For 1:1 electrolyte K_sp = S²
\(=[\frac{ K \times 10^{-3}}{Λ^o}]^2\)
\(=[\frac{1.85 \times 10^{-5} \times 10^{-3}}{14 \times 10^{-3}}]^2\)
= (0.1321 x 10^-3)²
= 0.01745 x 1010^-10
= 1.74 x 10^-12

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

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.

mass of 1 mole of CI₂ gas = 71
∴ mass of 0.1 mole of Cl₂ gas = 7.1 g mol^-1 
m = Zlt
t = m/ZI
\(= \frac{7.1}{\frac{71}{2 \times 96500} \times 3}\) (2Cl^- ➝ Cl₂ + 2e^-)
\(= \frac{2 \times 96500 \times 7.1}{71 \times 3}\)
= 6433.33s = 107.2 min

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

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

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

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

(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)

(i) In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200^oC and 20-40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes.
(ii) Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.
Oxidation occurs at the anode:
\(2 \mathrm{H}_{2(\mathrm{~g})}+4 \mathrm{OH}_{(a q)}^{-} \rightarrow 4 \mathrm{H}_{2} \mathrm{O}_{(l)}+4 \mathrm{e}^{-}\)
Reduction occurs at the cathode:
\(\mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(t)}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}_{(\mathrm{aq})}^{-}\)
(iii) The overall reaction is \(2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(1)}\)
(iv) The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H₂-O₂ fuel cell. Other fuel cells like propane -O₂ and methane O₂ have also been developed.

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

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.

Given : I = 1.608A;
t = 50 min = 50 \(\times\) 60 = 3000sec ; S = 250 mL;
C = 0.5 M ; \(\eta=100%\)%
i) Q = It     (I = Q / t)
= 1.608 x 3000 = 4824 Columb
No. of Faradays of electricity \(=\frac{4824}{96500}=0.04 \mathrm{~F}\)
ii) Electrolysis of CuSO₄
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})}\)
\(\therefore\) 2 F electricity will deposit 1 mole of Cu²⁺ to Cu ​​​​
\(\therefore\) 0.05 F electricity will deposit 
\(x=\frac{1 \times 0.05}{2}=0.025\) mole
Initial No. of moles of Cu²⁺ in 250 ml of solution \(=\frac{0.5}{1000} \times 250 \) mole = 0.125 moles
\(\therefore\) No. of moles of Cu²⁺ after elettrolysis = 0.125 - 0.025 = 0.1 mole
\(\therefore\) concentration of Cu²⁺
\(=\frac{0.1}{250} \times 1000=0.4 \mathrm{M}\)

(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.

(i) The Conductance of HCl will be more, because H⁺ ion has the maximum mobility of all the ions due to its smallest size and mass.
(ii) At \(25^{0} \mathrm{C}: \mathrm{H}^{+}=36.23 \mathrm{~m}^{2} \mathrm{~S}^{-1} \mathrm{~V}^{-1}\)
(iii) The conductance depends upon
1) Nature of electrolyte
2) Concentration
3) Mobility of ion
4) Temperature

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) 

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

(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.