# Negative potential of the oxidation of iron(II) to iron(III)

The oxidation potential which I found on the internet the following reaction

$$\ce{Fe^{2+} -> Fe^{3+} + e^-}$$

is $\pu{-0.77 V}$. But how can it be negative? Negative oxidation potential means oxidation tendency is very less or we can say negative oxidation potential means $\ce{Fe^2+}$ has very less tendency to lose electrons. But it doesn't make sense, because $\ce{Fe^3+}$ is in $\mathrm{3d^5}$ configuration, which is half-filled $\mathrm{d}$ orbital, which is very stable than $\ce{Fe^2+}$. So, $\ce{Fe^2+}$ should get readily oxidize to $\ce{Fe^3+}$. Then why do we have negative oxidation potential for this reaction?

As IanB2016 mentioned in his answer, the redox potential of a half-cell reaction is always in reference to another half cell. The usual "reference" half cell in electrochemistry and thermodynamics is the standard hydrogen electrode (SHE). Some key features of the SHE:

• the half cell reaction is $\ce{2 H+(aq) + 2e- <=> H2}$, i.e. the reductant is hydrogen gas and the oxidant is aqueous protons.

• the pressure of hydrogen gas for a standard hydrogen electrode is 1 atmosphere.

• the concentration of acid in the aqueous electrolyte is 1 molar, which corresponds to a pH of 0.

The oxidation potential of a species is related to its strength as a reducing agent. Things that have a negative oxidation potential are less reducing than hydrogen gas, and things that have a positive oxidation potential are more reducing than hydrogen gas. Lithium metal, for example, has an oxidation potential of more than three volts, i.e. $\gt+3\;\mathrm{V}$. It is an extremely powerful reducing agent, far more powerful than hydrogen gas.

As you note, $\ce{Fe^{2+}}$ has an oxidation potential of $-0.77\;\mathrm{V}$, meaning it is a less powerful reducing agent than hydrogen gas. This makes sense in several ways:

• aqueous solutions of iron(II) salts do not spontaneously reduce aqueous protons to hydrogen gas. If you mix iron(II) sulfate with water, for example, hydrogen bubbles do not spontaneously form.

• Hydrogen gas releases more energy when burned in oxygen than iron(II) salts do when they react with oxygen. Hydrogen reacts explosively, whereas iron(II) salts do not spontaneously explode or catch fire when they are oxidized by oxygen.

Finally, note that the oxidation potential of water (via the reaction $\ce{O2(g) + 4 H+ + 4 e− <=> 2H2O}$ at pH 0 at 1 atm of pressure) is -1.229 volts. Water is a very weak reducing agent, which means that oxygen is a strong oxidizing agent, stronger than iron(III). This means:

• iron(III) salts do not spontaneously oxidize water to produce oxygen.

• conversely, oxygen does spontaneously oxidize iron(II) to iron(III). Solutiosn of iron(II) salts are thermodynamically unstable under aerobic (oxygen-rich) conditions.

Redox (reduction - oxidation) Potentials are measured for a specific reaction half-cell [such as Fe2+/Fe3+] by comparing that half cell reaction to a Standard Reference Electrode potential, which is generally the Standard Hydrogen Electrode or cell (SHE) in [aqueous] solution. The SHE is the internationally accepted "Standard Electrode" in electro-chemistry and has arbitarily been asigned a reference potential of 0.0 Volts under 'standard conditions of temperature, pressure and species activities'. Note that there are many other different "practical" reference cells used in industry because the SHE is not very user friendly to use in a practical sense - the reasons for this are outside the scope of this question. Because of the arbitarily assigned value of 0.0 Volts for the SHE a specific half cell reaction can therefore have a value (reaction potential) which is either higher or lower than the SHE potential under the specific conditions previously mentioned. When this potential is higher than the SHE potential that potential is assigned a positive value and when it is less than the SHE potential it is assigned a negative value. Hence the reason the Fe2+/Fe3+ cell has negative sign before the 0.77V (actually +ve 0.79V in the British System when written as a Reduction Potential). Such positive and negative potentials are a direct measure of the half cell's ability to undergo reduction/oxidation as measured with respect to the SHE. Remember also that when considering reaction potentials you cannot simply consider one half cell reaction potential in isolation since both reduction / oxidation reactions occur together for any cell and therefore must be consider together. There are many more indepth explanations available but hopefully this answer will suffice as a "simplistic overview" for the question asked. One further thing I forgot to mention earlier is that you need to be careful about where the potentials you search on the internet originated from because different countries (eg, USA & England for example), have different ways of expressing the same Standard Potentials which can reverse the sign of the potential for the same half cell. For example in the British System these potentials are written has "Standard (Reduction) Potentials [at 25 Deg. C]" and the half cell in question is written as Fe3+ + e <=> Fe2+ which has a potential of +ve 0.79 Volt (note not -ve 0.79V)! It is crucial to stick with one system and do not mix them up and particularly so when using the Nernst Equation for solving oxidation / reduction type problems. A very helpful mnemnoic I used when I was at Uni studying Chemistry back in the late 1960's - 70's was [in the British System when using these Standard potential tables] to determine whether or not a reaction could be possible was: emphasized textThe oxidised form of the one above [more positive form] in the table would oxidise the reduced form of the one below in the table. Having the mnemonic table condition met for a particular RedOx cell doesn't necessarily mean that reaction will proceed either in part or in full because this approach does not include anything about reaction rates or kinetics - which enters into a whole different ball game! Hope this helps to understand these sometimes confusing aspects of electro-chemistry which can be complicated by different countries different systems of Standards.

As previously mentioned you need to be careful about the sign attributed to an electrode potential because it differs according to the particular sign convention being used; what I have used is the one recommended by the International Union of Pure and Applied Chemistry (IUPAC). So the best way to answer your query is as follows: if the system Mn+/M (where n+ is the valence of M) is a better reducing agent than Hydrogen, under standard conditions, the electrode potential is negative and if it is a poorer reducing agent, the potential is positive. Further, when the standard electrode potentials of the elements are arranged in order of the most negative [Li+e = Li -3.04V, at the top] to the to the most positive [F2 + 2e- = 2F +2.87V, at the bottom] constitutes the electrochemical series of the elements which may be used to interpret and also predict the reactivity of the elements. A cation-forming element will displace from aqueous solution another element which lies below it in THIS series and an anion-forming element will displace from aqueous solution another element which is above it in the series. Thus the order in this series is Li+/Li < ...... < ..... < 1/2F2/F-. trust this now clarifies this for you. [[Ref Bell & Lot 2nd Ed, Modern Approach to Inorganic Chemistry]]

• ya I understand, you said that the "When this potential is higher than the SHE potential that potential is assigned a positive value and when it is less than the SHE potential it is assigned a negative value." Ok, so do you mean its negative because it has less reducing power than hydrogen? – Krishna Behera Feb 5 '16 at 15:13
• Refer to the bottom of my original answer - added as an edit due to insufficient space available in this section. – IanB2016 Feb 18 '16 at 14:12