Calculate the equilibrium constant for the reaction
Fe²⁺ + Ce⁴⁺ $$\leftrightharpoons$$ Fe³⁺ + Ce³⁺
(given $$E_{C{e^{4 + }}/C{e^{3 + }}}^o$$ = 1.44 V; $$E_{F{e^{3 + }}/F{e^{2 + }}}^o$$ = 0.68 V)

The standard reduction potential for Cu²⁺|Cu is +0.34 V. Calculate the reduction potential at pH = 14 for the above couple. K_sp of Cu(OH)₂ is 1.0 $$\times$$ 10^-19

An excess of liquid mercury is added to an acidified solution of 1.0 $$\times$$ 10^-3 M Fe³⁺. It is found that 5% of Fe³⁺ remains at equilibrium at 25^oC. Calculate $$E_{Hg_2^{2 + }|\,Hg}^o$$, assuming that only reaction that occurs is
2Hg + 2Fe³⁺ $$\to$$ $$Hg_2^{2+}$$ + 2Fe²⁺
(Given $$E_{F{e^{3 + }}|\,F{e^{2 + }}}^o$$ = 0.77 V)

The Edison storage cells is represented as
Fe(s) | FeO(s) | KOH (aq) | Ni₂O₃(s) | Ni(s)
The half-cell reactions are:
Ni₂O₃ + H₂O (l) + 2e^- $$\leftrightharpoons$$ 2NiO(s) + 2OH^-; E^o = +0.40V
FeO(s) + H₂O(l) + 2e^- $$\leftrightharpoons$$ Fe(s) + 2OH^-; E^o = -0.87V
(i) What is the cell reaction?
(ii) What is the cell e.m.f? How does it depend on the concentration of KOH?
(iii) What is the maximum amount of electrical energy that can be obtained from one mole of Ni₂O₃?