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
= \(858 \mathrm{~S} \mathrm{~cm}{ }^{2} \mathrm{~mol}^{-1}\)

Hydrolysis of water
At anode:
\(2H_{2}O\rightarrow 4H^{+}+O_{2}+4e^{-}\) ..... (1)
At cathode:
\(2H_{2}O+2e^{-}\rightarrow H_{2}+2OH^{-}\) ....(2) 
Overall reaction 
\(6H_{2}O\rightarrow 4H^{+}+4OH^{-}+2H_{2}+O_{2}\)
(or)
Equation (1) +(2) \(x^2 \Rightarrow 2H_{2}O\rightarrow 2H_{2}+O_{2}\)
\(\therefore\) According to faradays Law of electrolysis, to electrolyse two mole of Water (36g ≃ 36 mL of H₂O), 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = \(4\times 96500\)C.
\(\therefore\) When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to \(\frac{4\times96500\quad C}{36 \quad mL}\times9\times10^{12}L\)
\(=\frac{4\times96500\times9\times10^{12}}{36\times10^{-3}}C\)
= \(96500\times10^{15}C\)
\(\therefore\) Given that in 1 second, \(2\times10^{6}\) C is generated therefore, the time required to generate \(96500 \times 10^{15}\) C is = \(\frac{1\quad S}{2\times 10^{6}C}\times 96500 \times10^{15}C\)
=\(48250 \times 10^{9} S\)
\(\therefore\) Number of years = \(\frac{48250 \times 10^{9}}{365 \times 24 \times 60 \times 60}\)   
=\(1.5299 \times 10^{6}\) years
1 year = 365 days
= 365\(\times\)24 hours
= 365\(\times\)24\(\times\)60 min
= 365\(\times\)24\(\times\)60\(\times\)60 sec.

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

By Faraday II law of electrolysis:
\(\frac{\mathrm{m}_{\mathrm{Ni}}}{\mathrm{E}_{\mathrm{Ni}}}=\frac{\mathrm{m}_{\mathrm{cr}}}{\mathrm{E}_{\mathrm{cr}}} \)
\(\mathrm{Ni}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}_{(\mathrm{s})}  \)
\(\mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}_{(\mathrm{s})}\)

\( \frac{2.935}{29.37}=\frac{\mathrm{x}}{17.33} \)
\(x =\frac{2.935 \times 17.33}{29.37} \)
=1.732 g

(i) Oxidation at anode:
\(2H_{2(g)}+4OH^{-}_{(aq)}\rightarrow 4H_{2}O_{(I)}+4e^{-}\)
(ii) 1 mole of hydrogen gas produces 2 moles of electrons at 25⁰C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres
\(\therefore \) no. of moles of hydrogen gas produced
= \(\frac{1 mole}{22.4 litres} \times 44.8 litres\) 
= 2 moles of hydrogen
(iii) \(\therefore \)  2 of moles of hydrogen produces 4 moles of electro i.e., 4F charge.
t = 10 min
t = 10 x 60 sec
t = 600s
We know that Q= It
\(I=\frac{Q}{t}\)
\(=\frac{4F}{10 mins}\)
\(=\frac{4\times96500\quad C}{10\times60\quad s}\)
I = 643.33 A
Electro deposition of copper
\(Cu^{2+}_{(aq)}+2e^{-}\rightarrow Cu_{(s)}\)
(iv) 2F charge is required to deposit
(v) 1 mole of copper i.e., 63.5 g
(vi) If the entire current produced in the fuel cell ie., 4 F is utilised for electrolysis, then \(2\times63.5\) i.e., 127.0 g copper will be deposited at cathode.

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.

(i) The half cell reactions are :
\(2Br^{-} \rightarrow Br_{2}+2e^{-}\)     \(E^{0}_{ox}=-1.09V\) ...(1)
\(2Fe^{3+}+2e^{-}\rightarrow2Fe^{2+}\)  \(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=+0.771V\) ..(2)
(ii) Adding (1) of (2) :
\(2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{2+}+Br_{2}\)   \(E^{0}_{cell}=?\) ...(3)
\(E^{0}_{cell}=E^{0}_{ox}+E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}\)
= (-1.09 + 0.771)V
= -0.319V
(iii) E⁰_cell is – ve; \(\Delta G\) is +ve and the cell reaction is non spontaneous.
(iv) Hence Fe³⁺ cannot oxidises Br^- to Br₂.

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.

(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.

1. \(\kappa=\mathrm{C}\left(\frac{1}{\mathrm{~A}}\right)\)
2. \(\operatorname{let} \frac{l}{A}=x\)
i) \(0.01 \mathrm{M} \mathrm{KCl}: \kappa=0.01 x=10^{-2} x\)
ii) \(0.005 \mathrm{M} \mathrm{KCl}: \kappa=0.005 x=5 \times 10^{-3} x\)
iii) \(0.1 \mathrm{M} \mathrm{KCl} \quad: \kappa=0.1 x=10^{-1} x\)
iv) \(0.25 \mathrm{M} \mathrm{KCl}: \kappa=0.25 x=2.5 \times 10^{-1} x\)
v) \(0.5 \mathrm{M} \mathrm{KCl}: \mathrm{K}=0.5 x=5 \times 10^{-1} x\)
\(\therefore 5 \times 10^{-1} x>2.5 \times 10^{-1} x>10^{-1} x>10^{-2} x>5 \times 10^{-3} x\)
\((ie) 0.5 \mathrm{M} \mathrm{KCl}>0.25 \mathrm{M} \mathrm{KCl}>0.1 \mathrm{M} \mathrm{KCl}>0.01 \mathrm{M} \mathrm{KCl}>0.005 \mathrm{M} \mathrm{KCl}\)

(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) Molar conductivity
Given : C = 0.01 M;
\(\kappa=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \)
\(=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1} \)
\(\lambda_{\text {cation }}^{0}=248.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\lambda_{\text {anion }}^{0}=51.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\Lambda_{m}^{0}=\frac{\kappa \times 10^{-3}}{C} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)
\(=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \)
\(=1.5 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)
ii) \(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)
\(\Lambda_{\mathrm{m}}^{0}=\lambda_{\text {cation }}^{0}+\lambda_{\text {anion }}^{0} \)
\(=(248.2+51.8) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \)
\(=300 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(=300 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}  \)
\(\alpha =\frac{1.5 \times 10^{-3}}{300 \times 10^{-4}}=0.05\)
iii) \(\mathrm{K}_{\mathrm{a}} =\frac{\alpha^{2} \mathrm{C}}{1-\alpha} \)
\(\mathrm{K}_{\mathrm{a}} =\frac{(0.05)^{2} \times(0.01)}{1-0.05}=2.6 \times 10^{-5} \)
(or) 
\(\mathrm{K}_{\mathrm{a}} =\alpha^{2} \mathrm{C}  \)
\(=(0.05)^{2} \times(0.01)  \)
\(\mathrm{K}_{\mathrm{a}} =2.5 \times 10^{-5}\)

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.

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

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

\((E^{0}_{ox})_{Fe^{2+}|Fe} = -0.44\)  and 
\((E^{0}_{red})_{Cu^{2+}|Cu} = 0.34\)
These +ve emf values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.