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.

Efficiency of a fuel cell ($$\phi $$) = $${{\Delta G} \over {\Delta H}} \times 100$$

Generally, fuel cells are expected to have an efficiency of 100 percent.

RT ln K = nFE°

ln K = $${{nFE^\circ } \over {RT}}$$

= $${{2 \times 0.46} \over {0.0591}}$$

$$ \Rightarrow $$ K = 4 $$ \times $$ 10¹⁵

From the given expression:

At anode : A $$ \to $$ A⁺ + e^–

At cathode : B⁺ + e^– $$ \to $$ B

Overall reaction is : A + B⁺ $$ \to $$ A⁺ + B