(i) First, find E⁰_cell
(ii) Then find \(E^{ 0 }_{ anode }\) by using the formula
\(E^{ 0 }_{ cell }=E^{ 0 }-_{ cathode }E^{ 0 }_{ anode }\)
Given E_cell = 0.059V; E⁰_cu2+/cu = +0.34V
[Ni²⁺] = 0.01M and [cu²⁺] = 0.1M
E_cell = E_⁰cell - \(\frac { 0.059 }{ 2 } log\frac { \left[ Ni^{ 2+ }(aq) \right] }{ \left[ Cu^{ 2+ }(aq) \right] } \)
\(0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { \left( 0.01 \right) }{ 0.1 } \left( \because n=2 \right) \)
\(0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { 1 }{ 10 } \left[ \because log10^{ -1 }=-1 \right] \)
\(\because 0.059=E^{ 0 }_{ cell }+0.0295 \times 1\)
\(E^{ 0 }_{ cell }=0.059-0.0295\)
= 0.0295 V = 0.03 V
\(E^{ 0 }_{ cell }=E^{ 0 }_{ cathode }E^{ 0 }_{ anode }\)
0.03 = 0.34 - E⁰_anode
or \(E^{ 0 }_{ anode }=E^{ 0 }_{ Ni2+/Ni }=0.34-0.03\)
_{= 0.31 V}

(a) (i) Molar conductivity of a solution at a given concentration is the conducting power of all the ions produced by 1 mol of an electrolyte.
(ii) Secondary battery can be recharged by passing current through it in opposite direction so that it can be used again.
(iii) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.
(b) (i) Faraday's first law of electrolysis states that the amount of chemical reaction which occurs at any electrode during electrolysis by current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). 
(ii) According to Kohlrausch law of independent migration of ions limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte

(i) Mira's father is right as it prevents the iron rods from corrosion.
(ii) Galvanization and cathodic protection.
(iii) Fe₂O₃.x H₂O
(iv) Scientific knowledge and care for his property.

(i) India is a tropical country and summer extend over a long period. This means solar power plants are very effective and can be a better substitute of electricity.
(ii) We must hold seminars with the help of NGO's and other organisations. Educate people about the use of solar heaters, solar generators, solar cookers and solar batteries. We can explain the advantages of the use of solar power over electricity.

Cu⁺ + e^-⇾ Cu; ∆G^o₃=?
(i)------(ii) gives the required result, i.e.,ΔG₃^° = ∆G⁰₁ - ΔG₂^° = [-0.674 - (- 0·153)] F = - 0·521 F
\(-nFE^0_{Cu^+/CU}=-0.5621F\ or\ E^0_{CU^+/Cu}=0.521V\)

Comparing oxidation potential of iron with those of metals A, B and C, iron has higher oxidation potential than C only. Hence, when coating of C is broken, rusting will become faster.

Tarnish will be removed if the following reaction takes place :
\(Al+{ { Ag }_{ 2 } }S\longrightarrow { Al }^{ 3+ }+2Ag+{ S }^{ 2- }\)
EMF of this cell reaction = \({ E }_{ { Ag }_{ 2 }S/{ 2Ag },{ S }^{ 2- } }^{ 0 }-{ E }_{ { Al }^{ 3+ }/{ Al } }^{ 0 }=-0.71-(-1.66)V=+0.95V\)
As EMF is positive, the reaction will take place and tarnish will be removed.

(a) Molar Conductivity: It is defined as the conducting power of all the ions produced by one gram mole of an electrolyte in a solution. It is denoted by Λ_m'
\(∧_m={k\over C}\times 1000\ S\ m^2\ mol^{-1}\)
where 'K' is electrolytic conductivity of solution and 'C' is concentration of the solution expressed in mol L^-l (or mol dm^-3).

Variation of conductivity and molar conductivity with concentration.
In case of strong electrolyte, A_m increases a little on dilution as shown in figure by straight line (decrease in cone.) because number of ions do not increase appreciably, only mobility of ions increases where as in case of weak electrolyte, number of ions as well as mobility of ions increase appreciably, therefore, A_m increases sharply as shown in the figure in form of a curve.
(b) R = 1500 Ω, K = 0.146 x 10^-3 S cm^-1
\(k={1\over R }\times{l\over A}\Rightarrow {l\over A}= K\times R = 0.146 x 10%^{-3} S cm^{-1} \times 1500 ohm\)
\({l\over A}={219\times10^{-3}cm^{-1}}=0.219\ cm^{-1}\)