Standard electrode potential values :

${\mathrm{Fe}}^{2+}/\mathrm{Fe}, \mathrm{E}°=-0.44 \mathrm{V}, {\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}, \mathrm{E}°=0.77 \mathrm{V}$

If ${\mathrm{Fe}}^{2+}, {\mathrm{Fe}}^{3+}$and $\mathrm{Fe}$ blocks are kept together, it will lead to 

On the basis of the following $\mathrm{E}°$ values, the strongest oxidizing agent is

$$\begin{gathered} {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}\to {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}+{\mathrm{e}}^{-1} ; \mathrm{E}°=-0.35 \mathrm{V} \\ {\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-1} ; \mathrm{E}°=-0.77 \mathrm{V} \end{gathered}$$

Standard reduction potentials of the half reactions are given below :

$$\begin{gathered} {\mathrm{F}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{F}}^{-}; \mathrm{E}°=+2.85 \mathrm{V} \\ {\mathrm{Cl}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}; \mathrm{E}°=+1.36 \mathrm{V} \\ {\mathrm{Br}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{Br}}^{-}; \mathrm{E}°= +1.06 \mathrm{V} \\ {\mathrm{I}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{I}}^{-}; \mathrm{E}°= +0.53 \mathrm{V} \end{gathered}$$

The strongest oxidising and reducing agents respectively are

(i) EMF of cell$=$(Oxidation potential of anode)$-$(Reduction potential of cathode)
(ii) EMF of cell$=$(Oxidation potential of anode)$+$(Reduction potential of cathode)
(iii) EMF of cell$=$(Reduction potential of anode)$+$(Reduction potential of anode)
(iv) EMF of cell$=$(Oxidation potential of anode)$-$(Oxidation potential of cathode)

Which of the above relations are correct?

Given: $$\begin{gathered} {\mathrm{Zn}}^{2+}| \mathrm{Zn}=-0.76 \mathrm{V}, \\ {\mathrm{Cu}}^{2+}| \mathrm{Cu} =+0.34 \mathrm{V}, {\mathrm{Fe}}^{2+}| \mathrm{Fe}=-0.41 \mathrm{V} \end{gathered}$$