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.

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

(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.

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.