Calculate the ${E_\mathrm{cell}}$ (not ${E^\circ})$ at $\pu{25 °C}.$

$$\ce{Cu(s) | Cu^2+ (\pu{0.10 M}) || Fe^2+ (\pu{0.0030 M}) | Fe(s)}$$ $$ \begin{align} E^\circ(\ce{Cu^2+/Cu}) &= \pu{0.339 V} \\ E^\circ(\ce{Fe^2+/Fe}) &= \pu{-0.440 V} \end{align}$$

I found that $\ce{Cu}$ gets oxidized and $\ce{Fe^2+}$ gets reduced and found $E^\circ_\mathrm{cell}$:

$$\ce{Cu(s) + Fe^2+(aq) -> Cu^2+(aq) + Fe(s)}$$

$$ \begin{align} E^\circ_\mathrm{cell} &= E^\circ(\ce{Cu^2+/Cu}) + E^\circ(\ce{Fe^2+/Fe}) \\ &= \pu{0.339 V} + (\pu{-0.440 V}) \\ &= \pu{-0.101 V} \end{align} $$

However, I'm not sure if I'm supposed to add them or subtract them because I've seen both done, which is confusing. In what situations do you add, and which situations do you subtract half reaction potential values?

I then plugged this calculated value into the Nernst equation

$$E = E^\circ -\frac{RT}{zF}\ln Q$$

using $z = 2$ and plugged in the correct constants. However, I am not sure if I calculated $Q$ correctly. I did [anode]/[cathode], but is it supposed to be [products]/[reactants]? It is unclear to me because my instructor did the latter, but I keep seeing the former everywhere else because that is how you would calculate it in an equilibrium problem. I know it doesn't make a difference in this specific problem, but I want to know for future problems when concentration of all components are given (such as if they are all aqueous).

From these steps, I calculated that $E_\mathrm{cell} = \pu{-0.164 V}.$ How to approach this question?

  • 1
    $\begingroup$ When you are using the standard electrode potentials for the two half reactions, check what those half reactions are. For example, the $\mathrm{E_{Cu2+|Cu}}$ represents the half reaction $\ce{Cu^2+ + 2e -> Cu}$. Then you take the total reaction, and write it as a sum of two half reactions. For this you can take the reverse of a half reaction (i.e. the opposite), and then you would have to change the sign of the std. electrode potential as well. At the end, you add the half-reactions and add the electrode potentials at the same time. $\endgroup$
    – S R Maiti
    Mar 27, 2021 at 23:21
  • 1
    $\begingroup$ @luckyschili Please don't add the $\pu{E}$ values. They should be subtracted. $\endgroup$
    – Maurice
    Mar 28, 2021 at 9:43

2 Answers 2


If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times \log0.10} = \pu{0.309 V}$$ $$\ce{E_{Fe} = \pu{-0.44 V} + \pu{0.0296 V} \times \log0.0030} = \pu{- 0.515 V}$$ So the overall performance of the cell is $$\ce{E_{cell} = E_{Cu} - E_{Fe}}= \pu{0.309 V} - \pu{(-0.515 V) = 0.824 V}$$

  • $\begingroup$ So, the Nernst equation can only be used for half reactions and not the whole cell? If not, I don't understand why what I did was not correct. And Q in the Nernst equation for a half reaction will always be the cation species, regardless of whether or not it is the oxidizing and reducing agent in its respective reaction? $\endgroup$ Mar 27, 2021 at 22:37

I am resubmtting an answer that was down voted, deleted, hidden and needs 5 votes to be reinstated but can get none because it is hidden. If the answer needs corrections state them.

The half reaction potentials are written as reductions one of them has to be reversed to an oxidation half reaction and its potential sign changed: Cu++ +2e- = Cu E[O] = +0.339. Fe++ +2e- = Fe E[O] = -0.440 lets change the iron to oxidation

Cu++ +2e- = Cu E[O] = +0.339. Fe = Fe++ + 2e- E[O] = +0.440 Add the half reactions the electrons are balanced [if not they must be balanced] and they subtract out. Cu++ + Fe = Cu + Fe++ E[O] = +0.779 Positive voltage means the reaction will proceed as written. Had I changed the other half reaction the voltage would be the same but negative and the reaction would proceed in the opposite direction. The standard concentrations for a solution are 1Molar activity for aqueous solutions for every ion in the equation and the activity of a pure solid is ONE not zero as was stated elsewhere. So to change conditions the complete Nernst equation is used There is a term for each reactant and product. In this case since pure solid metals are used their activities are 1 and it looks like they are omitted. They are not.[were the metals impure the mole fraction would suffice].

Increasing the concentration of the reactants over one increases the potential decreasing the products from one increases the potential. The products are the numerator in Q same as the equilibrium constant.


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