R = ρ.l/A
Cell constant = R/ρ
\(= k.R (l/ \rho = K)\)
= 1.25 x 10^-3 Ω^-1 cm^-1 x 800Ω
= 1 cm^-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

E^o_Zn²⁺[Zn] = 0.76V and E^o_Fe²⁺[Fe] = -0.44V
Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc

In general, specific conductance of an electrolyte decreases with dilution.So,0.002N solution has least specific conductance.

(i) The conductivity of an electrolytic solution is determined by using a wheatstone bridge arrangement in which one resistance is replaced by a conductivity cell filled with the electrolytic solution of unknown conductivity.
(ii) In the measurement of specific resistance of a metallic wire, a DC power supply is used. Here, if we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell. So, AC current is used for this measurement to prevent electrolysis

(iii) A wheatstone bridge is constituted using known resistances P, Q, a variable resistance S and conductivity cell (Let the resistance of the electrolytic solution taken in it be R) as shown in the figure. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C. Connect a suitable detector (Such as the telephone ear piece detector) between the junctions 'B' and 'D'.
(iv) The variable resistance (S) is adjusted until the bridge is balanced and in this conditions there is no current flow through the detector.
 Under balanced condition,
\(\frac { P }{ Q } =\frac { R }{ S } \)
\(\therefore R=\frac { P }{ Q } \times S\)
(v) The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured 'S' value under balanced condition using the above expression

 Sn⁴⁺ + 2e^- ⟶ Sn²⁺ E⁰ = 0.15V
Given: n = 2 electrons 
F = 96495 coulombs
Formula: ΔG = - nFE^O
Solutlon: ∴ ΔG = - 2 x 96495 x 0.15
= 28.948
Free energy = -28.948 kJ.

Given:
Specific conductance = K = 5.55 x 10^-3 mho cm^-1
Concentration C = 1M
Volume = 1 litre.
Formula: Equivalent conductance = K x V
= \(\frac { k\times 1000 }{ C } \)
Solution:
\(\frac { { 5.55\times 10 }^{ -3 }\times 1000 }{ 1 } =5.55\times { 10 }^{ -3 }\times { 10 }^{ -3 }\)

Given: Electrochemical
equivalent = Z = 2.35 g amp^-1 sec^-1
Time = t = 10 sec.
Current strength = 1 = 5 ampere
Formula: m = Zlt
Solution: 2.35 x 10 x 5 = 117.5 gm
The amount of the substance deposited
= 117.5 g

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

Kohlraush's law:
(i) At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
(ii) For a uni - univalent electrolyte such as NaCl, the Kohlraush's law is expressed as
\(\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}=\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Na}^{+}}+\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Cl}}^{-}\)
(iii) In general, according to Kohlraush's law, the molar conductivity at infinite dilution for a electrolyte represented by the formula A_x B_y, is given below.
\(\left(\Lambda_{m}^{0}\right)_{A_{x} B_{y}}=x\left(\lambda_{m}^{0}\right)_{A^{y+}}+y\left(\lambda_{m}^{0}\right)_{B^{x-}}\)
b) Calculation of molar conductance at infinite dilution for weak electrolytes experimentally.
(i) However, the same can be calculated using Kohlraush's Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.
\(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0} \) ..........(1)
\(\Lambda_{\mathrm{HCl}}^{0}=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0} \) .........(2)
\(\Lambda_{\mathrm{NaCl}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\) ...........(3)
(ii) Equation (1) + Equation (2) - Equation (3) gives,
\(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}\right)+\left(\Lambda_{\mathrm{HCl}}^{0}\right)-\left(\Lambda_{\mathrm{NaCl}}^{0}\right)=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}-}^{0} \)
\(=\Lambda_{\mathrm{CH}_{2} \mathrm{COOH}}^{0}\)

(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.

(i) Anode: Zinc container
(ii) Cathode: Graphite rod in contact with MnO₂
(iii) Electrolyte: Ammonium chloride and zinc chloride in water
(iv) Oxidation at anode:
\( { Zn }_{ (s) }\rightarrow { Zn }_{ (aq) }^{ 2+ }+2e^{ - }\)
(v) Reduction at cathode:
\({ 2NH }_{ 4(aq) }^{ + }+{ 2e }^{ - }\rightarrow { 2NH }_{ 3(aq) }+{ H }_{ 2(g) }\)

The galvanic cell is represented by a cell diagram, for example, Daniel cell is represented as
\({ Zn }_{ (s) }|{ Zn }_{ (aq) }^{ 2+ }||{ Cu }_{ (aq) }^{ 2+ }|{ Cu }_{ (s) }\)
(i) A single vertical bar (|) represents a phase boundary
(ii) The double vertical bar (||) represents the salt bridge.
(iii) The anode half cell is written on the left side of the salt bridge and the cathode half cell on the right side.
(iv) The anode and cathode are written on the extreme left and extreme right, respectively.
(v) The emf of the cell is written on the right side after cell diagram.

The galvanic cell is represented as
\({ Zn }_{ (s) }|{ Zn }_{ (aq) }^{ 2+ }||{ Cu }_{ (aq) }^{ 2+ }|{ Cu }_{ (s) }\)
The anode half cell is \({ Zn }_{ (s) }|{ Zn }_{ (aq) }^{ 2+ }\)
The cathode half cell is \({ C }u_{ (Aq) }^{ 2+ }|{ Cu }_{ (s) }\)

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.

Electrochemical cells are mainly classified into the following two types.
(i) Galvanic Cell ( Voltaic cell) : It is a device in which a spontaneous chemical reaction generates an electric current i.e., it converts chemical energy into electrical energy. It is commonly known as a battery.
(ii) Electrolytic cell : It is a device in which an electric current from an external source drives a nonspontaneous reaction i.e., it converts electrical energy into chemical energy.

Equivalent conductance is defined as the conductance of 'V' m³ of electrolytic solution containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are one meter apart.
\(\Lambda =\frac { k({ Sm }^{ -1 })\times { 10 }^{ -3 }(\text{ gram equivalent) }^{ -1 }{ m }^{ 3 } }{ N } \)

(i) Specific resistance is denoted by ρ(rho).
(ii) Specific resistance depends on the nature of the electrolyte.
(iii) Reciprocal is specific resistance \(\frac{1}{ρ}\) is specific conductance or conductivity denoted by K (Kappa).
(iv) ohm metre.

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

Cell reactions:
Oxidation at anode: \(Cd_{(s)}\rightarrow Cd^{2+}_{(aq)}+2e^{-}\);    (E⁰_ox)_cd|cd²⁺ = 0.40V ; (E⁰)_cd|cd²⁺ = -0.40V
Reduction at cathode: \(Cu^{2+}_{(aq)}+2e^{-}\rightarrow Cu_{(s)} \);  (E⁰_red)_cu²⁺|Cu = +0.34V
Adding: \(Cd_{(s)}+Cu^{2+}_{(aq)}\rightarrow Cd^{2+}_{(aq)}+Cu_{(s)}\)
E⁰_cell=(E⁰_ox)+(E⁰_red)
=(-0.4) + 0.34V
= 0.74V.
Emf is +ve, so \(\Delta G\) is -ve, the cell reaction is feasible.

\(\kappa=\mathrm{C}\left(\frac{l}{\mathrm{~A}}\right)\) (or) k = 1/R.l/A
Where, C - Conductance
\(\frac{l}{A} \Rightarrow\) Cell constant (or)
Sp. Conductance \(\propto\) Cell constant
Sp. Conductance \(\propto\) Concentration
1. Sp. Conductance = concentration x cell constant
i) \(\kappa=0.1 \times 0.5=0.05=5 \times 10^{-2} \mathrm{Scm}^{-1}\)
ii) \(\kappa=0.1 \times 0.25=0.025=2.5 \times 10^{-2} \mathrm{Scm}^{-1}\)
2. So the first case will have greater specific conductance with cell constant 0.05.

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) 

\({ E }_{ cell }={ E }_{ cell }^{ o }-\frac { Rt }{ nF } In\frac { [{ C] }^{ l }[{ D] }^{ m } }{ [{ A] }^{ x }{ [B] }^{ y } } \) (or)
\({ E }_{ cell }={ E }_{ cell }^{ o }-\frac { 2.303RT }{ nF } \log\frac { [{ C] }^{ l }[{ D] }^{ m } }{ [{ A }]^{ x }[{ B }]^{ y } } \)

(i) An electrochemical cell consists of two half cells. With an open circuit, the metal electrode in each half cell transfers its ions into solution.
(ii) Thus, an individual electrode develops a potential with respect to the solution
(iii) The potential of a single electrode in a half cell is called single electrode potential

The electrochemical equivalent is defined as the amount of substance deposited or liberated at the electrode by a charge of 1 coulomb.
m = z  (If I = 1 ampere and t = second).

The greater the E^o value means greater is the tendency shown by the species to accepts I electrons and undergo reduction. So higher the (E^o) values lesser is the tendency to undergo corrosion.

The cell voltage depends on the nature of the electrodes, the concentration of the electrolytes and the temperature at which the cell is operated.

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

Conductivity, molar conductance and equivalent conductance increases with dilution whereas Conductance (C) decreases.

l = 1.5 cm = 1.5 × 10^-2m 
A = 4.5 cm² = 4.5 × (10^-4)m²
R = 15Ω
\(\kappa = \frac{1}{R}(\frac{l}{A})\)
\(\kappa = \frac{1}{15\Omega}\times\frac{1.5\times10^{-2}m}{4.5\times10^{-4}m^2}\)
= 2.22 ohm^-1 m^-1
= 2.22 Sm^-1

The oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.