Given: ΔG = -64.84 x 10³J
n = 2
F = 96495 C
Formula: ΔG = -nF/E^o
Solution:
\({ E }^{ o }=-\frac { \triangle G }{ nF } =\frac { -(64.84\times { 10 }^{ 3 }) }{ 2\times 96495 } \)
= \(\frac { 64840 }{ 2\times 96495 } =0.3359V\)
E^o = 0.3359 V

Given:
E₁ = 0.5 V
Cr³⁺+ 3e^- ➝ Cr_(s).......(1)
E₂ = 0.41 V
Cr³⁺+ e^- ➝ Cr²⁺  .....(2)
The required reaction is,
Cr²⁺+ 2 e^- ➝ Cr_(s)
Then,
Formula:
\({ E }_{ 3 }=\frac { 3{ E }_{ 1 }+{ E }_{ 2 } }{ 2 } \)
Solution:
= \(\frac { 3(0.5)+(0.41) }{ 2 } =\frac { 1.5+0.41 }{ 2 } \)
= 0.955 V
E₃ = 0.955 V

Given: E = 0.301 V; [Cu²⁺] = 0.01M
Formula:
\({ E }_{ { Cu }^{ 2+ }/Cu }^{ o }={ E }_{ { Cu }^{ 2+ }/Cu }^{ }+\frac { 2.303Rt }{ nF } \log\frac { [{ Cu }^{ 2+ }] }{ [Cu] } \)
Solution:
\(=+0.301+\frac { 0.0591 }{ 2 } \log\frac { 0.01 }{ 1 } \)
\({ E }^{ o }=0.301+\frac { 0.059 }{ 2 } \times 2=0.3591V\)
E^o = 0.36 V

Given:
ΔG = -23.125 kJ = -23.125 x 10³J
F = 96495 coulombs
∴ n = 1
Formula: 
ΔG^o = -nFE^o
\(\therefore { E }^{ o }=-\frac { \triangle G }{ nF } \)
Solution: \(=-\left( \frac { -23.125\times { 10 }^{ 3 } }{ 1\times 96495 } \right) \)
\(=\frac { 23.125\times { 10 }^{ 3 } }{ 96495 } \)
= 0.239
E^o = +0.239 V

Given: Cell reaction is
Cd/CdCl₂, 25H₂O / AgCI_(s) Ag E^o is 0.675 V.
Cd ➝ Cd²⁺ + 2e^-
n = 2
F = 96495 coulombs
Formula: ΔG = - nFE
Solutlon: ΔG = - 2 x 96495 x 0.675
ΔG = - 2 x 96495 x 0.675
ΔG = -130.26 kJ

 Sn⁴⁺ + 2e^- ⟶ Sn²⁺ E⁰ = 0.15V
Given: n = 2 electrons 
F = 96495 coulombs
Formula: ΔG = - nFE^O
Solutlon: ∴ ΔG = - 2 x 96495 x 0.15
= 28.948
Free energy = -28.948 kJ.

Given:
λ_∞CH₃COONa = 91.0 ohm^-1 cm²g eq^-1
Formula:
λ_∞CH₃COONa = λ_∞CH₃COONa + λ_∞HCl - λ_∞NaCl
λ_∞HCl = 426.16 ohm^-1 cm²g eq^-1
λ_∞NaCl = 126.45 ohm^-1 cm²g eq^-1
Solution: ∴ λ_∞CH₃COOH
= 91.0 + 426.16 - (126.45)
= 517.16 - 126.45
 λ_∞CH₃COOH = 390.71 ohm^-1cm² g eq^-1

Given:
Specific conductance = K = 5.55 x 10^-3 mho cm^-1
Concentration C = 1M
Volume = 1 litre.
Formula: Equivalent conductance = K x V
= \(\frac { k\times 1000 }{ C } \)
Solution:
\(\frac { { 5.55\times 10 }^{ -3 }\times 1000 }{ 1 } =5.55\times { 10 }^{ -3 }\times { 10 }^{ -3 }\)

Given:
Length = 0.5 cm
Area = 5cm²
Formula: Cell constant = \(\frac { 1 }{ a } \)
Solution: Cell constant = \(\frac { 0.5cm }{ 5{ cm }^{ 2 } } \)
= 0.1 cm^-1
Cell constant =  0.1 cm^-1

Given: Initial concentration of the solution = 1 M
The concentration remaining after electrolysis = 0.667 M
Solution:
∴ The amount deposited = 1 - 0.667 M
= 0.333 M
1F = Faraday = 3 x 0.333 M
= 0.999M
= 1M
∴1 Faraday current is used.

Quantity of electricity, Q = 0.5 F
Concentration of solution, C = 5 M
Solution:
(i) For AgNO₃
1 mol of Ag⁺ = 1F
0.5F = 0.5 mol of AgNO₃
Concentration of AgNO₃ after electrolysis
= 5 - 0.5
= 4.5M
(ii) For CuSO₄
1 mol of CuSO₄ (or) Cu²⁺ = 2F
2F = 1M CuSO₄
0.5F = \(\frac12\) x 0.5 = 0.25 M
Concentration of CuSO₄ after electrolysis
= 5 - 0.25
= 4.75M
(iii) For AlCl₃
1 mol of AlCl₃ Br Al³⁺ = 3F
3F = 1M AlCl₃
0.5F = \(\frac13\) x 0.5
= \(\frac{0.5}{3}\) = 0.167 M
Concentration of AlCl₃ after electrolysis = 5 - 0.167
Concentration of AlCl₃ after electrolysis = 4.833 M

Initial concentration of
AgNO₃ = 1M= IN
Quantity of current 0.75 F
Formula:
1Faraday = 1equivalent mass
Solution:
For IF current In AgNO₃ will be liberated.
For 0.75 F current 0.75 N AgNO₃ will be liberated
The concentration of AgNO₃ remaining
= 1.0 - 0.75 = 0.25 N
ஃ The concentration of AgNO₃ remaining

Given: Electrochemical
equivalent = Z = 2.35 g amp^-1 sec^-1
Time = t = 10 sec.
Current strength = 1 = 5 ampere
Formula: m = Zlt
Solution: 2.35 x 10 x 5 = 117.5 gm
The amount of the substance deposited
= 117.5 g

Amount of the substance deposited,
m = 150 g
current strength = 1 = 10 ampere
Time = t = 1sec.
By Faraday's first law,
Formula: m = Zlt
Solution: \(\therefore Z=\frac { m }{ It } =\frac { 150 }{ 10\times 1 } \)

E = E^O- 2.303 \(\frac { RT }{ 2F } \) log Zn²⁺
E_Zn²⁺/Zn = E_{Zn²⁺/ Zn}- 2.303 \(\frac { RT }{ 2F } \)  log [Zn²⁺]