Deriving reduction potential from redox reactions

From a past USNCO test:



> 
> **40**. What is the standard reduction potential of $\ce{Hg^2+(aq)}$ to $\ce{Hg(l)}$?
> 
> 
> \begin{align}
> \ce{2 Hg^2+(aq) + 2 e- &-> Hg2^2+(aq)} &\quad E^\circ &= \pu{+0.90 V} \\
> \ce{Hg2^2+(aq) + 2 e- &-> 2 Hg(l)} &\quad E^\circ &= \pu{+0.80 V}
> \end{align}
> 
> 
> **(A)** $\pu{+1.70 V}$  
> 
> **(B)** $\pu{+0.85 V}$  
> 
> **(C)** $\pu{+0.10 V}$  
> 
> **(D)** $\pu{-0.10 V}$
> 
> 
> 


While I understand the method behind it (converting potentials to free energy first, adding, then converting back to $E),$ I don't understand why we have to do that in the first place, since when calculating a typical $E\_\text{cell}$, one just adds the potentials.


I've heard people say it's because the number of electrons transferred affects the potential, but it doesn't really make sense to me, as a typical potential is intensive. Why is this case the exception?


The answer is $\pu{0.85 V},$ as calculated from the $\Delta G$ method and $\pu{1.70 V}$ by simply adding the reduction potentials.




The electrode potential and the redox potential are closely related but different things.


The redox potential is just an alternative to Gibbs energy of particular redox reaction. It flips signs with flipping reaction direction. It generally needs conversion to Gibbs energy to have additivity to calculate redox potential of other reactions, because of not being additive for different electron counts.


For $$\ce{ox + n/2 H2 -> red + n H+}$$ is


$$\Delta G\_\mathrm{r}=-nFE$$


where F is the Faraday constant and E is the redox potential.


as from physics we know that


$$E=qU$$


where E is energy, q is passed charge and U is voltage.


Therefore, if we calculate reaction Gibbs energy as difference of other 2 reaction Gibbs energies:


$$\Delta G\_\mathrm{r,3}=\Delta G\_\mathrm{r,2} - \Delta G\_\mathrm{r,1}$$


then converted to redox potentials it looks this way:


$$-n\_3FE\_3 = -n\_2FE\_2 - (-n\_1FE\_1)$$


and divided by $-F$ :


$$n\_3E\_3 = n\_2E\_2 - n\_1E\_1$$


The electrode potential is the ordinary physical relative electrostatic potential. It's value is in an ideal case value of redox potential of ongoing half reaction in direction $\ce{ox -> red}$, but is affected by various, mostly kinetic factors.


The cell voltage is the difference of electrode potentials without dependency on electron counts of involved redox half reactions. It is not calculation of one redox potential from other two ones. The cell voltage is not a redox potential of a half reaction.


 2

### Comments


This solution is wrong following the discussion in the comments. I have bookmarked the answer and will revisit it soon.


### Definition


Let us start with the definition of $E^\circ = -\dfrac{\Delta G^\circ}{nF}$.


### Case of the same number of electrons


In the same cell, same number of electrons are being *released* and *captured* and the effect of electrons cancels out:


$$E\_1^\circ = -\dfrac{\Delta G\_1^\circ}{nF};E\_2^\circ = -\dfrac{\Delta G\_2^\circ}{nF}\text{.}$$


Since the denominators are the same, we can add the two as


$$E^\circ = E\_1^\circ + E\_2^\circ= -\dfrac{\Delta G\_1^\circ}{nF} - \dfrac{\Delta G\_2^\circ}{nF} = -\dfrac{(\Delta G\_1^\circ + \Delta G\_2^\circ)}{nF}\text{,}$$


which makes sense.


### Case of different number of electrons


Now consider the general case for two electrochemical preocesses:


$$E\_1^\circ = -\dfrac{\Delta G\_1^\circ}{n\_1F};E\_2^\circ = -\dfrac{\Delta G\_2^\circ}{n\_2F}\text{.}$$


In this case


$$E^\circ = E\_1^\circ + E\_2^\circ= -\dfrac{\Delta G\_1^\circ}{n\_1F} - \dfrac{\Delta G\_2^\circ}{n\_2F} = -\dfrac{n\_2\Delta G\_1^\circ + n\_1\Delta G\_2^\circ}{(n\_1 \times n\_2)F}\text{,}$$


 0

The title of your question is incorrect. You are determining a reduction potential from two related reduction potentials.


The fastest way to do this is to take a weighted average. To complete the reaction, electrons participate either in the first or the second reaction, in a 1:1 ratio.


A related question is [here](https://chemistry.stackexchange.com/q/49871), also from UNSCO. In this case, one step requires two electrons and the other just one, so you have to do a 2:1 weighted average.


In some cases, the species undergo [comproportionation](https://chemistry.stackexchange.com/q/105568) or [disproportionation](https://chemistry.stackexchange.com/a/31694). Here is the example for copper species:


$$\ce{Cu^2+ (aq) + e– <=> Cu+ (aq)}\ \ E^\circ = +\pu{0.15 V}$$
$$\ce{Cu+ (aq) + e– <=> Cu (s)}\ \ E^\circ = +\pu{0.52 V}$$


The +1 oxidation state is not stable, it falls apart into the zero and +2 oxidation state under these standard conditions.


 0

The electrode potential measures the energy change involving the transfer of one mole of electrons. The potential for Hg[II] to Hg[I] is one mole transferred; the potential for Hg[I] to Hg[0] is a second mole transferred. Concatenating the equations and adding the voltages gives the energy change for two [2] electrons so the composite voltage is divided by 2 to give the standard potential.


Should the electrons differ in each half reaction such as Fe to Fe[II] followed by Fe[II] to Fe[III] the 2 electron potential is multiplied by 2, the one electron potential added and the sum divided by three.


 0