ΔG^o = -nFE^o_cell
= -2 x 96500 x 0.24
= - 46320 J mol^-1
= -46.32 KJ mol⁻¹

α = Λ/Λ^o
= 6/400
K_a = α²C
\(= \frac{6}{400} \times \frac{6}{400} \times \frac{1}{36}\)
= 6.25 x 10^-6

Charging anode:
PbSO_4(s)  + 2e^-  ➝ Pb_(s) + SO₄^2-_(aq)
Cathode:
PbSO_4(s)  + 2H₂O_(I)  ➝ PbO_2(s) + SO₄^2-_(aq) + 2e^-

IF = 96500 C = charge of one mole of e^- charge of 6.022 x 10^-23 electron

(i) The ion which has lower reduction I potential will be oxidised first at the anode.
(ii) Among (0.035V) and (0.44V), Fe²⁺ / Fe oxidation is easy because it has the lower reduction potential (-0.44V).

The emf of a single cell is about 2V

(i) \({ \Lambda }_{ m }^{ o }{ BaCl }_{ 2 }={ \lambda }_{ { Ba }^{ 2+ } }^{ o }+2{ \lambda }_{ { Cl }^{ - } }^{ o }\)
(ii) \({ \Lambda }_{ m }^{ o }{ Al }_{ 2 }{ { (SO }_{ 4 }) }_{ 3 }=2{ \lambda }_{ { Al }^{ 3+ } }^{ o }+2{ \lambda }_{ { { So }_{ 4 }^{ 2- } } }^{ o }\)

Yes, the above given statement is true.
When the dilution increases, the ions are far apart and the attractive forces decrease. At infinite dilution the ions are so far apart, the interaction between them becomes insignificant and hence, the molar conductivity increases and reaches a maximum value at infinite dilution

\({ \Lambda }_{ m }={ \Lambda }_{ m }^{ o }-(-A+B{ \Lambda }_{ m }^{ o })\sqrt { C } \)

Unit of x
\(k=\frac { 1 }{ \rho } .\frac { l }{ A } \left( \frac { 1 }{ ohm } .\frac { m }{ { m }^{ 2 } } \right) \)
= ohm^-1 m^-1 = mho m^-1 (or) Sm^-1

Given that
κ = 1.3 Sm^-1 (for 0.1M KCl solution)
R = 190 Ω
\(\kappa = \frac{1}{R}(\frac{l}{A})\)
κ . R =\((\frac{l}{A})\) =  (1.3 Sm^-1) (190Ω)
= 247 m^-1
\(\kappa_{(NaCl)} = \frac{1}{R_{(NaCl)}} (\frac{l}{A})\)
\(= \frac{1}{6.3 K\Omega}(247 m^{-1})\)  (6.3KΩ = 6.3 x 10³Ω)
= 39.2 x 10^-3 Sm^-1
\(\Lambda_m = \frac{\kappa \times 10^{-3} mol^{-1} m^3}{M}\)
\(=\frac{39.2 \times 10 ^{-3}(Sm^{-1})10^-3 (mol^{-1} m^3)}{0.003}\)
\(\Lambda_m\) = 13.04 \(\times\) 10^-3  Sm² mol^-1

Metals having higher oxidation potential will liberate H₂ from H₂SO₄. Hence, the metal M₁ having +xV, oxidation potential will liberate H₂ from H₂SO₄.

On dilution the concentration decreases. Conductivity decreases with decrease in concentration (or dilution) as the number of ions per unit volume that carry the current in a solution decrease on dilution.

(i) Faraday is defined as the quantity of electricity required to liberate one gram equivalent of a substance and it is equal to 96,495 coulombs.
(ii) It is denoted by the symbol F Faraday is the quantity of electricity that contains 1 mole of electrons.
(iii) 1 Faraday = 96,495 coulombs = 1 Mole electron.

Oxidation occurs at the anode
\({ Pb }_{ (s) }\rightarrow { Pb }_{ (aq) }^{ 2+ }+{ 2e }^{ - }\)
The Pb²⁺ions combine with \({ SO }_{ 4(aq) }^{2-}\) to from PbSO₄ precipitate.
\({ Pb }_{ (aq) }^{ 2+ }+{ SO }_{ 4(aq) }^{ 2- }\rightarrow { PbSO }_{ 4(s) }\)
Reduction occurs at the cathode
\({ PbO }_{ 2(s) }+{ 4H }_{ (aq) }^{ + }+{ 2e }^{ - }\rightarrow { Pb }_{ (aq) }^{ 2+ }+{ 2H }_{ 2 }O(l)\)
The Pb²⁺ ions also combine with \({ SO }_{ 4(aq) }^{2-}\) ions from sulphuric acid to form PbSO₄ precipitate.
\({ Pb }_{ (aq) }^{ 2+ }+{ SO }_{ 4(aq) }^{ -2 }\rightarrow { PbSO }_{ 4 }\)

Zinc is oxidised to Zn^{2+ i}ons and the Cu²⁺ ions are reduced to metallic copper. The half reactions are represented as below.
\( { Zn }_{ (s) }\rightarrow { Zn }_{ (aq) }^{ + }+2e^{ - }\) (oxiation)
Loss of election oxidation
\({ Cu }_{ (aq) }^{ 2+ }+{ 2e }^{ - }\rightarrow { Cu }_{ (s) }\) (reduction)
Gain of electron oxidation.

Oxidation half cell : A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell : A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

The conductivity cell in which the electrodes are separated by 1m and having V m³ of electrolytic solution which contains 1 mole of electrolyte. The conductance of such a system is called the molar conductance (\({ \Lambda }_{ m }\))
\({ \Lambda }_{ m }=\frac { k({ Sm }^{ -1 })\times { 10 }^{ -3 } }{ M } { mol }^{ -1 }{ m }^{ 3 }\)

E = E^O- 2.303 \(\frac { RT }{ 2F } \) log Zn²⁺
E_Zn²⁺/Zn = E_{Zn²⁺/ Zn}- 2.303 \(\frac { RT }{ 2F } \)  log [Zn²⁺]

(i) The half cell reactions are :
\(2Br^{-} \rightarrow Br_{2}+2e^{-}\)     \(E^{0}_{ox}=-1.09V\) ...(1)
\(2Fe^{3+}+2e^{-}\rightarrow2Fe^{2+}\)  \(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=+0.771V\) ..(2)
(ii) Adding (1) of (2) :
\(2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{2+}+Br_{2}\)   \(E^{0}_{cell}=?\) ...(3)
\(E^{0}_{cell}=E^{0}_{ox}+E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}\)
= (-1.09 + 0.771)V
= -0.319V
(iii) E⁰_cell is – ve; \(\Delta G\) is +ve and the cell reaction is non spontaneous.
(iv) Hence Fe³⁺ cannot oxidises Br^- to Br₂.

(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.

Anodic oxidation: 
The electrode at which the oxidation occurs is called the anode. In  Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zn²⁺ ions and the electrons. The Zn²⁺ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (-ve).
\({ Zn }_{ (s) }\longrightarrow { Zn }^{ 2+ }{ _{ (aq) } }+ 2{ e }^{ - } \) (loss of electron-oxidation) 
Cathodic reduction:
The electrons flow through the circuit from zinc to copper, where the Cu²⁺ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve).
\({ Cu }^{ 2+ }_{ (aq) }+{ 2e }^{ - }\longrightarrow { Cu }_{ (s) } \)(gain of electron-reduction) 

(i) The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key as shown in the figure. The electrode which is attached to the negative end of the power supply is called the cathode, and the one which attached to the positive end is called the anode. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time pull the electrons from the anode.

Cell reactions:
Na⁺ ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.

Cathode (reduction)
\(N a_{(l)}^{+}+e^{-} \rightarrow N a_{(l)} \quad ; \quad E^{0}=-2.71 V\)
Similarly, Cl^- ions are attracted towards anode where they lose their electrons and oxidised to chlorine gas.
Anode (oxidation)
2CI^-_(l) ⟶ CI_2(g) + 2e^- E⁰ = -1.36V
The overall reaction is
2Na⁺_(l) + 2Cl^-_(l)➝ 2Na_(l) + Cl_2(g) ; E° = - 4.07V
(ii) The negative E° value shows that the above reaction is a non-spontaneous one.
(iii) Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCI.
(iv) In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell.
(v) But the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and anode is +ve.

(i) Current: The flow of electrons through a wire or any conductor.
(ii) Electrode: A metallic rod / barf strip which conducts electrons into and out of a solution.
(iii) Anode: The electrode at which oxidation occurs. It sends electrons into the outer circuit. It has negative charge and is shown as (-) in cell diagrams.
(iv) Cathode: The electrode at which lect ons are received from the outer circuit. It has a positive charge and is known as (+) in the cell diagrams.
(v) Electrolyte is the salt solunon in a cell
(vi) Anode compartment: The compartment of cell in which oxidation half-reaction occurs. It contains the anode.
(vii) Cathode compartment: The compartment of the cell in which reduction half-reaction occurs. It contains the cathode.
(vii) Half-cell: Each half of an electrochemical cell, where oxidaton occurs, is called the half cell.

According to Ostwald dilution Law,
\({ K }_{ a }=\frac { { \alpha }^{ 2 }C }{ (1-\alpha ) } \)  ...(1)
Substitute a value in the above expression (1)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\left( 1-\frac { { \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } \right) } \)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\frac { { \Lambda }_{ m }^{ o }-{ \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } } \)
⇒ \({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }{ (\Lambda }_{ m }^{ o }-{ \Lambda }_{ m }) } \)

E^o_cell = 1.56V [Zn²⁺]
= 0.1 M [Ag⁺] = 10 M
Formula:
\([{ E }_{ cell }={ E }_{ cell }^{ o }-\frac { RT }{ nF } In\frac { [{ Zn }^{ 2+ }] }{ [{ Ag] }^{ 2 } } ]\)
Solution:
= 1.56 - 0.02955 log 0.001;
= 1.56 - (- 0.08865)
= 1.56 + 0.08865 = 1.6486 V
E_cell = 1.6486 V.

Given:
E₁ = 0.5 V
Cr³⁺+ 3e^- ➝ Cr_(s).......(1)
E₂ = 0.41 V
Cr³⁺+ e^- ➝ Cr²⁺  .....(2)
The required reaction is,
Cr²⁺+ 2 e^- ➝ Cr_(s)
Then,
Formula:
\({ E }_{ 3 }=\frac { 3{ E }_{ 1 }+{ E }_{ 2 } }{ 2 } \)
Solution:
= \(\frac { 3(0.5)+(0.41) }{ 2 } =\frac { 1.5+0.41 }{ 2 } \)
= 0.955 V
E₃ = 0.955 V

Quantity of electricity, Q = 0.5 F
Concentration of solution, C = 5 M
Solution:
(i) For AgNO₃
1 mol of Ag⁺ = 1F
0.5F = 0.5 mol of AgNO₃
Concentration of AgNO₃ after electrolysis
= 5 - 0.5
= 4.5M
(ii) For CuSO₄
1 mol of CuSO₄ (or) Cu²⁺ = 2F
2F = 1M CuSO₄
0.5F = \(\frac12\) x 0.5 = 0.25 M
Concentration of CuSO₄ after electrolysis
= 5 - 0.25
= 4.75M
(iii) For AlCl₃
1 mol of AlCl₃ Br Al³⁺ = 3F
3F = 1M AlCl₃
0.5F = \(\frac13\) x 0.5
= \(\frac{0.5}{3}\) = 0.167 M
Concentration of AlCl₃ after electrolysis = 5 - 0.167
Concentration of AlCl₃ after electrolysis = 4.833 M

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) 
\(\left[\begin{array}{l} \text { Where } \\ \therefore \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ \mathrm{~T}=298 \mathrm{~K} \\ \mathrm{IF}=96500 \mathrm{C} \mathrm{mol}^{-1} \end{array}\right]\)

i) Molar conductivity
Given : C = 0.01 M;
\(\kappa=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \)
\(=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1} \)
\(\lambda_{\text {cation }}^{0}=248.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\lambda_{\text {anion }}^{0}=51.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\Lambda_{m}^{0}=\frac{\kappa \times 10^{-3}}{C} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)
\(=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \)
\(=1.5 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)
ii) \(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)
\(\Lambda_{\mathrm{m}}^{0}=\lambda_{\text {cation }}^{0}+\lambda_{\text {anion }}^{0} \)
\(=(248.2+51.8) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \)
\(=300 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(=300 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}  \)
\(\alpha =\frac{1.5 \times 10^{-3}}{300 \times 10^{-4}}=0.05\)
iii) \(\mathrm{K}_{\mathrm{a}} =\frac{\alpha^{2} \mathrm{C}}{1-\alpha} \)
\(\mathrm{K}_{\mathrm{a}} =\frac{(0.05)^{2} \times(0.01)}{1-0.05}=2.6 \times 10^{-5} \)
(or) 
\(\mathrm{K}_{\mathrm{a}} =\alpha^{2} \mathrm{C}  \)
\(=(0.05)^{2} \times(0.01)  \)
\(\mathrm{K}_{\mathrm{a}} =2.5 \times 10^{-5}\)