In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.

I was taught ($\approx 1970$) in the following manner. Given the reduction half cells:

$$ \begin{align} \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\ \ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\ \ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V} \end{align} $$

To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:

$$ \begin{align} \ce{Ag(s) &→ Ag+ + e- } &\quad E^\circ &= \pu{-0.799 V} \\ \ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V} \end{align} $$

Now you can write a balanced chemical reaction as such:

$$ \begin{align} \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\ \ce{4Ag(s) &→ 4Ag+ + 4e- } &\quad E^\circ &= \pu{-0.799 V} \\ \ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V} \end{align} $$

and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.

Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?

  • $\begingroup$ It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-) $\endgroup$ – Poutnik Apr 13 '19 at 17:05
  • 1
    $\begingroup$ For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $\ce{A_{Ag+} \ne 1.000 }$. $\endgroup$ – MaxW Apr 13 '19 at 17:09
  • $\begingroup$ I know it well. But it does not make sense to write the silver electrode potential like if silver had been better reduction agent than zinc. $\endgroup$ – Poutnik Apr 13 '19 at 17:12
  • 1
    $\begingroup$ @Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation. $\endgroup$ – MaxW Apr 13 '19 at 17:17
  • $\begingroup$ The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $\ce{Ag/Ag+}$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate. $\endgroup$ – Poutnik Apr 13 '19 at 17:20

I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.

The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.

Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.

| improve this answer | |
  • 1
    $\begingroup$ So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides. $\endgroup$ – MaxW Apr 13 '19 at 17:00
  • 1
    $\begingroup$ Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation. $\endgroup$ – M. Farooq Apr 13 '19 at 17:29

I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.

If the cell potential is calculated from reduction potential of the cathode half reaction minus that of the anode half reaction, there are more points of possible errors, including trouble with arithmetic.

You might argue that students learn more and gain deeper insight when given more opportunity to make mistakes, so maybe my approach is inferior. For example, students can use the approach I am using without knowing which electrode is the anode and which is the cathode.

| improve this answer | |
  • $\begingroup$ Just curious - Does the text book you use also employ that technique for balancing redox reactions? $\endgroup$ – MaxW Apr 13 '19 at 17:29
  • 2
    $\begingroup$ I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_\text{red} + E_\text{ox}$ while OpenStax Chemistry uses $E_\text{cathode} - E_\text{anode}$. $\endgroup$ – Karsten Theis Apr 13 '19 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.