The reactions at cathode for the deposit of Na and Al are \({ Na }^{ + }+{ e }^{ - }\longrightarrow Na\) and \({ Al }^{ 3+ }+3{ e }^{ - }\longrightarrow Al\)  Thus, 1 faraday deposits 1 mol of Na whereas 3 faradays are required for depositing one mol of Al.

\(Z n^{2+}(a q)+2 e^{-} \rightarrow Z n(s)\)
∵ 2 Faradays. i.e. 2 moles of electrons will deposit 65g of zinc.
∵ 1 Faradays, i.e. 1 mole of electrons will deposit 32.5 g of zinc.

\( \Lambda_m^{\infty} N a_2 S O_4=2 \lambda_m^{\infty} N a^{+}+\lambda_m^{\infty} S O_4^{2-} or\\ \Lambda^{\circ} \mathrm{Na}_2 \mathrm{SO}_4=2 \lambda^{\circ} \mathrm{Na}^{+}+\lambda^{\circ} \mathrm{SO}_4^{2-} \)

The process of slowly eating away of the metal due to the attack of the moisture and atmospheric gases on the surface of the metal resulting in the formation of a compound such as oxides, sulphides, carbonates, sulphates etc., is called corrosion.
The electrochemical phenomenon of rusting of iron can be described as :
At Anode: Fe(s) undergoes oxidation to releases electrons.
Fe_(s)\(\rightarrow\)Fe²⁺_(aq)+2e^-
At Cathode:
O_2(g)+4H⁺+4e^-\(\rightarrow\)2H₂O_(l)
Electrons released at anode move to another metal and reduce oxygen in presence of H⁺ . It is available from H₂CO₃ formed from the dissolution of CO₂ from air into water. H⁺ in water may be available also through the dissolution of other acidic oxides from the atmosphere.
This site behaves as a cathode.
Net Reaction:
Fe_(s)+2H⁺_(aq)+\(\frac{1}{2}\)O_2(g) \(\rightarrow\)Fe²⁺_(aq) 
2Fe²⁺_(s)+\(\frac{1}{2}\)O_2(g)+2H₂O_{(l)​​​​​​​} \(\rightarrow\)Fe₂O_3(s)+4H⁺ ​​​​​​​

K = 5.25 x 10^-5 S cm^-1, M = 2.5 x 10^-4 M.
\(\lambda_{\mathrm{HCOOH}}^{0}=\lambda_{\left(\mathrm{HCOO}^{-}\right)}^{0}+\lambda_{\mathrm{H}}^{0}+\) 
= 349.5 + 50.5 : 400 S cm² mol^-1.
\(\Lambda_{m}=\frac{1000 \kappa}{\mathrm{M}}=\frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}}\) 
= \(\frac{1000 \times 525}{10 \times 2.5 \times 100}\) 
\(\Rightarrow \quad \Lambda_{m}=\frac{525}{2.5}=\frac{5250}{25}=210 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) 
\(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{0}}=\frac{210}{400}=\frac{21}{40}=0.525\) 
\(\Rightarrow \ \alpha\) =  0.525 x 100% = 52.5%

1 mol = \(107.9g\)
1 mol of Ag is deposit by \(96500C\)
If charge of electron is \(1.60\times { 10 }^{ -19 }C\) 
No. of electrons in 1 mol of silver = \(\frac { 96500C }{ 1.60\times { 10 }^{ -19 }C } \) 
= \(6.03\times { 10 }^{ 23 }{ e }^{ - }\)
Avagadro number = \(6.03\times { 10 }^{ 23 }\)

\({ { \wedge }^{ ° } }_{ m }\left( { CaCl }_{ 2 } \right) ={ \lambda }^{ ° }\left( { Ca }^{ 2+ } \right) +2{ \lambda }^{ ° }\left( { Cl }^{ - } \right) \)
\(=119.0S{ cm }^{ 2 }{ mol }^{ -1 }+2\times 76.3S{ cm }^{ 2 }{ mol }^{ -1 }=271.6S{ cm }^{ 2 }{ mol }^{ -1 }\)
\({ { \wedge }^{ ° } }_{ m }\left( { MgSO }_{ 4 } \right) ={ \lambda }^{ ° }\left( { Mg }^{ 2+ } \right) +2{ \lambda }^{ ° }\left( { { SO }_{ 4 } }^{ 2- } \right) \)
\(=106 \ S{ \ cm }^{ 2 }{ \ mol }^{ -1 }+160 \ S{ \ cm }^{ 2 }{ \ mol }^{ -1 }=266 \ S{ \ cm }^{ 2 }{ \ mol }^{ -1 }\)

Nickel-cadmium cell:
t is another type of secondary cell which has longer life than lead storage cell but more expensive to manufacture. The overall reaction during discharge is 
\(Cd(s)+{ 2Ni(OH) }_{ 3 }(s)\rightarrow CdO(s)+{ 2Ni(OH) }_{ 2 }(s)+{ H }_{ 2 }O(l).\)
Merit:
It is easy to handle as it is less bulky than lead storage cell and has longer life than lead storage cell.
Demerit:
It is more expensive than lead storage cell.

Quantity of electricity passed = 9·65 x 60 x 60 C = 34740 C
∴ Ag⁺ ions deposited = 34740/96500 mole = 0.36 mole
Cu²⁺ions deposited = 34740/(2 x 96500) = 0.18 mole
[Ag⁺] left = 1 - 0.36 = 0.64 M
[Cu²⁺] left = 1 - 0.18 = 0.82 M
Cell reaction is: Cu + 2 Ag⁺ ➝ Cu²⁺+ 2 Ag
\(ΔE=E^0_{cell}-E_{cell}={0.0591\over 2}log{[Cu^{2+}]\over [Ag^+]^2}={0.0591\over 2}log{0.82\over (0.64)^2}=8.89\times10^{-3}V\)

(a) Lead storage battery: It is a secondary cell. It consists of a lead anode and a grid of lead dioxide as cathode. A solution of sulphuric acid (38% by mass) is used as electrolyte
The cell reactions when the battery is in use are given below:
Anode : Pb(s) + SO₄^2-(aq) ⇾ Pb504(s) + z«
Cathode: PbO₂ (s) +SO₄^2- (aq) + 4H⁺(aq) + 2e^- ⇾ PbSO₄+2H₂O(I) i.e. overall cell reaction consisting of cathode and anode reactions is :
Pb (s) + PbO₂ (s) + 2H₂SO₄ (oq) ⇾ 2PbSO₄ (s) + 2H₂O(I)
On recharging the battery, the reaction is reversed.
2PbSO₄ + 2H₂O ⇾ Pb(s) + PbO₂(s) + 4H⁺ + 2SO₄^2-
(b)\(E_{cell}=E^0_{cell}-{0.0591V\over 6}log{[Cr^{3+}]^2\over [Cr_2O_7^{2-}][H^+]^{14}}\)
\(=1.33V-{0.0591V\over 6}log{(0.2)^2\over (0.1)(1.0\times10^{-4})^{14}}\)
\(=1.33V-{0.0591V\over 6}log{4\times 10^{-2}\over 1.0\times10^{-57}}\)
\(=1.3V-{0.0591V\over 6}log4\times 10^{55}\)
\(=1.33V-{0.0591V\over6}\times 55.6021\)
\(=1.33V-{3.286V\over6}=1.33V-0.5476V=0.7824V\) 

Reduction potentials in the order Z > Y > X means that Z can be reduced most easily and X least easily, i.e., their oxidizing powers are : Z > Y > X.

Thus, Y will oxidize X but not Z.