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This is the problem: enter image description here

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

I found that the reaction is $$\ce{Cu(s) + Fe^2+(aq) -> Cu^2+(aq) + Fe(s)}$$ $\ce{Cu}$ gets oxidized and $\ce{Fe^2+}$ gets reduced.

I added the ${E°_\mathrm {Cu^{2+}/Cu}}$ and ${E°_\mathrm{Fe^{2+}/Fe}}$ to get ${E°_\mathrm{cell} = -.101}$ . But I'm not even sure if I'm supposed to add them or subtract them because I've seen both done, which is really confusing me. 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. enter image description here

I said 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 ${E_\mathrm{cell} = -.164 V}$

How to approach this question?

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    $\begingroup$ 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 at 23:21
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    $\begingroup$ @luckyschili Please don't add the $\pu{E}$ values. They should be subtracted. $\endgroup$
    – Maurice
    Mar 28 at 9:43
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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 $$\ce{E_{cell} = E_{Cu} - E_{Fe}}= \pu{0.309 V} - \pu{(-0.515 V) = 0.824 V}$$

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  • $\begingroup$ 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 at 22:37

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