‌ (a) Given : ‌A=0.625cm2,l=50‌cm
R=5×103‌ohm,ρ=?
m=0.05m,K=?
∧m=?
‌ Cell constant ‌=‌

ℓ

A

=‌

50‌cm

0.625cm2

=80cm−1
‌ Resistivity ‌=‌

R

‌ cell constant ‌

‌ or ‌‌

R×A

l

⇒‌

5×103×0.625

50

⇒‌

5×103‌ ohm ‌

80cm−1

⇒62.5‌ ohm ‌cm
⇒62.5‌ ohm cm. ‌
‌ Conductivity ‌=‌

1

‌ Resistivity ‌

×‌

l

A

‌‌

1

5×103

×‌

50

0.625

=‌

50

5×103×625×10−3

‌=‌

10

625

=0.016 scm−1
Molar conductivity (Λm)=‌

103K

M

Λm=‌

K

M

×1000=‌

10×1000

625×0.05

=320 sm2‌mol
(b) Given: E°Cu2+∕Cu+0.34V
‌E°(1∕2Cl2∕Cl−)=+1.36V
‌E°H+∕H2(g)′‌Pt=0.00V,E°(1∕2O2∕H2O)=+1.23V
At cathode:
‌Cu(aq)2++2e−→Cu( s);E∘=0.34
‌H(aq)++e−→‌

1

2

H2(g);E∘=0.000V
The reaction with a higher value of E∘ takes place at the cathode, so deposition of copper will take place at the cathode.
At anode: The oxidation reactions are possible at the anode.
Cl(aq)−→‌

1

2

Cl2(g)+e−;E∘=1.36V
2H2O(l)→O2(g)+4H+(aq)+4e−; E∘=+1.23V
At the anode the reaction with a lower value of E∘ is preferred. But due to the over potential of oxygen, Cl−gets oxidised at anode to produce Cl2 gas.
OR
(a)
‌Zn( s)∕Zn2+(0.1M)∥ (0.01M)Ag+∕Ag( s)
‌E°Zn2+∕Zn=−0.76V
‌E°Ag+∕Ag=+0.80V‌‌emf=?
‌E‌cell ‌=E°‌cell ‌−‌

0.0591

n

‌log‌

‌ [Anode ‌]

[‌ Cathode ‌]

‌E°‌cell ‌=E°‌cathode ‌−E°‌anode ‌
‌=E°Ag∕Ag−E°Zn∕Zn
‌=0.80−(−0.76)=1.56V
‌E‌cell ‌=1.56−‌

0.0591

2

‌log‌

[Zn2+]

[Ag+]2

‌=1.56−‌

0.0591

2

‌log‌

[0.1]

[0.01]2

‌=1.56−0.0295‌log‌1000
‌=1.56−3(0.0295)
‌=1.56−0.09=1.4715
(b) ϒ is a weak electrolyte as n dilution complete dissociation of weak electrolyte takes place and thus a sharp increase in molar conductivity while in strong electrolyte it has already dissociated completely. So on dilution molar conductivity does not rises much.