# Electrochemistry - reduction potentials - an IB Chemistry Question [closed]

I have re-read the Electrochemistry chapter from my textbook just now to remind my self of the exact formulae. I still don't understand how to approach this question, however - how can we work it out? I can either give you the mark scheme answers straight away and we can work backwards, or I can wait for an answer and then tell you if that is the correct solution. What ever is the norm on this forum! Thank you.

• The norm on this Q&A site is for you to show what you have tried (or are stuck on) and then help may arrive. No one is just supposed to post an answer to a homework problem or the like. – Ed V Feb 23 at 19:33
• @ Ed V - thank you for the clarification. As I state above, I have re-read the electrochemistry chapter in my textbook. So I know how to calculate the standard cell potential from individual reduction potentials. However, I have never come across this type of question before and so cannot offer any suggestions. I am happy to do some extra reading to find out, but didn't know what to google as, again, this wasn't something I had come across. If you could point me in the right direction, that would be very useful. Thank you. – vgupt Feb 23 at 19:43
• This is a pair of puzzle questions. For the first question, you need a voltaic cell with positive potential, so the reaction would be spontaneous, with the second half reaction in the top table being the cathode reduction. Now which of the 5 half reactions in the bottom table would make a viable anode oxidation reaction? Maybe several of them. But then you narrow the choices by using the reduction half reaction in the first line in the upper table. Think about it: it is mostly a logic problem, not chemistry per se. – Ed V Feb 23 at 19:54

First put your redox potentials on a horizontal line by order of increasing standard redox potentials. You will see that it starts on the left-hand side by $$\ce{Zn^{2+}/Zn}$$ at $$\pu{-0.76 V}$$. Then, going to the right, you find $$\ce{V^{III}/V^{II}}$$ at $$\pu{-0.26 V}$$. Now you find $$\ce{Pb^{2+}/Pb}$$, and so on.
Now try to learn how to use this scale. The reduced species of any couple can react with any oxidized species of any couple situated to its right. For example, $$\ce{Zn}$$ is the reduced species of the couple $$\ce{Zn^{2+}/Zn}$$. So $$\ce{Zn}$$ can reduce all oxidized species, and for example all oxidized forms of Vanadium ions. But $$\ce{Zn^{2+}}$$ cannot be reduced, because there is no reduced species at the left hand side of $$\pu{- 0.76 V}$$.
On the other end of the scale, you find $$\ce{Cl2/Cl-}$$ at $$\pu{+1.36 V}$$. Hopefully you admit that $$\ce{Cl2}$$ can oxidize all reduced species, and $$\ce{Cl-}$$ cannot reduce anything.
Now let's consider $$\ce{Pb^{2+}/Pb}$$, at $$\pu{-0.13 V}$$. $$\ce{Pb}$$ can reduce all oxidized species situated on the positive side of the axis, for example $$\ce{VO^{2+}}$$ which is at $$\pu{+0.34 V}$$. But it cannot reduce $$\ce{V^{3+}}$$ to $$\ce{V^{2+}}$$, because $$\ce{V^{3+}/V^{2+}}$$ is on tis left-hand side. And this is exactly your first question.