For 2015 A2 9701 CIE exams, you will use standard reduction potential, E^{0}, values. [This post will refer to info mentioned on page 285 of the CIE coursebook, ISBN: 978-0-521-12661-8]

These are values at 298K (25oC) 101 kPa (1 atmosphere) and 1 mole dm-3 (molar) concentration.

That value will change if you deviate from those conditions. You could use the Nernst equation to calculate what that value would be, but this isn’t on the syllabus. To my horror, this came up in **9701/41/O/N/12 (i.e. paper 41, winter, 2012)**. and also in **9701/51/M/J/14** ** (i.e. paper 51, summer, 2014)**.

**Warning… rant…**

*Although the Nernst equation was given in a slightly mild form, I still found it highly questionable that in a final exam, CIE are deliberately putting new material on there that most students have never even seen before or had any practice in using it. It takes time to “get to grips” with the new information, and one could say it’s plain mean to make students have to (in effect) learn new information in a final exam especially with all the stress and time constraints involved. But it’s what they do. Doubtless they are proud that doing things like this, probably justifying it with things like “it helps separate and identify the very top students – those who have read widely around the subject” but it will hamper many other students who would probably NEVER EVER expect that they’d have to learn something new in a exam. What CIE think a syllabus is for I don’t really know…ANY information to be covered in a final exam SHOULD be mentioned in the syllabus. They have done similar things in paper 5, for example Fractional distillation in 9701/53/O/N/13 and Integrated first order rate law on 9701/51/O/N/12, molality / colligative properties on 9701/51/O/N/10, Arrhenius equation on 9701/51/O/N/11, Graham’s law (effusion) on 9701/53/O/N/11. Granted some of them are “old syllabus” papers but I’m pretty sure those things like molality never appeared on the syllabus for those exams (you can guess by the way it it mentioned in the paper). Any boast that it helps identify the stop students is negated somewhat by the fact that they would have to “mathematically adjust” the students grades when the majority of them do badly in a paper because of these new things.*

**End of rant…**

Anyway…. getting back to the topic at hand.

The Nernst equation isn’t on the syllabus, but you are expected to know how the electrode potential will **change, i.e. does it become more positive or less positive**, if the concentration conditions are non standard

For example a typical standard reduction potential half-equation may be given:

1 mol dm-3 of Zn2+(aq) ions + 2 e- in equilibrium with Zn(s) gives E^{0} value of -0.76V

*(p.s. I advise you to KNOW this equation and the E ^{0} value, also learn the Cu2+ +2e- <–> Cu(s) E^{0} = +0.34V)*

So what happens to the electrode potential if the conc. of Zn2+(aq) was **less than** 1 mol dm-3?

Well, honestly, I have done poorly in advising students how to do this. For some reason I kept getting caught on some weird same single flawed reason in trying to give the steps needed to get the answer. I won’t repeat the flawed step (which gave inconsistent results) here, to hopefully stop implanting (/reinforcing) it. Instead I’ll state the correct method…

So, say for Zn2+(aq) + 2e- <–> Zn(s) E^{0} = – 0.76V then

if the [Zn(aq)]2+ was 0.05 mol dm-3, how would the reduction potential change?

These standard reduction potentials are **equilibria, so you apply Le Chatelier’s principles.**

**Identify which direction the equilibrium will shift.**** That direction must become more feasible.** (i.e. more energetically favourable, think of it as “enables” the shift).**. Use that “more feasible direction” to decide what must happen in the reduction half-equation which goes to the right? **

The reduction potential value, whose change we must predict, describes the reaction going from left to right i.e. *from Zn2+(aq) ions* we go *to Zn(s)* and the standard potential for that was -0.76V.

In the case of 0.05 mol dm-3 Zn2+(aq), the equilibrium will shift to the left, i.e. going to the left must becomes more feasible and of course going **to the right** (the direction of the reduction equation) must therefore be **less feasible**. We already said going right was not very feasible, the value was negatovre. So the reaction for the lower concentration of Zn2+(aq) has a new reduction potential (Caution: Not a standard reduction potenial, E^{0}, as we are no longer under standard conditions !!!) being a value of MORE NEGATIVE than -0.76V (maybe perhaps -0.87V at a guess)

Try this…

Cu2+(aq) + e- <–> Cu+(aq), E^{0} = +0.15V

A 1 mol dm-3 solution of Cu2+(aq) is mixed with a 0.1 mol dm-3 solution of Cu+. Predict the reduction potential for that half cell.

Answer:

In effect, the [Cu+(aq)] has been lowered from standard conditions. Therefore eqm shifts to the right, i.e. progressing to the right hand side is more feasible than before (i.e. somewhat more energetically favourable), so the new reduction potential will be more positive than E^{0}.

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Fe3+(aq) + e- <–> Fe2+(aq), E^{0} = +0.77V

What will the E(red) value be if concentration of Fe3+(aq) is greater than 1 mol dm-3?

Answer: Eqm will shift to the right. Going to the right is already +’ve so E(redn) it must become more +’ve than +0.77V

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Cr3+(aq) + e– ⇌ Cr2+(aq) E^{0} = -0.41V

What will the E(redn) value be if concentration of Cr2+(aq) is less than 1 mol dm-3?

Answer: Eqm will shift to the right. Going to the right is -‘ve. But now as going to the right is more feasible, E(redn) will be less negative than -0.41V

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MnO4-(aq) + 8H+(aq) + 5e- <–> Mn2+(aq) + 4H2O(l) E^{0} = +1.52V

What will the E(red) value be if concentration of Mn2+(aq) is greater than 1 mol dm-3?

Answer: Eqm will shift to the left. Going to the right is already +’ve. To lower the feasibility the E(redn) must become less positive. E(redn) will be less than +1.52V

CIE could also ask you about what happens to the E(redn) value under different temperatures IF they gave a delta H value for the reaction – so beware, and try and work it out (similar use of Le Chatelier logic will apply, but take note: This method gives you the right answer which way the E(redn) changes under different temperatures, but doesn’t explain the actual physical reason it happens. For that you need to consider entropy – which only appears in the 2016 syllabus.

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Note in the course of making this post I discovered an error in the book. Page 299 the data for the five reduction half-equations. The copper equation shows only one e- being accepted by Cu2+(aq) turning it into Cu(s).** It should be two electrons.**