Electrode potentials and feasibility of reaction

The amount of copper in brass can be found as follows:

· react a known mass of the brass with concentrated nitric acid (in the fume cupboard);

· make the resulting solution to a known volume;

· pipette a known volume of the solution into a conical flask;

· add sodium carbonate solution drop by drop until a hazy precipitate appears, and then add dilute ethanoic acid until the precipitate just disappears;

· add an excess of 10% potassium iodide solution;

· titrate the liberated iodine with standard sodium thiosulphate solution.

There are three redox processes in this scheme: the oxidation of copper to copper(II) ions with concentrated nitric acid; the reduction of copper(II) ions to solid copper(I) iodide; the reduction of the liberated iodine to iodide ions with thiosulphate ions.

Standard electrode potentials are widely used to predict the feasibility of redox processes. In general if the electrode potential for the reaction is positive, it is regarded as being feasible. However, the over-riding conditions that affect this statement are often forgotten, namely:

· The value of E^o says something about the feasibility of the reaction under standard conditions only;

· The value of E^o says something about the feasibility of the reaction, but does not say anything about the rate of the reaction.

The value of E^o is related to the equilibrium constant for the reaction. What a positive value says is that the equilibrium constant K is greater than 1, a negative value of E^o says that it is less than 1.

Consider the three redox reactions mentioned but under standard conditions rather than those used in the experiment. My favoured method of finding overall E^o values for reactions is to turn round one of the reduction half-reactions and change the sign of its E^o, then add the two half reactions so as to get the overall reaction and its E^o. The purists who function only in terms of reduction potentials, and those adherents of the clockwise or anticlockwise rule (whatever they are), will not like this method. Too bad. It always gives the right answer.

Two feasible reactions....

The reaction of nitric acid and copper metal.

			E^o/V
Cu²⁺ + 2e^–	⇔	Cu	+ 0.34
NO₃^– + 2H⁺ + e^–	⇔	NO₂ + H₂O	+ 0.80

Invert the first equation and change the sign of E^o, double the second (but not the E^o) and add, and the final equation for the reaction becomes:

Cu + 2NO₃^– + 4H⁺	→	Cu²⁺ + 2NO₂ + 2H₂O	+ 0.46

So the reaction is feasible under standard conditions. The actual conditions are far from standard, but both work to make the reaction even more feasible, i.e. to drive the reaction to the right. The concentration of H⁺ ions in concentrated nitric acid is some 10 times that of the standard conditions; and the NO₂ is lost as a gas, so this pulls the equilibrium over to the right-hand side.

The reaction of sodium thiosulphate with iodine. This is the titration reaction:

			E^o/V
I₂ + 2e^–	⇔	2I^–	+ 0.54
S₄O₆^{2 –} + 2e^–	⇔	2S₂O₃^{2 –}	+ 0.09

Invert the second equation, change the sign of E^o and add to get the overall equation:

I₂ + 2S₂O₃^{2 –}	⇔	2I^– + S₄O₆^{2 –}	+ 0.45

The reaction is feasible under standard conditions; again the conditions are not standard, in particular the concentration of the thiosulphate ion is low, but the difference is not enough to prevent the reaction from occurring.

...and one that apparently isn't.

The reaction of iodide ions with copper(II) ions. This is the reaction that liberates the iodine for titration: it is

Cu²⁺ (aq) + 2I^–(aq) → CuI(s) + ½ I₂(aq)

			E^o/V
Cu²⁺ + e^–	⇔	Cu⁺	+ 0.15
½ I₂ + e^–	⇔	I^–	+ 0.54

As before, turn round the second reaction and add to get the reaction that occurs:

Cu²⁺ + I^–	→	Cu⁺ + ½ I₂	– 0.39

The reaction is not spontaneous under standard conditions – but it does occur in this analytical method, and very quickly too.

The value of E^o can be used to give the equilibrium constant for the reaction, and it comes to around 10 ^{– 7}. This is very small, and indicates little reaction. But this is under standard conditions. The factor that makes the reduction reaction possible is the fact that Cu⁺ is not in solution at a concentration of 1 mol dm^{– 3}, indeed is not in solution at all.

The key to the 'problem': the insolubility of copper(I) iodide.

Copper(I) iodide is an extremely insoluble solid. The equilibrium constant for the reaction

Cu⁺ (aq) + I^–(aq) ⇔ CuI (s)

is around 10¹². Thus as soon as any Cu⁺ appears it immediately precipitates, and this is sufficient to drive the reaction over to the right hand side. Indeed it is not hard to show that the equilibrium constant for the production of solid copper(I) iodide as distinct from copper(I) under standard conditions is of the order of 10⁵. This reaction is complete by anyone's standards, and certainly is complete enough for the reaction to be used in quantitative analysis.

So beware; the unthinking use of standard electrode potentials can give highly misleading information about reactions that are not under standard conditions.