Anodic Oxidation: (Reverse the given reaction)
(E^o_ox ) = 0.76 V Cathodic reduction
E^o_cell = (E^o_ox )+ (E^o_red) 
= 0.76 + 0.34 = 1.1 V

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
2.303 x 10^-2 hour^-1 = \(\frac { 2.303 }{ 1806 min } log\frac { \left[ { 0.25 }_{ } \right] }{ \left[ A \right] } \)
\(=\left(\frac{2.303 \times 10^{-2} hour^{-1 }\times 1806 min}{2.303}\right) = \log \left(\frac{0.25}{A}\right)  \)
\(=\left(\frac{ 1806 \times 10 ^{-2}}{60}\right) = \log \left(\frac{0.25}{A}\right)  \)
\(= 0.301 = \log \left(\frac{0.25}{A}\right)  \)
\(=\log2 = \log \left(\frac{0.25}{A}\right)  \)
\(2 = \log \left(\frac{0.25}{A}\right)  \)
\([A] = \log \left(\frac{0.25}{2}\right)  = 0.125 M\)

 

rate₁ = k[0.1]ⁿ [0.1]^m ....(1)
rate₂ = k[0.2]ⁿ [0.1]^m ....(2)
(2) (1)
\(=\frac{2x}{x} = \frac{k[0.2]^n [0.1]^m}{k[0.1]^n [0.1]^m}\)
\(= \frac{2x}{x} = 2^n\)
n = 1
rate₃ = k[0.1]ⁿ [0.2]^m ....(3)
rate₄ = k[0.2]ⁿ [0.2]^m ....(4)
\(=\frac{8x}{2x} = \frac{k[0.2]^n [0.2]^m}{k[0.2]^n [0.1]^m}\)
= 8/2 = 2m
m = 2
rate = k[A] [B]²

\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope.

Number of A ions = N_c/8 = 8/8 = 1
Number of B ions = N_f/2 = 6/2 = 3
Simplest formula = AB₃

a) Fe²⁺  b) Fe³⁺  c) Fe⁰

As we move from La to Lu, their metallic behaviour because almost similar to that of aluminium.

Silicates are classified into various types based on the way in which the tetrahedral units, [SiO₄]^4- are linked together.
(i) Ortho silicates (Neso silicates):
The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or nesosilicates.
Examples: Phenacite - Be₂SiO₄ (Be²⁺ ions are tetrahedrally surrounded by O^2- ions)
(ii) pyro silicate (or) Soro silicates: Silicates:
Which contain [Si₂O₇]^6- ions are called pyro silicates (or) Soro silicates.
Example: Thortveitite - Sc₂Si₂O₇
(iii) Cyclic silicates (or Ring silicates): 
Silicates which contain (SiO₃)₃^2n- ions which are formed by linking three or more tetrahedral SiO₄^4- units cyclically are called cyclic silicates.
Example: Beryl [Be₃Al₂ (SiO₃)₆] (an aluminosilicate with each aluminium is surrounded by 6 oxygen atoms octahedrally)
(iv) Inosilicates: Silicates which contain 'n':
number of silicate units liked by sharing two or more oxygen atoms are called inosilicates.
Example: They are further classified as chain silicates and double chain silicates.
(v) Chain silicates (or pyroxenes):
These silicates contain [(SiO₃)_n]^2n- ions formed: by linking 'n' number of tetrahedral [SiO₄]^4- units linearly. Each silicate unit shares two of its oxygen atoms with other units.
Example: Spodumene - LiAl(SiO₃)₂·
(vi) Double chain silicates (or amphiboles): 
These silicates contains \(\left[ { Si }_{ 4 }{ O }_{ 11 } \right] _{ n }^{ 6n- }\) ions. In these silicates there are two different types of tetrahedra:
(a) Those sharing 3 vertices
(b) those sharing only 2 vertices.
Example:
Asbestos: These are fibrous and non-combustible silicates.
(vii) Sheet or phyllo silicates: 
Silicates which contain \(({ Si }_{ 2 }{ O }_{ 5 })_{ n }^{ 2n- }\) are called sheet or phyllo silicates. In these, Each [SiO₄]^4- tetrahedron unit shares three oxygen atoms with others and thus by forming two dimensional sheets.
Example: Talc, Mica etc.
(viii) Three dimensional silicates (or tectosilicates): 
Silicates in which all the oxygen atoms of [SiO₄]^4- tetrahedra are shared with other tetrahedra to form three dimensional network are called three dimensional or tectosilicates.
Example: Quartz.

(i) Order of the reaction =1; \(\mathrm{t}_{1 / 2}=60 \mathrm{~s} ; \mathrm{k}=?\)
\(\mathrm{k}=\frac{0.6932}{\mathrm{t}_{\frac{1}{2}}} \)
\(=\frac{0.6932}{60}  \)
\(k =1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
(ii) \(\left[\mathrm{A}_{0}\right]=100 \% ; \mathrm{t}=180 \mathrm{~s} ;[\mathrm{A}]=? ; \mathrm{k}=1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
For first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]} \)
\(1.155 \times 10^{-2} =\frac{2.303}{180} \log \left(\frac{100}{[A]}\right)  \)
\(\frac{0.01155 \times 180}{2.303} =\log \left(\frac{100}{[A]}\right)  \)
\(0.9027 =\log 100-\log [\mathrm{A}]  \)
\(\log [\mathrm{A}] =\log 100-0.9027 \)
\(\log [A]=2-0.9027 \)
\(\log [A]=1.0972 \)
[A] = antilog  of (1.0972)
[A] =12.51 %

1. The crude metal is refined by electrolysis. It is carried out in an electrolytic cell
Anode : Impure metal to be refined with dilute acid.
Cathode : Thin strips of pure metal
Electrolyte : Aqueous solution of the salts of the metal with dilute acid.
2. The metal dissolves from the anode, pass into the solution.
3. At the same amount of metal ions from the solution will be deposited at the cathode.
4. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Example: Electrolytic refining of silver.
Cathode: Pure silver
Anode: lmpure silver rods
Electrolyte: Acidified aqueous solution of silver nitrate
5. When a current is passed through the electrodes the following reactions will take place
(a) Reaction at anode: \({ Ag }_{ (s) }\longrightarrow { Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\)
(b) Reaction at cathode: \({ Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\longrightarrow { Ag }_{ (s) }\)
6. During electrolysis, at anode silver loses electrons and form silver ions and the silver ions migrate towards the cathode and get discharged and deposited on the cathode.
7. Copper, Zinc etc can also be refined by this process.

(i) 3 - cyanobutanoic acid
(ii) 2 - methyl propanenitrile
(iii) phenyl carbylamine

	Formaldehyde	acetaldehyde
1.	It does not undergo iodoform reaction with I2+NaOH	It undergo iodoform reaction with I2+NaOH
2.	It undergoes Cannizzaro reaction with alkaline solution.	It undergoes Aldol condensation with alkaline solution.

(i) 2-amino-2-methyl propane
\(\quad { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { NH }_{ 3 } } }{ C } } -{ CH }_{ 3 }\)
(ii) 2-(N, N- dimethyl) amino butane
\({ CH }_{ 3 }-{ CH }_{ 2 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ CH } -\overset { \underset { | }{ { CH }_{ 3 } } }{ N } -{ CH }_{ 3 }\)

Vitamin K.

(i) Oil dispersed in water (O/W type)
(ii) Water dispersed in oil (W/O type).

Emulsion can be separated into two separate layers. The process is called Deemulsification.

Alcohols are acidic enough to react with sodium but ethers are inert.

\(\mathrm{Hg}_{2} \mathrm{Cl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Hg}_{2}^{2+} \text { (aq) }+2 \mathrm{Cl^-}_{(\mathrm{aq})}\\ s \quad \quad \quad \quad \quad \quad s \quad \quad \quad \quad \quad 2s\)
\(\mathrm{K}_{\mathrm{sp}} =\left[\mathrm{Hg}_{2}^{2+}\right]{\left[\mathrm{Cl}^{-}\right]^{2}}  \)
\(=(\mathrm{s})(2 \mathrm{~s})^{2} \)
\(\mathrm{~K}_{\mathrm{sp}} =4 \mathrm{~s}^{3}\)

Chromium (24), the electronic configuration is 3d⁵4s¹. It ready to lose its outer most electron (4s¹) to get exactly half-filled stable electronic configuration. The electronic configuration of Zinc is 3d¹⁰4s².ie., It has completely filled stable configuration. From this configuration, the removal of 1e^- from 4s orbital is very difficult & it required more ionisation enthalpy. Due to this reason Zn has higher first Ionisation enthalpy (1.E₁) than that of chromium.

When boric acid or borate salt is heated with ethyl alcohol in presence of conc. sulphuric acid, an ester, trialkylboranes is formed.
The vapour of this ester burns with a green edged flame
This reaction is used to identify the presence of triethyl borate.
\({ H }_{ 3 }{ BO }_{ 3 }+{ 3C }_{ 2 }{ H }_{ 5 }OH\overset { Conc }{ \underset { { H }_{ 2 }{ SO }_{ 4 } }{ \longrightarrow } } B\left( { OC }_{ 2 }{ H }_{ 5 } \right)_3 +3{ H }_{ 2 }O\)
                       Ethyl borate
                        (Green edged flame)

(i) Repl cementby-OH: When the aqueous solution is boiled, phenol is obtained

This is an example of SN₁ reaction in which C₆H₅N₂CI initially gives C₆H₅⁺ and water is the nudeophile.
(ii) Replacement of RO- (or) RCOO- groups Similarly -N₂CI can be replaced acyloxy group by boiling with carboxylic acids.

(iii) Diazonlum coupling reaction: Diazonium salt reacts with aromatic amine and phenols to give azo compounds of the general formula.
Ar - N = N - Ar'
This reaction is known as Coupling reaction since all these compounds are intensely coloured and used as dyes, thousands of azodyes have been synthesised by this procedure.

During oxidation of unsymmetric ketones with oxidising agent which brings about the cleavage of C-C bond, the smaller alkyl group goes preferentially with the carbonyl group resulting in the carboxylic acids.
\({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }\overset { (O) }{ \underset { Con.HNO_{ 3 } }{ \longrightarrow } } \underset { Propanoic\ acid }{ { CH }_{ 3 }{ CH }_{ 2 }-COOH } +\underset { acetic\ acid }{ { CH }_{ 3 }COOH } \)

(i) The secondary structure elements (α-helix & βsheets) further folds to form the three dimensional arrangement. This structure is called tertiary structure of the polypeptide (protein).
(ii) Tertiary structure of proteins are stabilised by the interactions between the side chains of the amino acids.
(iii) These interactions include the disulphide bridges between cysteine residues, electrostatic, hydrophobic, hydrogen bonds and van der Waals interactions.

Proteins are classified based on their structure (overall shape) into two major types. They are fibrous protein and globular proteins.
(i) Fibrous proteins: These proteins are linear molecules. These are generally insoluble in water and are held together by disulphide bridges and weak intermolecular hydrogen bonds. The Example: Keratin, Collagen etc
(ii) Globular proteins: These proteins have an overall spherical shape. The polypeptide chain is folded into a spherical shape. These proteins are usually soluble in water and have many functions including catalysis.

(i) The pores of ordinary filter papers permit the passage of colloidal solutions.
(ii) In ultra filtrations, the membranes are made by using collodion cellophane or visiking.
(iii) When a colloidal solution is filtered using such a filter, colloidal particles are separated on the filter and the impurities are removed as washings.
(iv) This process is quickened by application of pressure.
(v) The separation of sol particles from electrolyte by filteration through an ultrafilter is called ultrafiltration.

The maximum work that can be obtained from a galvanic cell is
(W_max)_cell = - nFE_cell      ...(1)
Here the (-) sign is introduced to indicate that the work is done by the system on the surroundings. According to Second Law of thermodynamics, the maximum work done by the system is equal to the change in the Gibbs free energy of the system.
i,e, W_max= ∆G  .......(2)
From (1) and (2),
∆G = - nFE_cell   ....(3)
For a spontaneous cell reactions, the ∆G should be negative. The above expression (3) indicates that E_cell should be positive to get a negative ∆G value.
When all the cell components are in their standard state, the equation becomes
∆G^o = -nFE^o_cell

Hydrolysis of water
At anode:
\(2H_{2}O\rightarrow 4H^{+}+O_{2}+4e^{-}\) ..... (1)
At cathode:
\(2H_{2}O+2e^{-}\rightarrow H_{2}+2OH^{-}\) ....(2) 
Overall reaction 
\(6H_{2}O\rightarrow 4H^{+}+4OH^{-}+2H_{2}+O_{2}\)
(or)
Equation (1) +(2) \(x^2 \Rightarrow 2H_{2}O\rightarrow 2H_{2}+O_{2}\)
\(\therefore\) According to faradays Law of electrolysis, to electrolyse two mole of Water (36g ≃ 36 mL of H₂O), 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = \(4\times 96500\)C.
\(\therefore\) When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to \(\frac{4\times96500\quad C}{36 \quad mL}\times9\times10^{12}L\)
\(=\frac{4\times96500\times9\times10^{12}}{36\times10^{-3}}C\)
= \(96500\times10^{15}C\)
\(\therefore\) Given that in 1 second, \(2\times10^{6}\) C is generated therefore, the time required to generate \(96500 \times 10^{15}\) C is = \(\frac{1\quad S}{2\times 10^{6}C}\times 96500 \times10^{15}C\)
=\(48250 \times 10^{9} S\)
\(\therefore\) Number of years = \(\frac{48250 \times 10^{9}}{365 \times 24 \times 60 \times 60}\)   
=\(1.5299 \times 10^{6}\) years
1 year = 365 days
= 365\(\times\)24 hours
= 365\(\times\)24\(\times\)60 min
= 365\(\times\)24\(\times\)60\(\times\)60 sec.

(i) All radioactive transformations follow first order kinetics. For example,
₉₂U²³⁸ ⟶ ₉₀U²³⁴ +₂He⁴
(ii) Decomposition of sulphuryl chloride in the gas phase proceeds by first order kinetics.
SO₂Cl_2(g) ⟶ SO_2(g) + Cl_2(g)
(iii) Inversion of sucrose in acidic aqueous medium follows first order reaction.
C₁₂H₂₂O₁₁ +H₂O \(\overset { H+ }{ \longrightarrow } \) C₆H₁₂O₆ +C₆H₁₂O₆

Elemental copper is more stable than Cu²⁺. The electronic configuration of copper is 3d¹⁰ 4s¹ completely filled 3d orbital with stable configuration.
But Cu²⁺ has configuration as 3d⁹. Hence \(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{0}\) value is positive for Cu²⁺.

For cubic closed packed structure
\(r =\frac{a \sqrt{2}}{4}  \)
\(\therefore a =\frac{4 r}{\sqrt{2}}  \)
\(=\frac{4 \times 1.25 \times 10^{-8}}{1.414}=3.53 \times 10^{-8} \mathrm{~cm} \)
= 353 pm