Why do we sometimes change the sign for standard electrode potentials?

Question

Question:

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 $\pu{+0.81 V}$?

Answer 1

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,

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

Answer 2

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.

Answer 3

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.