Redox (II): Standard Electrode Potential Eo(6) The electrochemical series

Edexcel A level Chemistry (2017)

Topic 14: Redox (II): Standard Electrode Potential Eo(6)

Here is a further learning objective:

14/13 To know that standard electrode potentials can be listed as an electrochemical series

Electrode potentials and the electrochemical series

So what do you think this list is really telling you?

You will see lists like this is data books and examination questions but unless you are aware of the significance of the values then much of the meaning of electrode potential will pass you by.

So I thought I would just take this opportunity to re–emphasise the meaning of the values and use this learning objective to lay more stress on you understanding the significance of these values.

In fact, I can say that I know from my own personal experience of trying to get my head round the significance of electrode potentials and lists of them that it is not easy, at least it was not for me.

Here is a typical list:

Best oxidant	Half-Reaction	E° V	Best reducing agent
	Li+   +  e—  ⇌  Li	−3,04
	K+   +  e—  ⇌  K	−2,92
	Ba2+   +  2e—  ⇌  Ba	−2,90
	Ca2+   +  2e—  ⇌  Ca	−2,87
	Na+   +  e—  ⇌  Na	−2,71
	Mg2+   +  2e—  ⇌  Mg	−2,37
	Al3+   +  3e—  ⇌  Al	−1,66
	Mn2+   +  2e—  ⇌  Mn	−1,18
	2H2O +  2e— ⇌  2OH—   +   H2(g)	−0,83
	Zn2+   +  2e—  ⇌  Zn	−0,76
	Cr2+   +  2e—  ⇌  Cr	−0,74
	Fe2+   +  2e—  ⇌  Fe	−0,44
	Cr3+   +  3e—  ⇌  Cr	−0,41
	Cd2+   +  2e—  ⇌  Cd	−0,40
	Co2+   +  2e—  ⇌  Co	−0,28
	Ni2+   +  2e—  ⇌  Ni	−0,25
	Sn2+   +  2e—  ⇌  Sn	−0,14
	Pb2+   +  2e—  ⇌  Pb	−0,13
	Fe3+   +  3e—  ⇌  Fe	−0,04
	2H+   +  2e—  ⇌  H2(g)	0,00
	S + 2H+  +2e—   ⇌  H2S (g)	+0,14
	Sn4+   +  2e—  ⇌  Sn2+	+0,15
	Cu2+   +  e—  ⇌  Cu+	+0,16
	SO42—  +   4H+  +  2e—   ⇌  SO2 (g)  +   2H2O	+0,17
	Cu2+   +   2e—  ⇌  Cu	+0,34
	2H2O +  O2  + 4e—   ⇌  4OH—	+0,40
	Cu+   +  e—  ⇌  Cu	+0,52
	I2  +2e—   ⇌  2I—	+0,54
	O2 + 2H+ + 2e— ⇌  H2O2	+0,68
	Fe3+   +  e—  ⇌  Fe2+	+0,77
	NO3—  +   2H+  +  e—   ⇌  NO2   +   H2O	+0,78
	Hg2+   +  2e—  ⇌  Hg(l)	+0,78
	Ag+   +  e—  ⇌  Ag	+0,80
	NO3—  +   4H+  +  3e—   ⇌  NO   +   2H2O	+0,96
	Br2  +2e—   ⇌  2Br—	+1,06
	O2 + 4H+ + 4e— ⇌  2H2O	+1,23
	MnO2  +  4H+ +  2e—  ⇌  Mn2+  +  2H2O	+1,28
	Cr2O72— + 14H+ + 6e— ⇌  2Cr3+ + 7H2O	+1.33
	Cl2  +2e—   ⇌  2Cl—	+1,36
	Au3+   +  3e—  ⇌  Au	+1,50
	MnO4—  +  8H+ +  5e—  ⇌  Mn2+  +  4H2O	+1,52
	Co3+   +  e—  ⇌  Co2+	+1,82
	F2  +2e—   ⇌  2F—	+2,87

We call the list an electrochemical series.

But what does it all really mean?

First, notice how the redox equilibria are written.

Each one is written as a gain of electrons, a reduction. 

So for example the equation:

Li⁺ (aq) +  e^—  ⟶    Li (s)

represents the half–cell:

Li⁺_(aq) | Li_(s)

And the electrode potential was measured when the half–cell was connected to the standard hydrogen electrode; the reference electrode. Here is the cell diagram:

Pt | [H_{2 (g)} ,2H⁺_(aq)] || Li⁺_(aq) | Li_(s)

So this is how things are set up to measure these standard electrode potential values

but what does —3.04v mean?

It means this: of all the species on the right hand side of the list, lithium (Li) metal at the top is the best reducing agent or reductant. Or you can say it is most easily oxidized.

Similarly, fluoride ions (F^—) near the bottom of the list with an electrode potential of +2.87v are the poorest reductants.  Those ions are least easily oxidized.  Fluoride ions do not give up the electron that completes the fluorine outer shell at all easily, as we know.

Now you can look at the other left hand species on the list and make similar statements about them too.

So for example at E^o = +1.52v manganate (VII) ions (MnO₄^—) are very good oxidizing agents, they easily gain electrons and are reduced to manganese (II) ions (Mn²⁺) in the presence of dilute acid.

So looking at that list again, the best oxidizing agents (oxidants) are at the bottom right.  And you can see there there is Fluorine, Manganate (VII) ions and Chlorine—all good oxidizing agents. 

You can see I hope the implications of this layout of half–cell reactions.

If a reaction is going to be feasible then only top right species (e.g. Lithium (Li), Sodium (Na) etc.) will reduce bottom left species such as manganate (VII) ions (MnO₄^—) or cobalt (III) ions (Co³⁺).

This conclusion is vital to understand if you are going to progress using electrode potential values as I have posted earlier here.