For the reaction,

Cu²⁺ + 2e^$$-$$ $$ \to $$ Cu,  E^o = 0.337 V

$$\Delta $$G^o = - nFE^o

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu²⁺ + e^$$-$$ $$ \to $$ Cu⁺,  E^o = 0.153 V

$$\Delta $$G^o = - nFE^o

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu²⁺ + e^$$-$$ $$ \to $$ Cu⁺

$$\Delta $$G^o = –0.521 F = –nFE°

$$ \Rightarrow $$ E° = 0.52 V

At Anode:

C₅H₁₂ + 10H₂O $$ \to $$ 5CO₂ + 32H⁺ + 32e^-

At Cathode:

8O₂ + 32H⁺ + 32e^- $$ \to $$ 16H₂O
-------------------------------------------------
C₅H₁₂(g) + 8O₂(g) $$ \to $$ 5CO₂(g) + 6H₂O(l)

$$\Delta $$G = 5×$$\Delta $$G_CO2 + 6 $$\Delta $$G_(H2O) – [$$\Delta $$G_(C5H12) +8 × $$\Delta $$G_O2 ]

= 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

= – 3387 kJ mol^–1

= – 3387 × 10³ J mol^–1

$$\Delta $$G = - nFE^o_cell

From the overall equation we find n = 32

– 3387 × 10³ = -32 $$ \times $$ 96500 $$ \times $$ E^o_cell

E^o_cell = $${{ - 3387 \times {{10}^3}} \over { - 32 \times 96500}}$$ = 1.0968 V

Kohlrausch’s law states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the component ions.

$$\lambda $$_{$$\infty $$} = $$\lambda $$_a + $$\lambda $$_c

where, $$\lambda $$_a = equivalent conductance of the anion

$$\lambda $$_c = equivalent conductance of the cation.

Substances which have higher reduction potential are stronger oxidizing agent.

[Fe(CN)₆]^4$$-$$ $$ \to $$ [Fe(CN)₆]^3$$-$$ + e^$$-$$;  E^o = $$-$$0.35 V

Fe²⁺ $$ \to $$ Fe³⁺ + e^$$-$$;  E^o = $$-$$0.77 V

Higher the +ve reduction potential, stronger will be the oxidising agent. Oxidising agent oxidises other compounds and gets itself reduced easily.