In this cell, the copper is oxidized and $E_{cell} = \pu{+1.15 V}$. $$\begin{align} \ce{Cu^2+(aq) + 2e− -> Cu(s)} & &(E &= \pu{+0.34 V})\\ \ce{Mn^3+(aq) + e− -> Mn^2+(aq)} & &(E &= ??)\\ \end{align}$$

Calculate the value of the standard electrode potential for the $\ce{Mn^3+(aq)|Mn^2+(aq)}$ half-cell.

So the answer is given as $\pu{+1.49V}$.

The answer stated that the equation would be $(+1.15)-(-0.34) = 1.49$, but why are we changing the sign for the electrode potential for copper? Shouldn't it remain as $\mathrm{+0.34}$ so the manganese half cell would be $\pu{+0.81 V}$?

  • $\begingroup$ Try to search the site and internet for the keywords, I bet you will find multiple interesting hits. In a way, it is analogical to question "Which river bank is the left bank?". On your left, or on your left if looking downstream? $\endgroup$
    – Poutnik
    Jun 11 at 10:46
  • $\begingroup$ I am very curious to see Mn3+(aq) /Mn2+(aq) half cell. $\endgroup$
    – Poutnik
    Jun 11 at 10:48
  • $\begingroup$ It is given that copper is oxidized, so the required electrode potential will become -0.34 V. Electrode potential is an intensive property and Gibbs energy an extensive one. So we can equate the change in Gibbs energy for the net cell reaction to that for the sum of oxidation and reduction cell reactions. $\endgroup$
    – Apurvium
    Jun 12 at 7:39

3 Answers 3


The answer is right, but I find the double negative pretty unintuitive. The question tells you that

$$\begin{align} \ce{2Mn^3+ + Cu &-> 2 Mn^2+ + Cu^2+} & E^\circ_\mathrm{cell} &= \pu{+1.15 V} \end{align}$$

(This is the balanced reaction between $\ce{Mn^3+}$ and $\ce{Cu}$; it specifies that $\ce{Cu}$ is the species being oxidised. The balancing isn't hugely important, it's just there to make the equation correct.)

The $E^\circ_\mathrm{cell}$ for this reaction is the difference between the two half-cell reduction potentials. This much you should be familiar with:

$$E^\circ_\mathrm{cell} = E^\circ(\ce{Mn^3+}/\ce{Mn^2+}) - E^\circ(\ce{Cu^2+}/\ce{Cu})$$

so rearranging,

$$\begin{align} E^\circ(\ce{Mn^3+}/\ce{Mn^2+}) &= E^\circ_\mathrm{cell} + E^\circ(\ce{Cu^2+}/\ce{Cu}) \\ &= \pu{1.15 V} + \pu{0.34 V} \\ &= \pu{1.49 V}. \end{align}$$

  • $\begingroup$ But this method will not work in case the electrons are not cancelling... Because electrode potential is an intensive property. E.g., in this link chemistry.stackexchange.com/a/155900/55479 related to Cu. $\endgroup$
    – Apurvium
    Jun 12 at 7:45
  • 1
    $\begingroup$ @Apurvium sorry but I really don't get what you are saying. The fact that they are intensive is precisely what lets us just subtract one reduction potential from the other, without accounting for stoichiometry. If you use Gibbs energies to work it out, you will reach the same answer just with a lot more hassle. I am curious to see how you would come up with something different. $\endgroup$
    – orthocresol
    Jun 12 at 10:01
  • $\begingroup$ My point was, when the combination of two half-cell reactions gives a full cell reaction, then the stoichiometry will not matter because $E$ is an intensive property. But in case, two half-cell reactions gives another half-cell reaction, then we have to go by Gibbs energy method. E.g., link in my previous comment. $\endgroup$
    – Apurvium
    Jun 13 at 10:42
  • $\begingroup$ @Apurvium Oh, I see: yup, that’s correct. But that’s not the case here. $\endgroup$
    – orthocresol
    Jun 13 at 12:11

The standard redox/electrode potential for a half-reaction

$$\ce{ox^{m} + n e- -> red^{m-n}}$$

is defined within thermodynamics as

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

for a formal reaction

$$\ce{ox^{m} + }\frac n2 \ce{ H2 -> red^{m-n} + n H+ }$$

"American school" flips the sign of the standard potential value when the opposite direction of reaction is considered, to formally honour the thermodynamic relations. By other words, it is an alternative equivalent way to express the Gibbs energy of redox reactions, that flips the sign with direction too.

"European school" considers the potential value conventionally always for the forward reaction and such values are tabelized as standard redox potentials.

The former approach is cleaner from the thermodynamic point of view.

OTOH, the latter approach is in agreement with measured electrode potentials against SHE, with tendency to be oxidized/reduced and is less vulnerable to confusions.

It can be reconciliated if there are formally strictly distinguished redox and electrode potentials. But even then.....

The problem has 2 analogies:

  • Left river banks are those on your left

    • always (naive sense)
    • if looking downstream (geographic convention).
  • Anodes/cathodes are electrodes where oxidation/reduction occurs

    • always(physical chemistry)
    • if in galvanic mode in context of rechargable cells (engineering cell design convention). For Li-ion cells, cathodes are always the oxide ones and anodes the graphite ones.

This is how I applied what I learnt in my undergraduate freshman chemistry class to answer the given question:

In the given cell, the copper has been oxidized and the $\ce{Mn^3+}$ has been oxidized. The given $E^\circ_\mathrm{cell} =\pu{ +1.15 V}$. The reduction half-reactions can be given as: $$\begin{align} \ce{Cu^2+ (aq) + 2e− -> Cu(s)} & & E_\ce{Cu^2+/Cu} = \pu{+0.34 V}\\ \ce{Mn^3+(aq) + e− -> Mn^2+(aq)} & & E_\ce{Mn^3+/Mn^2+} = x \ \pu{V }\\ \end{align}$$

Although both half-reactions are given as reductions to show the correct reduction potentials, in the total redox reaction, $\ce{Cu}$ is actually oxidized in anode. However, the $E^\circ_\mathrm{cell}$ is defined as: $$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$$ where both $E^\circ_\mathrm{cathode}$ and $E^\circ_\mathrm{anode}$ are given as reduction potentials with correct signs. Thus, $$E^\circ_\mathrm{cathode} = E_\ce{Mn^3+/Mn^2+} = x \ \pu{V } \ \text{ and } E^\circ_\mathrm{anode} = E_\ce{Cu^2+/Cu} = \pu{+0.34 V}$$ $$\therefore \ E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \ \Rightarrow \ \pu{1.15 V} = x - \pu{0.34 V} \\ x = \pu{1.15 V} + \pu{0.34 V} = \pu{1.49 V} $$

In this way, we avoid both double negative and sign reverse.


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