E_cell = E^o_cell - \(\frac{0.0591}{2} \log [\frac{Zn^{2+}}{Cu^{2+}}]\)
E_l = E^o_cell - \(\frac{0.0591}{2} \log [\frac{10^{-2}}{1}]\)
E_l = E^o_cell + 0.0591 ...(1)
E₂ = E^o_cell - \(\frac{0.0591}{2} \log [\frac{1}{10^{-2}}]\)
E₂ = E^o_cell - 0.0591 ...(2)
E₁ > E₂
Zn_(s)| ⟶ Zn²⁺_(aq) + 2e^-
Cu²⁺_(aq) + 2e^- ⟶ Cu_(s)
Zn_(s) + Cu²⁺_(aq)  ⟶ Zn²⁺_(aq) + Cu_(s)

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

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

a) Equivalent conductance
 \(\lambda_{\infty} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} =\frac{1}{3} \lambda_{\infty} \mathrm{Al}^{3+}+\frac{1}{2} \lambda_{\infty} \mathrm{SO}_{4}^{2-}  \)
\(=\left(\frac{1}{3} \times 189\right)+\frac{1}{2}(160)  \)
\(=63+80=143 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~g} \mathrm{eq}^{-1} \)
 b) Molar conduçtance
\(\mu_{\infty} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} =2 \mu_{\infty} \mathrm{Al}^{3+}+3 \mu_{\infty} \mathrm{SO}_{4}^{2-} \)
=2(189) + 3(160) 
=378 + 480
= \(858 \mathrm{~S} \mathrm{~cm}{ }^{2} \mathrm{~mol}^{-1}\)

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⁺
\(=\frac{0.1}{250} \times 1000=0.4 \mathrm{M}\)

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

(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

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

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.

Cathodic protection:
In this technique, unlike galvanising the entire surface of the metal to be protected need not be covered with a protecting metal. Instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded. This known as sacrificial protection. (or) Cathodic protection.

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