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Calculate $E_\mathrm{cell}$ for an electrochemical cell consisting of SCE and a platinum wire indicator electrode dipped in a titration vessel containing $\pu{25 ml}$ $(\pu{0.01 M})$ solution of $\ce{Fe^2+}$ ions after addition of $\ce{Ce^4+}$ solution of concentration $\pu{0.01 M}$ of

A) $\pu{5 ml}$
B) $\pu{25 ml}$
C) $\pu{50 ml}$

SCE is attached to the negative terminal of voltmeter. Given data:

$$ \begin{align} E^\circ(\ce{Fe^2+/Fe^3+}) &= \pu{0.77 V} \\ E^\circ(\ce{Ce^4+/Ce^3+}) &= \pu{1.61 V}\\ E(\text{SCE}) &= \pu{-0.24 V} \end{align} $$

I tried doing this, but one thing I don't get over with is that isn't iron ions and cerium ions both oxidising and reducing? Why is there another electrode? Please clarify as much as possible.

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    $\begingroup$ The initial solution is $0.01$ M and contains $250$ micromole $\ce{Fe^{2+}}$. If you add $5$ mL $\ce{Ce^{4+}}$ $0.01$ M, which is $ 50$ micromole, a chemical reaction will occur : $$\ce{Ce^{4+} + Fe^{2+} -> Ce^{3+} + Fe^{3+}}$$ The obtained mixture contains only $250 - 50 = 200$ micromoles of $\ce{Fe^{2+}}$ and $50$ micromoles of $\ce{Fe^{3+}}$. With this you can calculate the redox potential of the couple $\ce{Fe^{2+}/Fe^{3+}}$ using Nernst's law. Go on ! We will not make the whole calculations for you ! Hopefully you have understood my steps... $\endgroup$ – Maurice Sep 7 '20 at 16:34
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    $\begingroup$ what is the role of Ce's Ecell here? also do you know any good books to get some basic practice for electrochemisry? i am struggling with some concepts and would appreciate any insight on how to approach this $\endgroup$ – napstablook Sep 7 '20 at 17:07
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    $\begingroup$ Yes, cerium and iron ions are both oxidizing and reducing. Does it matter ? // Why the 2nd electrode ? Try it without it. $\endgroup$ – Poutnik Sep 7 '20 at 17:27
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    $\begingroup$ Consider it as the cerium redox system ( rather than being a cerium cell), existing aside of the iron redox system. Platinum electrode is inert, having redox potential of the dissolved redox systems ( which are not surprisingly equal ). $\endgroup$ – Poutnik Sep 7 '20 at 18:54
  • $\begingroup$ yes using cerium that way works in bit A but in bit B? i feel like bit C is replacing iron reduction with cerium reduction. but i have no idea about bit B $\endgroup$ – napstablook Sep 8 '20 at 5:06
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Let's start at the beginning, with a solution where $\ce{[Fe^{2+}}] = 0.01$ M. In order to apply Nernst's law, we also need $\ce{[Fe^{3+}}]$, which is not known, but must be very very low, but cannot be zero. We will see that its exact value is without importance. Let's suppose it is $\ce{[Fe^{3+}}] = 10^{-6}$ M. The redox potential of the solution is : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77 + 0.059 log(10^{-4}) = 0.53 V}$$ Now this potential cannot be directly measured. The only thing that can be measured is the difference of potentiel between this electrode and a reference, which is here SCE, at - $0.24$ V. So if you dip a platinum electrode and a SCE electrode in this solution, you will measure a voltage between them to be : $$\ce{E_{cell} = 0.53 V -(- 0.24) V = 0.77 V}$$ This not a reliable value, as it supposes that we know the amount of $\ce{Fe^{3+}}$ in the solution. Anyway, let's go on.

First addition of Cerium(IV).

Now you add $5$ mL $\ce{Ce^{4+} 0.001 M}$, that is $50~\mu$moles. This cerium ion will react with $\ce{Fe^{2+}}$ ions according to the equation $$\ce{Ce^{4+} + Fe^{2+} -> Ce^{3+} + Fe^{3+}}$$ The initial solution contained $250~ \mu $moles $\ce{Fe^{2+}}$. Adding $~50~\mu$mole $\ce{Ce^{4+}}$ consumes $~50~\mu$mole $\ce{Fe^{2+}}$ and produces $~50~\mu$mole $\ce{Ce^{3+}}$ and $~50~\mu$mole $\ce{Fe^{^3+}}$. The final amount of $\ce{Fe^{2+}}$ is $250 - 50 = 200~ \mu$mol. The potential of this solution is, according to Nernst's law : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77V + 0.059 log\frac{50}{200} = 0.77V - 0.036 V = 0.734 V} $$ The potential of the whole cell (solution + SCE) is : $$\ce{E_{cell} = 0.41 V - (-0.734 V) = 1.144 V}$$ It is worth the trouble mentioning that here we could also have got the same result using the Cerium potential and Nernst's law. In principle. But, to apply Nernst's law to the Cerium couple we would need to know the residual concentration in $\ce{Ce^{4+}}$ which must be very small but not zero. So we have to use the couple $\ce{Fe^{3+}/Fe^{2+}}$.

Second addition of Cerium(IV).

Here you add $25$ mL $\ce{Ce^{4+} 0.001 M}$, that is $250~\mu$moles. The same reaction happens. But here, the amount of Cerium added is equal to the amount of original Iron in the solution. So at the end, there is in solution $250~\mu$moles of $\ce{Ce^{3+}}$ and the same amount of $\ce{Fe^{3+}}$. The residual amount of $\ce{Ce^{4+}}$ and $\ce{Fe^{2+}}$ are very small, but unknown. The potential of the solution must be the average of $\ce{E°(Fe^{3+}/Fe^{2+})}$ and $\ce{E°(Ce^{4+}/Ce^{3+})}$, and this is $(0.77$ V + $1.61$ V)/$2$ = $1.19$ V. The potential of the cell after this $2$nd addition is then $$\ce{E_{cell} = 1.19 V - (-0.24V) = 1.43 V}$$

Third edition of Cerium(IV). Here there is a huge excess of Cerium(IV). There is exactly twice the amount of cerium(IV) necessary to oxidize all Iron(II). The final solution contains the same amount of both Cerium ions. The potential of the solution cannot be calculated with the Iron couple, as the amount of remaining $\ce{Fe^{2+}}$ is unknown and very low. So the potential of the cell must be calculated by Nernst's law applied to the Cerium couple. As the final amount of both ions $\ce{Ce^{3+}}$ and $\ce{Ce^{4+}}$ are equal, the potential of the solution is equal to the standard redox potential, which is $1.61$ V. And the measured difference of potential between the solution and the SCE is : $$\ce{E_{cell} = 1.61 V - (-0.24 V) = 1.85 V}$$

To summarize all these calculations, the measured potential in the cell (solution + SCE) increases steadily from about $0.77$ V, to $1.14$ V, then to $1.43$ V, then to $1.85$ V.

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