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

(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.

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

Ni	Cr
\(\mathrm{m}_{\mathrm{Ni}_{\mathrm{i}}} =2.935 \mathrm{~g} \) \(\mathrm{E}_{\mathrm{Ni}} =\frac{58.74}{2} \) \(=29.37 \mathrm{~g} \mathrm{eq}^{-1}\)	\( \mathrm{m}_{\mathrm{cr}} =x \) \(\mathrm{E}_{\mathrm{cr}} =\frac{52}{3} \) \(=17.33 \mathrm{~g} \mathrm{eq}^{-1}\)

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

1. Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is,
xA + yB ⇌ IC + mD
2. The reaction quotient Q for the above reaction is given below
\(Q=\cfrac { \left[ C \right] ^{ ' }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)  .....(1)
3. We have already learnt that,
ΔG = ΔG + RT In Q .........(2)
4. The Gibbs free energy can be related to the cell emf as follows.
[ஃ equation (1) and (2)]
ΔG = - nFE_cell; ΔG^o = - nFE^o_cell
5. Substitute these values and Q from (1) in the equation (2)
(2) ⇒\(-nF{ E }_{ cell }=-n{ FE }^{ o }_{ cell }+RT\quad In\cfrac { \left[ C \right] '\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) ....(3) 
6. Divide the whole equation (3) by (-nF)
(4) ⇒\({ E }_{ cell }={ E }_{ cell }-\cfrac { RT }{ nF } In\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) 
(or)
\({ E }_{ cell }={ E }_{ cell }-\cfrac { 2.303RT }{ nF } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) .....(4)
7. The above equation (4) is called the Nernst equation
8. At 25°C (298 K), the above equation (4) becomes,
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 2.303\times 8.314\times 298 }{ n(96500) } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)    
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 0.0591 }{ n } log\cfrac { { \left[ C \right] }^{ l }{ \left[ D \right] }^{ m } }{ { \left[ A \right] }^{ x }\left[ B \right] ^{ y } } \) ......(5) 
\(\left[\begin{array}{l} \text { Where } \\ \therefore \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ \mathrm{~T}=298 \mathrm{~K} \\ \mathrm{IF}=96500 \mathrm{C} \mathrm{mol}^{-1} \end{array}\right]\)

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.