If I am given the individual oxidation and reduction half-equations, how can I find the emf of the combination? For example:
$$ \begin{align} E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) &= \pu{0.77 V}\\ E^\circ(\ce{Fe^2+}/\ce{Fe^0}) &= \pu{-0.44V} \end{align} $$
How can I find the emf of this redox pair, i.e. $E^\circ(\ce{Fe^3+}/\ce{Fe^0})?$
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:
$$\ce{Fe^3+ + 3/2 H2 -> Fe + 3H+}\tag{1}$$
$$\ce{Fe^3+ + 1/2 H2 -> Fe^2+ + H+}\tag{2}$$
$$\ce{Fe^2+ + H2 -> Fe + 2H+}\tag{3}$$
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:
$$n_1 E^\circ_1 = n_2 E^\circ_2 + n_3 E^\circ_3$$
$$E^\circ_1 = \frac{n_2 E^\circ_2 + n_3 E^\circ_3}{n_1}$$
$$ = \frac{1 \cdot \pu{0.77 V} + 2 \cdot (\pu{-0.44 V})}{3}$$
$$ = \pu{-0.367 V}$$
