Mn²⁺ + 2e^- ➝ Mn E^o = -1.18V
Mn²⁺ ➝ Mn³⁺ + e^- E^o = -1.51V
3Mn²⁺➝ Mn³⁺ + 2Mn³⁺ E^o_cell = ?
E^o_cell = (E^o_ox )+ (E^o_red) 
= 1.51 - 1.18 and non spontaneous
= -2.69V
Since E°is ve ΔG is +ve and the given forward cell reaction is non- spontaneous

t_1/2 α \(\frac{1}{[A_{0}]^{n-1}}\)...(1)
If [A₀ = 2[A₀]; then t_1/2 = 2t_1/2
2t_1/2 α \(\frac{1}{[2A_{0}]^{n-1}}\)...(2)
(2)/(1) = \(2= \frac{1}{[2A_{0}]^{n-1}} \times \frac{1}{[A_{0}]^{n-1}}\)
\(2= \frac {[2A_{0}]^{n-1}} {[A_{0}]^{n-1}}\)
\(2= \frac {1} {2}^{n-1}\)
2 = (2^-1)^n-1
2¹ = (2^-n+1)
n = 0

\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
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.

1. The acidic nature of the alcohol is due to the polar nature of O-H bond.
2. When an electron with drawing -I groups such as - Cl,-F etc, is attached to the carbon bearing the OH group it withdraws the electron density towards itself and thereby facilitating the proton donation. In contrast, the electron releasing group such as alkyl group increases the electron density on oxygen and decreases the polar nature of O-H bond, Hence it results in the decrease in acidity.
\(\begin{array}{lll}1^{0} \text { alcohol }> & 2^{0} \text { alcohol }> & 3^{\circ} \text { alcohol. } \\ \text { Only one alkyl } & \text { two alkyl } & \text { three alkyl } \\ \text { group } & \text { groups } & \text { groups }\end{array}\)

1. To calculate the molar conductance at infinite dilution of a weak electrolyte
2. To calculate the degree of dissociation of a weak electrolyte
3. To calculate the solubility of a sparingly soluble salts

i) It is a zero order reaction as the graph is satisfying the equation \([\mathrm{A}]=\left[\mathrm{A}_{0}\right]-\mathrm{kt}.\)
ii) The slope of the curve is the negative of the rate constant that is denoted by -k.
iii) Unit of rate constant is Ms^-1 or mol L^-1 S^-1.

(i) Chloro benzene:
When Chlorobenzene is hydrolysed with 6-8% NaOH at 300 bar and 633K in a closed vessel, sodium phenoxide is formed which on treatment with dilute HCl gives phenol.

(ii) isopropyl benzene:
A mixture of benzene and propene is heated at 523K in a closed vessel in presence of H₃PO₄ catalyst gives cumene (isopropylbenzene). On passing air to a mixture of cumene and 5% aqueous sodium carbonate solution, cumene hydro peroxide is formed by oxidation. It is treated with dilute acid to get phenol and acetone. Acetone is also an important byproduct in this reaction.

(a) If we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell.
(b) So, AC current is used for this measurement to prevent electrolysis.

First law:
The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.
m α Q \(\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}} \Rightarrow \mathrm{Q}=\mathrm{It}\right]\)
m α It
m = ZIt
Where Z = electro chemical equivalent of the substance
I = current
t = time of passage of current
Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.
m α Z
\(\frac{m_{1}}{Z_{1}}=\frac{m_{2}}{Z_{2}}\)
m = mass of the metal deposited
Z = electro chemical equivalent

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentration is called a zero order reaction.
Let us consider the following by hypothetical zero order reaction.
 A⟶ product 
The rate law can be written as,
Rate = k[A]⁰ 
\(\frac { -d\left[ A \right] }{ dt } =k(1)\ \ \ \therefore \left( { \left[ A \right] }^{ 0 }=1 \right) \)
\(\Rightarrow -d\left[ A \right] =kdt\)
Integrate the above equation between the limits of [A₀] at zero time and [A] at some later time 't',
\(-\int _{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }{ d\left[ A \right] } =k\int _{ 0 }^{ t }{ dt } \)
\(-{ \left( \left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
[A₀] - [A] = kt   
\(k=\frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \)
Straight line equation y = mx + c
ie., \(\left[ A \right] =-kt+\left[ { A }_{ 0 } \right] \)
⇒ y = c + mx
A plot [A] vs time gives a straight line with a slope of -k and y intercept of [A₀]

Phenol couples with benzene diazonium chloride in an alkaline solution to form p-hydroxy azo benzene (a red orange dye)

a) \(Z n_{(s)}\left|\mathrm{Zn}_{(a q)}^{2+} \| \mathrm{Cu}_{(\mathrm{aq})}^{2+}\right| \mathrm{Cu}\)
b) \(\mathrm{Cr}_{(\mathrm{s})}\left|\mathrm{Cr}_{(\mathrm{aq})}^{3+} \| \mathrm{Cu}_{(\mathrm{aq})}^{2+}\right| \mathrm{Cu}_{(\mathrm{s})}\)

Integrated rate equation for zero order reaction:
\(k=\frac{[A]_{0}-[A]}{t} ; t_{\frac{1}{2}}=\frac{[A]_{0}}{2 k}\)
Integrated rate equation for first order reaction.
\(k=\frac{2.303}{t} \log \frac{a}{a-x} ; t_{\frac{1}{2}}=\frac{0.693}{k}\)

(i) Benzylchloride to benzylalcohol
\(\underset { Benzyl \ chloride }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }Cl } + NaOH { \longrightarrow } \underset { Benzyl \ alcohol }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }OH } +NaCl\)
(ii) benzyalalcohol to benzoic acid
 \(\underset { Benzyl \ alcohol }{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH} }\stackrel{\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{+}}{\longrightarrow}\underset { Benzaldehyde } {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}}\stackrel{\mathrm{Na}_{2} \mathrm{CrO}_{7} / \mathrm{H}^{+}}{\longrightarrow} \underset { Benzoic \ acid } {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}}\)

Cathodic protection:
In this technique, unlike galvanising the entire surface of the metal to be protected need not be covered with a protecting metal. Instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded. This known as sacrificial protection. (or) Cathodic protection.

(i) Let A = 100M, [A₀] - [A] = 20M,
For the zero order reaction
\(k=\left( \frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \right) \)
(i) 20% completion  \(k=\left( \frac { 20M }{ 20min } \right) \) = 1 mol L^-1 min^-1 
(ii) 80% completion
\(\mathrm{K}=1 \mathrm{~mol} \mathrm{~L}{ }^{-1} \mathrm{~s}^{-1} ;\left[\mathrm{A}_{0}\right]=100 \mathrm{M} ;\left[\mathrm{A}_{0}\right]-[\mathrm{A}]=80 \mathrm{M} ; \mathrm{t}=?\)
\(\therefore t=\left(\frac{\left[A_{O}\right]-[A]}{K}\right)=\frac{80}{1}=80 \mathrm{mins}\)

(i) This test is based on the behaviour of the different nitro alkanes formed by the three types of alcohols with nitrous acid and it consists of the following steps.
i) Alcohols are converted into alkyl iodide by treating it with I₂ / P.
ii) Alkyl iodide so formed is then treated with AgNO₂ to form nitro alkanes.
iii) Nitro alkanes are finally treated with HNO₂ (mixture of NaNO₂ / HCl) and the resultant solution is made alkaline with KOH.
 Result:
1. Primary alcohol gives red colour.
2. Secondary alcohol gives blue colour.
3. No colouration will be observed in case of tertiary alcohol.

Anode: Porus graphite
Cathode: Transition metal oxide such as CoO₂
Electrolyte: Lithium salt in an organic solvent
(i) Oxidation at anode
\(\mathrm{Li}_{(\mathrm{s})} \rightarrow \mathrm{Li}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-}\)
(ii) Reduction at cathode
\(\mathrm{Li}^{+}+\mathrm{CoO}_{2(\mathrm{~s})}+\mathrm{e}^{-} \rightarrow \mathrm{LiCoO}_{2(\mathrm{~s})}\)
(iii) Overall reactions
\(\mathrm{Li}_{(\mathrm{s})}+\mathrm{CoO}_{2} \rightarrow \mathrm{LiCoO}_{2(\mathrm{~s})}\)
(iv) Both electrodes allow Li⁺ ions to move in and out of their structures.
(v) During discharge, the Li⁺ ions produced at the anode move towards cathode through the non - aqueous electrolyte.
(vi) When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li⁺ ions move from cathode to anode where they become embedded on the porous graphite electrode. This is known as intercalation.

Uses:
Used in cellular phones, laptops, computers, digital cameras, etc...

According to Arrhenius, activation energy of the reaction is the minimum energy that a molecule must have to posses to react.
\(\mathrm{k}=\mathrm{A} e^{-\left(\frac{E_{a}}{R T}\right)}\)
Where A is the frequency factor
R is the gas constant
E_a is the activation energy of the reaction
T is the absolute temperature. (in k)
Taking logarithm on both side of the equation (1)
\(\ln \mathrm{k}=\ln \mathrm{A}+\ln \mathrm{e}^{-\left(\frac{E_{a}}{R T}\right)}  \)
\(\ln \mathrm{k}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\right) \quad(\therefore \ln \mathrm{e}=1) \)
\(\ln \mathrm{k}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\right)\left(\frac{1}{\mathrm{~T}}\right)\)
At temperature T = T₁; the rate constant k = k₁
\(\ln \mathrm{k}_{1}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{1}}\right)\)
At temperature T = T₂; the rate constant k = k₂
\(\ln \mathrm{k}_{2}=\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{2}}\right)\)
(4)-(3)
\(\ln \mathrm{k}_{2}-\ln \mathrm{k}_{1}=-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{2}}\right)+\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{1}}\right) \)
\(\Rightarrow \ln \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right) \)
\(2.303 \log \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}\right) \)
\(\log \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}\right) \)
\(\ln \mathrm{k}_{2}-\ln \mathrm{k}_{1}=-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{2}}\right)+\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_{1}}\right)\)
This equation can be used to calculate E_a from rate constants k₁ and k₂ at temperatures T₁ and T₂.

1. Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is,
xA + yB ⇌ IC + mD
2. The reaction quotient Q for the above reaction is given below
\(Q=\cfrac { \left[ C \right] ^{ ' }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)  .....(1)
3. We have already learnt that,
ΔG = ΔG + RT In Q .........(2)
4. The Gibbs free energy can be related to the cell emf as follows.
[ஃ equation (1) and (2)]
ΔG = - nFE_cell; ΔG^o = - nFE^o_cell
5. Substitute these values and Q from (1) in the equation (2)
(2) ⇒\(-nF{ E }_{ cell }=-n{ FE }^{ o }_{ cell }+RT\quad In\cfrac { \left[ C \right] '\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) ....(3) 
6. Divide the whole equation (3) by (-nF)
(4) ⇒\({ E }_{ cell }={ E }_{ cell }-\cfrac { RT }{ nF } In\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) 
(or)
\({ E }_{ cell }={ E }_{ cell }-\cfrac { 2.303RT }{ nF } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) .....(4)
7. The above equation (4) is called the Nernst equation
8. At 25°C (298 K), the above equation (4) becomes,
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 2.303\times 8.314\times 298 }{ n(96500) } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)    
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 0.0591 }{ n } log\cfrac { { \left[ C \right] }^{ l }{ \left[ D \right] }^{ m } }{ { \left[ A \right] }^{ x }\left[ B \right] ^{ y } } \) ......(5) 
\(\left[\begin{array}{l} \text { Where } \\ \therefore \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ \mathrm{~T}=298 \mathrm{~K} \\ \mathrm{IF}=96500 \mathrm{C} \mathrm{mol}^{-1} \end{array}\right]\)

Given data: Time required for 99.9% completion is about 10 times its half life period. Prove that for a first order reaction the time required for 99.9% completion is about 10 times its half life period.
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution:
\({ t }_{ 99.9 }=\frac { 2.303 }{ k } \log\frac { 100 }{ 100-99.9 } \)
\(=\frac { 2.303 }{ k } \log\frac { 100 }{ 0.1 } \)
\(=\frac { 2.303 }{ k } \log1000\)
\({ t }_{ 99.9 }=\frac { 2.303\times 3 }{ k } \)
\({ t }_{ 99.9 }=\frac { 6.909 }{ k } ;\)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)  \(\therefore k=\frac { 0.693 }{ { t }_{ \frac { 1 }{ 2 } } } \)
\(\frac { { t }_{ 99.9\% } }{ { t }_{ \frac { 1 }{ 2 } } } =\frac { 6.909 }{ k } \times \frac { k }{ 0.693 }\)
\({ t }_{ 99.9 }=\frac { 6.909 }{ 0.693 } \times { t }_{ \frac { 1 }{ 2 } }\)
t_99.9 = 10 x t_1/2
Time required for 99.9% completion of the reaction = 10t_1/2