Given: K = 0.335;
Eo = (Ag+, Ag) = 0.7991 V
n = 1, F = 96495 C
Formula:
\({ E }^{ o }=\frac { -2.303RT }{ nF } \log K\)
Solution:
\(=\frac { -0.591 }{ 1 } \log 0.335\)
Standard emf of the cell
= - 0.0280 V
Eocell = EoF - EoL; EoR = ?
∴ EoR = Eocell - EoL
= -0.0280 + 0.7991 = 0.771 V
Eo(Fe3+,Fe2+) = 0.771 V.

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: 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: 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.