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








**This question already has an answer here**:
 
 




[Deriving a reduction potential from two other reduction potentials](/questions/9454/deriving-a-reduction-potential-from-two-other-reduction-potentials)

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Closed 4 years ago.



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:


$$n\_1 E^\circ\_1 = n\_2 E^\circ\_2 + n\_3 E^\circ\_3$$


$$E^\circ\_1 = \frac{n\_2 E^\circ\_2 + n\_3 E^\circ\_3}{n\_1}$$


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


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


[![answer to the question](https://i.stack.imgur.com/RUh54.jpg)](https://i.stack.imgur.com/RUh54.jpg)


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