Which is the best oxidizing agent and the best reducing agent from the species listed below?


I determined that oxidizing agent as $\ce{Zn^2+}$ because it appears on the table with the most positive reduction potential and determined that $\ce{Ba^2+}$ as the best reducing agent because it has the most negative reduction potential. I am not sure of my answer.

enter image description here

  • $\begingroup$ Welcome to Chemistry.SE! This appears to be a homework question. Please share your thoughts and attempts towards the solution, otherwise your question may be closed. $\endgroup$ – ringo Apr 8 '16 at 1:40
  • 3
    $\begingroup$ I added my attempt to the question. $\endgroup$ – Jenna Maiz Apr 8 '16 at 2:02
  • 1
    $\begingroup$ You have done good job. $\endgroup$ – Jaroslav Kotowski Apr 8 '16 at 5:45
  • 1
    $\begingroup$ Not that good, how you could get second wrong, while having this chart? $\endgroup$ – Mithoron Apr 8 '16 at 10:07
  • $\begingroup$ The chart says that reducing fluorine $\ce{F2}$ requires very little work, while reducing lithium cation $\ce{Li+}$ requires a lot of work. $\endgroup$ – Curt F. Dec 2 '16 at 14:03

It is a good idea to recap the the definitions of oxidation and reduction when approaching this kind of question, and also what it means to be an oxidising or a reducing agent:

  • Oxidising is defined as the loss of electrons, reduction as the gain.

  • An oxidising agent is good at oxidising other compounds while being reduced itself; and vice-versa.

Take another quick look at your list; all you have are neutral elements or the most common cations of said elements. If you want to oxidise or reduce, then we would always be going back and forth between either. Therefore, only sodium, barium and silver are potential reducing agents on your list (being oxidiseable themselves) while only zinc(II) and barium(II) can be oxidising agents. Here are the relevant equations:

$$\begin{align}\textbf{Reducing agents:}\\ &&\ce{Na &-> Na+ + e-}\\ &&\ce{Ag &-> Ag+ + e-}\\ &&\ce{Ba &-> Ba^2+ + 2 e-}\\ \textbf{Oxidising agents:}\\ &&\ce{Ba^2+ + 2 e- &-> Ba}\\ &&\ce{Zn^2+ + 2 e- &-> Zn}\end{align}$$

Using your table or our chemical intuition, we can then decide which reduction or oxidation is the easiest. The best oxidising agent without doubt is zinc(II) as it is much more noble than barium (although still not a noble metal). The best reducing agent is a slightly more tricky call without data, but barium(0) being further down in the periodic table allows us to assume it to be stronger. Luckily, this is supported by the data.


You have determined the correct oxidizing agent. It does however appear that you have determined the wrong reducing agent (if it is not a typo). $\ce{Ba^2+}$ is a poor reducing agent because it requires a lot of energy to remove a third electron from Barium (as it breaks the noble gas electron structure of $\ce{Ba^2+}$). The correct answer for reducing agent is barium (Ba), as can be read from the table you provided. Barium releases a lot of energy when oxidized to $\ce{Ba^2+}$ ($2.9~\mathrm{eV}$ per elevtron). This means it will easily undergo oxidation => strong reducing agent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.