, in the external supply, electrons flow from cathode to anode. In the solution, only ions flow and not the electrons.

(a) \(\text { (i) } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}: \mathrm{HBr} \stackrel{\text { reflux }}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}\)
                 1-Propanol                                         1-Bromopropane
\(\text { (ii) } \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Cl}+\mathrm{KOH}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{KCl}\)
      1-Chloropropane                                                        1-Propanol

(i)

(ii)

(iii)

(iv)

(a) E^o value of silver (E^o_Ag⁺ /Ag = +0.80V) is lower than Ag lAg that of gold, hence silver displaces gold
( E^o_Au3+ _/Au= + 1.40V), which gets deposited on the silver Object
E^o value of copper (E°_cu²⁺/cu = +0.34) is lower than that  of silver, hence silver cannot displace copper from its solution. 
(b) (i) Electrons flow from Zn to Ag plate.
(ii) Zn acts as anode and Ag acts as cathode
(iii) Cell will stop functioning 
(iv) Concentration of Zn²⁺ions will increase and that of Ag⁺ ions will decrease.
(v) No change 

(a) (i) Molar conductivity of a solution at a given concentration is the conducting power of all the ions produced by 1 mol of an electrolyte.
(ii) Secondary battery can be recharged by passing current through it in opposite direction so that it can be used again.
(iii) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.
(b) (i) Faraday's first law of electrolysis states that the amount of chemical reaction which occurs at any electrode during electrolysis by current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). 
(ii) According to Kohlrausch law of independent migration of ions limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte

(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.

A lead storage battery consists of anode of lead, cathode of a grid of lead packed with lead dioxide (PbO₂ ) and 38% solution of sulphuric acid as electrolyte. When the battery is in use, the following reactions take place
At anode:
\(Pb(s)+{ { So }_{ 4 } }^{ 2- }(aq)\longrightarrow { PbSO }_{ 4 }(s)+2{ e }^{ - }\)
At cathode
\({ PbO }_{ 2 }(s)+{ { So }_{ 4 } }^{ 2- }(aq)+{ 4H }^{ + }(aq)+2{ e }^{ - }\longrightarrow { PbSO }_{ 4 }(s)+2{ H }_{ 2 }O(l)\)
Overall reaction
\(Pb(s)+{ PbO }_{ 2 }(s)+2{ H }_{ 2 }{ SO }_{ 4 }(aq)\longrightarrow 2{ PbSO }_{ 4 }(s)+2{ H }_{ 2 }O(l)\)
On charging the battery, the reverse reaction takes place, i.e., PbSO₄ deposited on the electrodes is converted back into Pb and PbO₂ and H₂SO₄ is regenerated.
Reverse of reaction (i) will be reduction and hence will take place at cathode. Reverse of reaction (ii) c will be oxidation and hence will take place at anode.

(a) Both (A) and (R) are correct, (R) is the correct explanation of (A).

(b): The tendency to show H-bonding decreases with increasing hydrophobic character of carbon chain.

(d) o-nitrophenol is more volatile due to intramolecular hydrogen bonding, while p-nitrophenol is less volatile due to intermolecular hydrogen bonding which causes the association of molecules. Hence, (A) is incorrect but (R) is correct.

(a)Solid NaCI does not conduct electricity due to the absence of free ions. Thus, both (A)and (R)are correct and (R) is the correct explanation of (A).

(a) It is because neither they nor their ions have incompletely filled d-orbitals.
(b) Scandium (Sc) and Zinc (Zn).
(c) It is because they show variable oxidation state, can form intermediate complexes and have large surface area for adsorption of gases.
(d) It is due to strong interatomic forces of attraction due to presence of unpaired electrons.
(e) It is because Cu^{2 +} has one unpaired electron and undergoes d-d transition by absorbing light from visible region and radiate blue colour, where as Zn^{2 +} is colourless due to absence of unpaired electron.

(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.