Chapter 12 Electrochemical cells and electrode potentials
Reduction and Oxidation Reactions Predict what might happen when a piece of copper wire in a solution of 2% AgNO 3. If you try this experiment, you will initially see that the copper is a shiny copper color and the solution is clear. In less than one hour the solution is light blue and the wire is covered with shiny silver needles. What happened? Copper metal became copper ions in solution and silver ions became silver metal. Cu (s) + Ag + (aq) Cu2+ (aq) + Ag (s) (unbalanced) The Cu (s) loses electrons to become Cu 2+ (aq) ions and the Ag+ (aq) ions gain electrons to become Ag (s). Reactions that involve the exchange of electrons are called reduction and oxidation (redox) reactions. When a chemical species loses electrons we say that it is oxidized, and when a chemical species gains electrons we say that it is reduced. The Cu (s) loses electrons to be oxidized to Cu 2+ (aq). The Ag + (aq) gain electrons to be reduced to Ag (s).
We can break the reaction into the following two half reactions. Cu (s) Cu 2+ (aq) + 2e- Ag + (aq) + e- Ag (s) What would you predict if you placed a piece of Ag metal in a solution of Cu 2+? Since we observed that the reaction of Ag + and Cu is spontaneous, we would not expect the reverse reaction to be spontaneous. So no reaction occurs between Ag metal and Cu 2+. Or we call it nonspontaneous reaction
Terminology Reduction: gaining of electrons (decrease in the oxidation state Oxidation: loss of electrons (increase in the oxidation state) Reducing agent (reductant): species that donates electrons to reduce another reagent. (The reducing agent get oxidized.) Oxidizing agent (oxidant): species that accepts electrons to oxidize another species. (The oxidizing agent gets reduced.)
Oxidation-reduction reaction (redox reaction): a reaction in which electrons are transferred from one reactant to another. For example, the reduction of cerium(iv) by iron(ii): Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ The reduction half-reaction is given by... Ce 4+ + e - Ce 3+ The oxidation half-reaction is given by... Fe 2+ e - + Fe 3+ The half-reactions are the overall reaction broken down into oxidation and reduction steps. Half-reactions cannot occur independently, but are used conceptually to simplify understanding and balancing the equations
Rules for Balancing Oxidation-Reduction Reactions Write out half-reaction "skeletons." Balance the half-reactions by adding H +, OH - or H 2 O as needed, maintaining electrical neutrality. Combine the two half-reactions such that the number of electrons transferred in each cancels out when combined. For example, consider the following reaction of the peroxydisulfate ion with manganese ion: S 2 O 8 2- + Mn 2+ SO 4 2- + MnO 4 -
1. The reduction step is: S 2 O 8 2- + 2 e - 2 SO 4 2- (Each sulfur atoms goes from +7 to +6 oxidation state.) 2. The oxidation step is: Mn 2+ + 4 H 2 O 5 e - + MnO 4 - + 8 H + (Manganese(II) loses 5 electrons, going from +2 to +7.) 3. In combining the two equations, the oxidation step must be multiplied by "2," and the reduction step must be multiplied by "5" to cancel out the electrons: 2x[Mn 2+ + 4 H 2 O 5 e - + MnO 4 - + 8 H + ] 5x[S 2 O 8 2- + 2 e - 2 SO 4 2- ] 4. Adding these two equations together: S 2 O 8 2- + 2 Mn 2+ + 8 H 2 O 10 SO 4 2- + 2 MnO 4 - + 16 H + 5. Note that the half-reactions are charge-balanced before adding them together.
Consider the next example: Cr 2 O 7 2- + I - Cr 3+ + I 3 - The reduction step is given by... Cr 2 O 7 2- + 6 e - + 14 H + 2 Cr 3+ + 7 H 2 O (Cr(IV) in the dichromate on is reduced to Cr(III).) The oxidation step is give by... 3 I - 2 e - + I 3 - Multiplying the oxidation half-reaction by 3x and adding the two half-reactions together: Cr 2 O 7 2- + 9 I - + 14 H + 2 Cr 3+ + 3 I 3 - + 7 H 2 O
Oxidation-Reduction Reactions in Electrochemical Cells It is possible to separate the half-reactions of an oxidationreduction reaction in an electrochemical cell. Consider the following reaction: 2 Ag + + Cu (s) 2 Ag (s) + Cu 2+ The reduction half-reaction is given by... Ag + + e - Ag (s) The oxidation half-reaction is given by... Cu (s) 2 e - + Cu 2+
Redox reactions can be "separated" in a galvanic cell (also called a voltaic cell or battery):
Charactristics of the of the galvanic cell (battery) The left container (half cell) contains of 0.020 M CuSO 4. The right cell contains 0.0200 M AgNO 3. A Cu electrode is immersed in the CuSO 4 solution. An Ag electrode in immersed in the AgNO 3 solution. The two solutions "communicate" via a salt bridge which consists of a saturated KCl solution in a tube with glass frits in both ends. At the anode, oxidation takes place: Cu (s) 2 e - + Cu 2+ At the cathode, reduction takes place: 2 Ag + + 2 e - 2 Ag (s)
Chloride ions move into the CuSO 4 solution to maintain electrical neutrality. Potassium ions move into the AgNO 3 solution to maintain electrical neutrality. The volt meter reads 0.412 V at the instant the connection is made between the two electrodes. (This represents the difference is voltage ( E cell ) between the two electrodes. a. The copper electrode has an initial voltage of 0.2867 V. b. The silver electrode has an initial voltage of 0.6984 V. c. Voltage Difference = Voltage cathode - Voltage anode Voltage Difference = 0.6984-0.2867 = + 0.412 V This initial voltage drops as the reaction proceeds toward equilibrium as soon as connection is made. At equilibrium, the voltage read zero volts. The potential difference (voltage) between the anode and cathode is a measure of the tendency of the reaction to proceed from nonequilibrium to equilibrium
Operation of an electrolytic cell An external voltage source with a voltage larger than that of the battery is connected to the galvanic cell - positive pole to the silver electrode; negative pole to the copper electrode. The external voltage source reversed the direction of electron flow (and reverses the direction of the reactions at each electrode): The silver electrode switches from the cathode to the anode (the site of oxidation). The copper electrode switches from the anode to the cathode (the site of reduction. The voltage meter will read "negative" if still connected in the original fashion. The battery will "recharge.
Schematic Representation of Cells The copper(ii) sulfate/silver nitrate systems described above would be symbolized: Cu CuSO 4 (0.02 M) AgNO 3 (0.02 M) Ag Each vertical line (" ") represents a phase boundary or interface where a potential develops. Each double vertical line (" ") represents two phase boundaries (e.g., the salt bridge). The direction of electron flow is from left to right: Cu--->Ag.
The original, initial voltage for the cell The original, initial voltage for the cell is given by... E cell = E right - E left = E cathode - E anode = E Ag - E Cu The potentials (absolute voltages) at the two electrodes cannot be determined experimentally. Only the differences between electrodes (via a volt meter) can be measured. Potentials at electrodes are assigned relative values, based upon comparison to a standard. By convention, the standard hydrogen electrode (SHE) is used as the "agreed upon" reference half-cell against which all others are compared. The voltage or cell potential is related to the free energy of the reaction driving the cell: G = -nf E cell = -2.303RTlog(K eq )
Standard Hydrogen Electrode (SHE) The half-cell reaction is given by: 2 H 1+ + 2 e - H 2(g) The SHE (standard hydrogen electrode) is the reference point for determining relative electrode potentials. The SHE is symbolized: Pt, H 2 ( hydrogen = 1.00 atm) H+ (a H+ = 1.00) Ag + (a Ag+ = 1.00) Ag) The platinum electrode is a platinized platinum electrode (platinum coated with finely divided platinum called platinum black). The aqueous acid solution has an activity of 1.00 (i.e., approximately 1.00 M hydrogen ion) and is saturated with hydrogen gas, bubbled in at 1.00 atmosphere. The SHE is connected via the salt bridge and connecting wire to the other half-cell of the battery.
The SHE acts as either the anode or cathode, depending upon whether electrons are given up to the other half-cell or taken in. By convention, the SHE is assigned an absolute potential (voltage) = 0.00 volts at ALL temperatures. All other half-cells (and half-reactions) are measured relative to the SHE. By definition, the electrode potential is the potential of a cell with the standard hydrogen electrode acting as the anode and the other halfcell acting as the cathode. For example, consider the cell formed with the SHE as one half-cell and an Ag/Ag + half-cell on the other side consisting of a silver electrode and an a Ag+ = 1.00 silver nitrate solution. At the anode (oxidation): SHE 2 H + + 2 e - H 2(g) At the cathode (reduction): Ag/Ag + Ag + + e - Ag (s) As the instant the connection between half-cells is made, the experimentally observed difference in potential between the two electrodes measured by the volt meter (i.e., the battery voltage) is +0.799 V. Electrons flow from the SHE to the Ag electrode. By convention, the sign of the voltage is positive if electrons leaves the SHE; negative if electrons are taken in.
Consider what happens if the Ag/Ag 1+ half-cell is replaced with a Cd/Cd 2+ half-cell (a Cd+2+ = 1.00) attached to an SHE. 1) At the anode (oxidation): Cd/Cd 2+ Cd Cd 2+ + 2 e 1-2) At the cathode (reduction): SHE 2 H 1+ + 2 e 1- H 2(g) 3) At the instant the connection between half-cells is made, the experimentally observed difference in potential between the two electrodes measured by the volt meter (i.e., the battery voltage) is a negative voltage (-0.403 V). 4) Electrons flow to the SHE from the Cd electrode. 5) By convention, the sign of the voltage is negative if electrons are taken in by the SHE. This means that the half-cell reaction opposite the SHE is more reducing than the SHE. 6) A voltage greater than 0.403 V would have to be applied to reverse the flow of electrons.
Accepted IUPAC Conventions and Electrode Potentials Half-reactions are always written as reductions (i.e., electrodes potentials are by definitions reduction potentials). The sign of the electrode potential is determined relative to the SHE. Positive (+) means electrons flow out of the SHE to the other electrode (i.e., the SHE acts an the anode). Negative (-) means electrons flow to the SHE from the other electrode (i.e., the SHE becomes the cathode). The sign of the electrode potential signifies whether the net reaction of the battery is spontaneous "to the right" or "to the left." For example: The Ag/Ag + half-reaction has an electrode potential of +0.799 V, meaning electrons flow to the Ag electrode. The Cd/Cd 2+ half-reaction has an electrode potential of -0.403 V, meaning electrons flow to the SHE (and the reactions occur opposite to the directions written above).
Nernst equation The effect of species concentration of electrode potentials is described by the Nernst equation. For the reversible halfreaction: aa + bb +... ne 1- cc + dd +... where... E = the electrode potential E o = the standard electrode potential (i.e., the potential observed when species in the half-cell are at a = 1.00 or pressure = 1.00 atm) R = the gas constant (8.314 JK -1 mol -1 ) T = Kelvin temperature n = number of moles of electrons in balanced half reaction F = Faraday's constant (96,485 coulombs/mole; the charge on a mole of electrons).
Note that the standard electrode potential (E o ) is measured under standard conditions (a = 1, pressure = 1 atm). The Nernst equation corrects for nonstandard concentrations. 2.303RT/nF simplifies to 0.0592/n. E = E o if all species are at a = 1 and pressure = 1 atm.
Calculating battery voltages The battery voltage is given by: E cell = E right - E left = E cathode - E anode Each half-cell electrode potential is calculated, and the difference between electrode determined.
Applications of electrode potentials Redox titrations utilized to measure analytes via oxidation-reduction titrations will be discussed. This will include: Equilibrium constants for redox reactions. Titrations curves - shapes, endpoints, calculations, etc. Indicators for oxidation-reduction reactions.
Measuring Equilibrium Constants with Redox Reactions Consider the reaction: 2 Ag + + Cu (s) 2 Ag (s) + Cu 2+ with associated half-reactions: 2 Ag + + 2 e - 2 Ag (s) (reduction step) Cu(s) 2 e - + Cu 2+ (oxidation step) The equilibrium constant for this reaction is given by: The reaction above is the same as the galvanic cell... Cu Cu 2 + (?? M) Ag + (?? M) Ag
When this galvanic cell completely discharges to reach equilibrium, E cell = E right - E left = E cathode - E anode = E Ag - E Cu = 0 which means: E Ag = E Cu Substituting for each from the Nernst equation: Note that K eq is so large, the equilibrium lies heavily toward the right
Oxidation-Reduction Titrations
Oxidation-Reduction Titrations When a oxidation-reduciton (redox) reaction is used to measure an analyte, the titrations usually follow the electrode potential as a function of titrant (or analyte) concentration. Since the electrode potential is a log function of the concentrations, it behaves as a "p" function.
We will discuss the redox titrations in terms of a specific reaction and look at the calculations for each of the different regions of the titration curve. This would include: Initially (before any titrant was added) Before the equivalence point At the equivalence point After the equivalence point
Consider titration of iron(ii) with cerium(iv) Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ Oxidation half-reaction: Fe 2+ Fe 3+ + e - Reduction half-reaction: e - + Ce 4+ Ce 3+ The reactions must be fully reversible (i.e., the system must be at equilibrium at all times through the titration). To be fully reversible means that: E system = E Ce = E Fe = E indicator and E cell = 0 The equilibrium concentration ratios of the oxidized and reduced forms of the two species are such that their attraction for electrons are identical. The Nernst equation applies. Data from the titration can be used to calculate the titration curve using the Nernst equation for either the cerium(iv) or the iron(ii) half-reactions.
In practice: Before equivalence, the Fe(III) and Fe(II) concentrations are used to calculate the E system. After equivalence, Ce(III) and Ce(IV) concentrations are used to calculate the E system. At equivalence, simplifying assumptions are made based upon the stoichiometric relationships to calculate the E system. Example: 50.00 ml of 0.05000 M Fe 2+ is titrated with 0.1000 M Ce 4+ in a medium containing 1.0 M H 2 SO 4 Derivation/calculation of the titration curve 1. Initial region: no Ce(IV) has been added Since the amount of Fe 3+ cannot be calculated (and is essentially zero), and the amount of Ce 3+ is zero, the potential cannot be calculated. Some amount of reaction must occur before numbers can be plugged into the Nernst equation.
2. Before equivalence assume 5.00 ml of Ce 4+ solution have been added
3. At equivalence All the iron(ii) is converted to iron(iii); all the cerium(iv) is converted to cerium(iii) Note that at the equivalence point: [Fe 3+ ] = [Ce 3+ ] [Fe 2+ ] = [Ce 4+ ] ==> a very small amout!! The system potential can be calculated by combining the Nernst equations for both species as follows:
4. Beyond the equivalence point assume 25.10 ml of cerium(iv) have been added The iron(ii) is completely titrated. b. The cerium(iii) and cerium(iv) concentrations must be used to to calculate the potential of the system. Note that the number of moles of cerium(iii) is equal to the number of moles of iron(ii) originally present in the sample. Cerium(IV) present is due to the excess cerium(iv) added past equivalence. The system potential is calculated as follows:
Substituting into the Nernst equation:
Summary Of Redox Titration Curve 1. Titration begins beyond E o for reactant 2. 50% titrated E = E o for reactant 3. EP 200% titrated E = E o 2
Oxidation-Reduction Indicators Two types of indicators are commonly used: 1. Specific indicators. 2. Oxidation-reduction indicators ("true" redox indicators). Specific indicators are species that react with one of the titration species to produce a color change. "True" redox indicators respond to the system potential to produce a color change. Redox indicators are more versatile than specific indicators, since they depend on the system potential, not the specific reaction. The half-reaction for the redox indicator is: In (oxid) + n e - In (red)
Like ph indicators, the color of redox indicators must change by about 10-fold to be seen. This means to see a full color transition, ratio of the oxidized and reduced indicator species must change from 1:10 to 10:1. If these ratios are plugged into the Nernst equations for various "n" values, the change in potential at the endpoint should correspond to E In o +/- 0.0592/n to get an appropriate color change. For n = 1, the range around the endpoint is E In o +/- 0.0592/1 = +/- 0.0592 V (meaning the equivalence point must span this "window" for the indicator to work). For n = 2, the range around the endpoint is +/- 0.0296 V. For n = 3, the range around the endpoint is +/- 0.0197 V In practice, the equivalence point of a titration is first calculated, then a redox indicator whose E In o most closely matches the equivalence point is chosen. Consideration must also be given to the titration "break" at endpoint. If the standard electrode potentials of the analyte and titrant are too close to each other (<~0.40 V), it is nearly impossible to titrate the species. (The reaction is not complete enough to give a clear endpoint transition, and an indicator can not be "fit" to the titration.)
Examples of Specific Indicators Starch-I 3 - complex In the presence of I 3-, starch forms a deep blue/black color due to the formation of a complex between the starch and the I 3-. I 2(s) + I - (aq) I 3 - (aq) + Starch I 3 - (aq) I 3 - (aq). Starch (blue color) The presence/absence of I 3 - (aq) in solution can be used as an indicator. For example, consider the following titration of copper(ii): 2 Cu 2+ + 5 I - 2 CuI (s) + I 3 - If excess iodide is added, the reaction is driven to completion, and the I 3 1- is back titrated with S 2 O 3 2- : I 3 - + 2 S 2 O 3 2-3 I - + S 4 O 6 2- (tetrathionate ion)
I 3 - appears in solution as a yellowish color, and is visible to approximately 5 x 10-6 M concentration. During back-titration, as the yellow disappears, starch may be added to produce a blue color that disappears when the last I 3 1- is titrated (i.e., an endpoint). The blue color of the I 3 - (aq). starch indicator is visible to approximately 2 x 10-7 M concentration.
Thiocyanate. Iron(III) Complex Thiocyanate reacts with iron(iii) to produce a deep, red color: Fe 3+ + SCN - FeSCN 2+ (red) The red color can be used to detect the presence of Fe 3+ in titrations.
Example of redox indicators 1,10-phenanthrolines
Applications Oxidizing Agents KMnO 4 - titrant for Fe, Sn, and Oxalate Ce 4+ - Very stable in strong sulfuric acid, Titration of Fe and organics K 2 Cr 2 O 7 - little used now due to hazards and waste disposal of Cr I 2 - add KI to increase solubility - Vit C, wide range of applications Karl Fischer Detn of H 2 O
Reducing Agents Not too many of these. Fe 2+ not very stable Fe(NH 3 ) 2 (SO 4 ) 2 Iodide Thiosulfate - S 2 O 3 2- ===> S 4 O 6 2- + 2e - Many common REDOX procedures use a reducing agent to reduce the sample and then use an oxidizing agent titrant for analysis rather than a reducing agent titrant. eg. reduce Fe 3+ to Fe 2+ then titrate with Ce.