Given : $$2Hg(l) + 2F{e^{3 + }}(aq) \to Hg_2^{2 + }(aq) + 2F{e^{2 + }}(aq)$$

$$[F{e^{3 + }}] = 1.0 \times {10^{ - 3}}(M)$$

$$\therefore$$ [Fe³⁺] at equilibrium = 5% of 1.0 $$\times$$ 10^$$-$$3 (M)

$$ = {5 \over {100}} \times 1.0 \times {10^{ - 3}}(M) = 5 \times {10^{ - 5}}(M)$$

$$\therefore$$ $$[F{e^{2 + }}] = (1.0 \times {10^{ - 3}}) - (5 \times {10^{ - 5}}) = 0.95 \times {10^{ - 3}}$$

$$\therefore$$ $$[Hg_2^{2 + }]$$ at equilibrium $$ = {{0.95 \times {{10}^{ - 3}}} \over 2}M$$

$$\therefore$$ $${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[Hg_2^{2 + }]{{[F{e^{2 + }}]}^2}} \over {{{[F{e^{3 + }}]}^2}}}$$

At equilibrium, $${E_{cell}} = 0$$

$$\therefore$$ $$E_{cell}^0 = {{0.059} \over 2}\log {{\left[ {{{0.95 \times {{10}^{ - 3}}} \over 2}} \right]{{[0.95 \times {{10}^{ - 3}}]}^2}} \over {{{[5 \times {{10}^{ - 5}}]}^2}}}$$

or, $$E_{cell}^0 = {{0.0591} \over 2}\log {{{{[95]}^3} \times {{10}^{ - 5}}} \over {50}} = - 0.0276$$

Also, $$E_{cell}^0 = E_{F{e^{3 + }}|F{e^{2 + }}}^0 - E_{Hg_2^{2 + }|Hg}^0$$

or, $$ - 0.0276 = 0.77 - E_{Hg_2^{2 + }|Hg}^0$$

or, $$E_{Hg_2^{2 + }|Hg}^0 = 0.77 + 0.0276 = 0.7976\,V$$