An Introduction To Electrochemistry Reduction-Oxidation Titration

Oxidation-Reduction processes A- In oxidation-reduction (or redox) reaction, one of the reacting species is converted to a higher oxidation state and as a consequence is oxidized; the other reactant suffers a decrease in oxidation state and is thus reduced. Therefore, a redox reaction is one that occurs between a reducing and an oxidizing agents. Fe 2+ + Ce 4+ ↔ Fe 3+ + Ce 3+ B-1:an oxidizing agent will tend to take on an electron/s (i.e. electron acceptor) and be reduced to a lower oxidation state (i.e. more negative). Ce 4+ + e - → Ce 3+ -2:a reducing agent will tend to give up an electron/s (i.e. electron donor) and be oxidized (i.e. more positive) Fe 2+ → Fe 3+ + e - The condition of equilibrium in a redox reaction depends upon the relative tendencies of the reactants to increase or decrease in oxidation number. The mixing of a strong oxidizing agent with a strong reducing agent will result in an equilibrium, in the other hand, reactants that have less strong yield less favorable equilibrium.

C- Half-Reaction: Separation of an oxidation-reduction reaction into its component parts (i.e. into half-reaction) is a convenient way of indication clearly the changes that occur in the course of the reaction. The overall reaction: 5Sn 2+ +2MnO H + ↔ 5Sn 4+ +2Mn 2+ +8H 2 O is obtained by combining the half-reaction for the oxidation of tin (II), Sn 2+ ↔ Sn 4+ +2e With that for the reduction of permanganate, MnO e + 8H + ↔ Mn 2+ 4H 2 O A balanced equation for the overall process can be obtained by multiplication of the half-reactions, the first half reaction is multiplied by 5 and the second by 2. It is important to note that half-reactions provide no information as to the mechanism by which oxidation and reduction occurs. The electrons that appear in half-reactions are needed to account for the changes in oxidation numbers between the participants; the actual mechanism by which this change occurs can be only be established by experiment.

Oxidation-reduction reactions in electrochemical cells There are two kinds: Galvanic (voltaic) and Electrolytic. In galvanic cell: a chemical reaction spontaneously occurs to produce an electrical energy, e.g. lead storage battery. While in electrolytic cell: an electrical energy is used to fore a nonspontanous chemical reaction to occur e.g. electrolysis of water.

Cell components: Anode and cathode: The cathode is the electrode at which the reduction occurs, while the anode is the electrode where the oxidation takes place. The salt bridge: it allows charge transfer through the solutions but prevents mixing of the solutions. Liquid junctions: Cells with a liquid junction, such as that shown at the fritted porous disk in the figure above, are ordinarily employed to avoid direct reaction between the components of the tow half-cells. The platinum wires can be considered electrodes. Each will adopt an electrical potential that is determined by the tendency of the ions to give off or take on electrons, and this is called the electrode potential. A Voltmeter (uA) placed between the electrodes will indicate the difference in the potentials between the tow electrodes.

The Reference Electrode The electrode potential of the half-reaction has an arbitrarily been assigned a value of v. this is called the normal hydrogen electrode (NHE) or the standard hydrogen electrode (SHE). Potentials are dependent on concentrations and all standard potentials refer to conditions of unit activity for all species (or 1 atm).

Other types of reference electrodes: Calomel or Saturated Calomel Electrode (SCE). Silver/Silver Chloride Electrode.

Factors affecting Nernst equation: 1.It is a temperature dependent, therefore, it should be maintained at 25  c. 2.Matrix effect i.e. the solution of the interested analyte. 3.Presence of additional potentials in the electrochemical cells. This is called *liquid junction potential. It could minimized by a salt such as KCl. *liquid junction potential: it develops at the interface between tow ionic solutions that differ in composition and for which the mobility of the ions differs.

Detection of the end point 1- Self indication: If the titrant is highly colored, this color may be used to detect the end point. For example; potassium permanganate is deep purple, the production of its reduction Mn2+ is nearly colorless, being a very faint pink. 2- Starch indicator: this indicator is used for titrations involving iodine. Starch forms a complex color with I2, That is a very dark blue color. The color is sensitive to very small amounts of iodine. In titration of reducing agents with iodine, the solution remains colorless up to the equivalent point. A fraction of a drop of excess titrant turns the solution a definitive blue.

3- Redox indicators: most types of redox titrations are detected using redox indicators. These are highly colored dyes that are weak reducing or oxidizing agents that can be oxidized or reduced, the colors of the oxidized and reduced forms are deferent. The oxidation state of the indicator and its color will depend on the potential of the solution. A half-reaction and Nernst equation can be written for the indicator Ox indicator + ne - ↔ Red indicator So the ratio [Red in ]/[Ox in ], and therefore the color will change as the potential of the solution changes. The redox indicator reaction must be rapid and eversible, if the reaction is slow or is irreversible, the color change will be gradual and a sharp end point will be not detected.

Titration Involving Iodine: Iodimetry And Iodometry Redox titrations are among the most important types of analysis performed in many areas of application. Example in clinical laboratories are rare, since most analyses are for traces, but these titrations are still extremely useful for standardizing reagents. A- Iodimetry iodine is a moderately strong oxidizing agent and can be used to titrate reducing agents. Titration with I 2 are called Iodimetric methods. These titrations are usually performed in neutral or mildly alkaline (pH 8) to weakly acid solutions. If the pH is too alkaline, I 2 will disproportionate o hypoiodate and iodide: I OH = IO - + I - +H 2 O

There are three reasons for keeping he solution from becoming strongly acidic: 1-the starch used for the end point detection tends to hydrolyze or decompose in strong acid which will affect the EP. 2-the reducing power of several reducing agents is increased in neutral solution. 3-the I - produced in the reaction tends to be oxidized by dissolved oxygen in acid solution: 4I - +O 2 + 4H + → 2I 2 +2H 2 O Iodine has a low solubility in water but the complex,I 3 -, is very soluble. So iodine solutions are prepared by dissolving I 2 in a concentrated solution of potassium iodide: I 2 + I - → I 3 - I 3 - IS Therefore the actual species used in the reaction.

Iodometry Iodide ion is a weak reducing agents and will reduce strong oxidizing agents. It is not used, however, as a titrant mainly because of lack of convenient visual indicator system, as well as other factors such as speed of the reaction. When an excess of iodide is added to a solution of an oxidizing agent I2 is produced in an amount equivalent to the oxidizing agent present. This I2 can therefore be titrated with a reducing agent and the result will be the same as if the oxidizing agent were titrated directly. The titrating agent used is SODUIM THIOSULFATE. Analysis of an oxidizing agent in this way is called an Iodometric method. Consider, for example, the determination of dichromate: Cr 2 O I - (excess) + 14H + → 2Cr 3+ +3I 2 + 7H 2 O I 2 + 2S 2 O 3 2- → 2I - + S 4 O 6 2-

Why not titrate the oxidizing agents directly with the thiosulfate? Because strong oxidizing agents oxidize thiosulfate to oxidation states higher than that of terathionate, but the reaction is generally not stochiometric. The end point for iodometric titrations is detected with starch. The disappearance of the blue starch-I2 color indicates the end of the titration. The starch is not added at the beginning of the titration when iodine concentration is high. instead, it is added just before the EP when the dilute iodine color becomes pale yellow. There are tow reasons for such timing: 1- the iodine-starch complex is only slowly dissociated, and a diffuse end point would result if a large amount of the iodine were adsorbed on the starch. 2- most iodometric titrations are performed in strongly acid medium and the starch has a tendency to hydrolyze in acid solution.

The titration should be performed rapidly to minimize air oxidation of the iodide. Stirring should be efficient in order to prevent local excesses of thiosulfate, because it is decomposed in acid solution. In iodometric methods, a large excess of iodide is added to promote the reaction. The unreacted iodide dose not interfere, but it may be air-oxidized if the titration is not performed immediately. Sodium thiosulfate solution is standardized iodometrically against a pure oxidizing agent such as K 2 Cr 2 O 7, KIO 3, KBrO 3, or metallic copper (dissolved to give Cu 2+ ). With potassium dichromate, the deep green color of the resulting chromic ion makes it a little more difficult to determine the iodine-starch end point.

Common Titrant for Oxidation Reactions Iodine (Solution of I 2 + I - ) I 3 - is actual species used in titrations with iodine Either starch of Sodium Thiosulfate (Na 2 S 2 O 3 ) are used as indicator K = 7 x 10 2 Before endpoint Before endpoint At endpoint I StarchI3-I3- I S 2 O 3 2-