How can I find the emf of a cell given the individual reduction and oxidation reactions and their respective potentials? [duplicate]

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. $$E^\circ(\ce{Fe^3+}/\ce{Fe^0})?$$

marked as duplicate by Mithoron, Todd Minehardt, andselisk♦May 19 at 5:34

• Determine the Gibbs energy of reaction for each against the standard hydrogen electrode, add them up and determine the standard potential from the Gibbs energy of the summed reactions. – Karsten Theis May 18 at 4:05
• You can’t combine the two in a cell because both are reduction half reactions. You could do a disproportionation or comproportionation, maybe. – Karsten Theis May 18 at 4:17

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:

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

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

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

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

• Technically, a half reaction does not have a Gibbs energy of reaction. I think the numerical answer is correct, though, – Karsten Theis May 18 at 4:20