How can I find the emf of a cell given the individual reduction and oxidation reactions and their respective potentials?
This is my first answer in stack exchange if some mistake comment on.
$$E^\circ_\ce{Fe^3+/Fe} \neq E^\circ_\ce{Fe^3+/Fe^2+} + E^\circ_\ce{Fe^2+/Fe}$$

but

$$\Delta_r G^\circ_1 = \Delta_r G^\circ_2 + \Delta_r G^\circ_3$$

for the three reactions:

$$\ce{Fe^3+ + 3/2 H2 -&gt; Fe + 3H+}\tag{1}$$

$$\ce{Fe^3+ + 1/2 H2 -&gt; Fe^2+ + H+}\tag{2}$$

$$\ce{Fe^2+ +  H2 -&gt; Fe + 2H+}\tag{3}$$

With $$\Delta_r G^\circ = - n F E^\circ , $$ we can establish a relationship between the reduction potentials:

$$n_1 F E^\circ_1 = n_2 F E^\circ_2 + n_3 F E^\circ_3$$

Cancelling F and solving for $E^\circ_1$ allows us to calculate the reduction potential of the iron/iron(III) half reaction:

$$n_1 E^\circ_1 = n_2 E^\circ_2 + n_3 E^\circ_3$$

$$E^\circ_1 = \frac{n_2 E^\circ_2 + n_3 E^\circ_3}{n_1}$$

$$ = \frac{1 \cdot \pu{0.77 V} + 2 \cdot (\pu{-0.44 V})}{3}$$

$$ = \pu{-0.367 V}$$