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

mg = ZIt
\(= \frac{40 \times 3.86 \times 2500}{2 \times 96500} = 2g\)
(∵ t = 41 min 40 sec = 2500 seconds & \(Z = \frac{m}{n \times 96500} = \frac{40}{x \times 96500})\)

Λ = k/W x 10⁻³ mol⁻¹ dm³
\(=\frac{ 5.76 \times 10^{−3} S cm^{−1} \times 10^{−3}}{0.5} = mol^{-1} dm^{3}\)
\(=\frac{ 5.76 \times 10^{−3} S cm^{−1} \times 10^{−3}}{0.5}\) S cm^-1 mol⁻¹ dm³
= 11.52 S cm²mol^-1

(i) For a strong electrolyte, at high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high.
(ii) Moreover the ions also experience a viscous drag due to greater solvation.
(iii) These factors attribute for the low molar conductivity at high concentration.

k = \(\frac{1}{R}.\frac{1}{a}=\frac{1}{ρ}\)
Where K is specific conductance
R is resistance and \(\frac1a\) is cell constant.
ρ is specific resistance.

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

Electrochemical reaction at cathode is Ag⁺ + e^-  → Ag (reduction)
m = ZIT
Z = \(\frac{\text {molar mass of Ag}}{(96500)}\) = \(\frac{108}{1 \times 96500} \)
I = 2A
t = 20 x 60S = 1200 S
It = 2A  x 1200S = 2400C
m = \(\frac{108 gmol^{-1}}{96500 C mol^{-1}} \times\) 2400C
m = 2.68g

(i) Anode: The electrode at which the oxidation occurs is called anode. (loss of electrons)
(ii) Cathode: The electrode at which the reduction occurs is called cathode. (gain of electrons)

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.

For a weak electrolyte, at high concentration, the plot is almost parallel to concentration axis with slight increase in conductivity as the dilution increases. When the concentration approaches zero, there is a sudden increase in the molar conductance and the curve is almost parallel to \(\Lambda \)_m axis. This is due to the fact that the dissociation of the weak electrolyte increases with the increase in dilution (Ostwald dilution law)

a) Oxidation at anode :
\(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} ; \mathrm{E}_{\mathrm{Ox}}^{0}=2.37 \mathrm{~V}\) ...(1) 
Reduction at cathode:
\(A \mathrm{~g}^{+} +\mathrm{e}^{-} \rightarrow \mathrm{Ag} ; \mathrm{E}_{\text {red }}^{0}=+0.80 \mathrm{~V}  \) ...(2)
\(E_{\text {Cell }}^{0} =\left(\mathrm{E}_{\text {ox }}^{0}\right)_{\text {anode }}+\left(\mathrm{E}_{\text {red }}^{0}\right)_{\text {cathode }} \)
= 2.37 + 0.80 = 3.17V
Overall reaction : (1) + 2 x (2)
\(\mathrm{Mg} \rightarrow M g^{2+}+2 e^{-}  \)
\(2 \mathrm{Ag}^{+}+2 \mathrm{e}^{-} \rightarrow 2 A g \)
__________________
\(M g+2 A g^{+} \rightarrow M g^{2+}+2 A g\)
 b) \( \therefore \Delta G^{0}=-n F E^{0} \)
\(=-2 \times 96500 \times 3.17 \)
\(=-611810=-6.12 \times 10^{5} \mathrm{~J} \)
\(W=6.12 \times 10^{5} \mathrm{~J}\)
c) \(\Delta \mathrm{G}^{0}=-2.303 \mathrm{RT} \log \mathrm{K}_{c} \)
\(\log \mathrm{K}_{\mathrm{c}}=-\frac{\Delta G^{0}}{2.303 R T} \)
\(\log \mathrm{K}_{c}=-\frac{\left(-6.12 \times 10^{5}\right)}{2.303 \times 8.314 \times 298}=107.2 \)
\(K_{c}=A . \log 107.2 \)
\(\mathrm{K}_{c}=\text { Antilog of (107.2) }\)
= 1.58 x 10¹⁰⁷

Given: [Cu²⁺] = 0.1M; E⁰_Cu²⁺|Cu = +0.34V
Cell reaction is: \(Cu^{2+}_{(aq)}+2e^{-}\rightarrow Cu_{(s)}\)
\(E_{cell}=E^{0}-\frac{0.0591}{n}\log\frac{[Cu]}{[Cu^{2+}]}\)
\(= 0.34\frac{0.0591}{2}\log\frac{1}{0.1}\)
E_cell = 0.34 - 0.0296 = +0.31V

\(\kappa=\mathrm{C}\left(\frac{l}{\mathrm{~A}}\right)\) (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.

(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

Substances like AgCl, PbSO₄ etc., are sparingly soluble in water. The solubility product of such substances can be determined using conductivity measurements.
Let us consider AgCl as an example
AgCl(s) ⇌ Ag⁺ + Cl^-
K_sp = [Ag⁺][Cl^-]
Let the concentration of [Ag⁺] be 'C' mol L^-1.
As per the stoichiometry, if [Ag⁺] = C, then [Cl^-] also equal to 'C' mol L^-1.
Ksp = C.C
⇒ K =C²
We know that the concentration (in mol dm^-3) is related to the molar and sepcific conductance by the following expressions
\({ \Lambda }_{ 0 }=\frac { k\times { 10 }^{ -3 } }{ C(in \ mol \ { L }^{ -1 }) } \)
(or)
\(C=\frac { k\times { 10 }^{ -3 } }{ \Lambda } \)
Substitute the concentration value in the relation K_sp =C²
\({ K }_{ sp }={ \left( \frac { k\times { 10 }^{ -3 } }{ \Lambda } \right) }^{ 2 }\)

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:
\(\frac { { 5.55\times 10 }^{ -3 }\times 1000 }{ 1 } =5.55\times { 10 }^{ -3 }\times { 10 }^{ -3 }\)