It has higher capacity and longer life. Used in pacemakers, electronic watches, cameras etc.

It is an Galvanic cell.

For a uni - univalent electrolyte such as NaCl, the Kohlrausch's law is expressed as
\({ ({ \Lambda }_{ m }^{ o }) }_{ NaCl }=({ { \lambda }_{ m }^{ o })_{ { Na }^{ + } }+{ ({ \lambda }_{ m }^{ o }) }_{ { Cl }^{ - } } }\)

\(\alpha =\frac { { \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } \)

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

Molar conductance = Λ_{m =} \( \frac{k\ (Sm^{-1})\times10^{-3}}{M} mol^{-1}m^{3} \)
\(= \frac{(12.04 \times 10^{-2} Sm^{-1}) \times 10^{-3} (mol^{-1}m^{3})} {0.025}\)
= 481.6 x 10^-5 Sm²mol^-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

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) 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)

1. Daniel cell is a galvanic cell. This is a voltaic cell also.
(a) The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.
(i) Oxidation half cell: A metallic zinc strip  that dips into an aqueous solution of zinc sulphate taken in a beaker, as shown in Figure
(ii) Reduction half cell: A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker, as shown in Figure

(iii) Joining the half cells:
(a) The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolyte such as KCI, Na₂SO₄ etc.,
(b) The ions of inert electrolyte do not react with other ions present in the half I cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.
(c) When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.
(iv) Anodic oxidation:
(i) zinc strip acts as the anode.
(ii) Here,oxidation occurs.
The electrode at which the oxidation occur 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).
\(Zn_{ (s) }\longrightarrow { { Zn }^{ 2+ }_{ (aq) }+{ 2e }^{ - } } \)  (loss of electron-oxidation)  
(v) Cathodic reduction:
As discussed earlier. 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).
\({ Cu }_{ (aq) }^{ 2+ }+{ 2e }^{ - }\longrightarrow { { Cu }_{ (s) } } \)(gain of electron - reduction)
b) When a Zinc metal strip is placed in a copper sulphate solution, the blue colour of the solution fades and the copper is deposited on the zinc strip as red - brown crust due to the following spontaneous chemical reaction.
\(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{CuSO}_{4(\mathrm{aq})} \rightarrow \mathrm{ZnSO}_{4(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})}\)
The energy produced in the above reaction is lost to the surroundings as heat.
In the above redox reaction, Zinc is oxidised to Zn²⁺ ions and the Cu²⁺ ions are reduced to metallic copper. The half reactions are represented as below.
\(\mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \text {(oxidation) }  \)
\(\mathrm{Cu}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})} \text { (reduction) }\)
If we perform the above two half reactions separately in an apparatus as shown in figure, some of the energy produced in the reaction will be converted into electrical energy.

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) 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.