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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.

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  • $\begingroup$ It seems to me like the current question wording makes it a different question, compared to the OP's one. $\endgroup$
    – Poutnik
    Apr 17, 2023 at 13:59

4 Answers 4

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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.

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  • $\begingroup$ Hello Poutnik. Maybe you forgot to put a minus sign in the last equation? Thereby, you would have $-n_3 E_3 = -n_2 E_2 - (-n_1E_1)$? $\endgroup$ Apr 17, 2023 at 15:47
  • $\begingroup$ Check again. Remember that redox potential sign as very formal parametr depends on half reaction direction and Gibbs energy sign. $\endgroup$
    – Poutnik
    Apr 17, 2023 at 16:53
  • $\begingroup$ I agree it doesn't change anything. But it seemed to me that Ananta saw it without the - sign and replied. Indeed what is additive is $\Delta_r G$, and you get $E_3^0 =(n_2/n_3)E_2^0 - (n_1/n_3)E_1^0$ for the unknown $E_3^0$. The additivity of standard reduction potentials to get an unknown one, in the other post, is a mistake. $\endgroup$ Apr 17, 2023 at 17:02
  • $\begingroup$ That is why I distinguish redox and electrode potentials. $\endgroup$
    – Poutnik
    Apr 17, 2023 at 17:17
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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

$$E^\circ = E_1^\circ + E_2^\circ= -\dfrac{\Delta G_1^\circ}{n_1F} - \dfrac{\Delta G_2^\circ}{n_2F} = -\dfrac{n_2\Delta G_1^\circ + n_1\Delta G_2^\circ}{(n_1 \times n_2)F}\text{,}$$

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  • $\begingroup$ @Poutnik didn't quite understand what you are trying to say here. $\endgroup$
    – ananta
    Apr 17, 2023 at 11:24
  • $\begingroup$ @Poutnik is there any problem with my analysis? $\endgroup$
    – ananta
    Apr 17, 2023 at 11:28
  • $\begingroup$ @Poutnik I have shown how this calculation is performed in the question. Are you saying this type of calculation is correct or wrong? Please be clear. Also, here, I believe you have to consider $H_2 \rightarrow 2H^+ + 2e^-$ (oxidation reaction) with negative of the reduction potential to obtain the combustion reaction. $\endgroup$
    – ananta
    Apr 17, 2023 at 11:36
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – ananta
    Apr 17, 2023 at 11:46
  • $\begingroup$ @ananta Hello Ananta. If you see the OP's question, the answer is $E^0 = 0.85 \; \pu{V}$. If the additivity of standard reduction potential was correct, as you claim, then the answer would be $E^0 = 0.90 \; \pu{V} + 0.80 \; \pu{V} = 1.7 \; \pu{V}$, which is not the answer. How can you explain this? $\endgroup$ Apr 17, 2023 at 17:18
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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, 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 or disproportionation. 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.

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  • $\begingroup$ The point is, it is not the title of his question. The question was somewhat hijacked. $\endgroup$
    – Poutnik
    Apr 18, 2023 at 19:29
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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.

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  • $\begingroup$ Yes, it is for one coulomb of electrons; allow the poetic license because we all have forgotten physics. $\endgroup$
    – jimchmst
    Apr 18, 2023 at 22:32

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