(a) (i) When the concentration of the electrolyte approaches zero, the molar conductivity is termed as limiting molar conductivity. It is represented by Λm.
(ii) Fuel cell: Fuel cells are the galvanic cells or electrochemical cells that transform the chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc.
Given,
For 0.1‌mol‌L−1‌KCl solution
Resistance (R)=100Ω
Conductivity (k)=1.29×10−2Ω−1cm−1
Cell constant (G*)=k×R
‌=1.29×10−2Ω−1cm−1×100Ω
‌=1.29cm−1
For 0.02‌mol‌L−1‌KCl solution
Resistance (R)=520Ω
‌ Conductivity ‌=‌

‌ Cell constant ‌(k)

‌ Concentration ‌(C)

K‌=‌

1.29cm−1

520Ω

K‌=2.48×10−3Ω−1cm−1
(C)‌=0.02‌mol‌L−1
‌=0.02×10−3‌mol‌cm−3
Concentration (C)=0.02‌mol‌L−1
=0.02×10−3 ‌mol‌ ‌cm−3
Molar conductivity (Λm) ‌=‌

‌ Conductivity ‌(k)

‌ Concentration ‌(C)

‌=‌

2.48×10−3Ω−1cm−1

0.02×10−3‌mol‌cm−3

Λm‌=124Ω−1cm2mol−1
OR
(a) Faraday's first law of electrolysis states that "The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed."
m=Z×Q
where, m= mass of a substance deposited or liberated at an electrode.
Q= amount of charge passed through it and
Z= electrochemical equivalent
The reduction of one mol of Cu2+ to Cu can be represented as:
Cu2++2e−→Cu
Since, 2‌mol of electrons are involved in the reduction, so the amount of charge required is 2F.
(b) The given cell reaction can be represented as
Mg( s)+Cu2+(aq)→Mg2+(aq) +Cu( s)
E‌cell ‌=E∘−‌

2.303‌RT

nF

‌log‌

Mg2+

Cu2+

E‌cell ‌=2.71−‌

2.303×0.0831×298

2×96500

log‌

0.1

0.01

E‌cell ‌=2.71−‌

0.0591

2

‌log‌10
E‌cell ‌=2.68V