1. 1 Calculating the Cell Potential The process of calculating the cell potential is simple and involves calculation of the potential of each electrode separately, then the overall cell potential can be determined from the simple equation: E cell = E cathode - E anode Sometimes, this relation is written as: E cell = E right – E left This is true since the convention is to place the cathode to the right of the cell while the anode is placed to the left of the cell

2. 2 For example, look at the following cell representation: Cu(s)| CuSO 4 (0.100 M) || ZnCl 2 (0.200 M)|Zn(s) This cell is read as follows: a copper electrode is immersed in a 0.100 M CuSO 4 solution (this is the first half-cell, anode), 0.200 M ZnCl 2 solution in which a Zn(s) electrode is immersed (this is the second half-cell, cathode). If the E cell is a positive value, the reaction is spontaneous and if the value is negative, the reaction is nonspontaneous in the direction written and will be spontaneous in the reverse direction.

3. 3 Effect of Concentration on Electrode Potential The IUPAC convention for writing half-cell reactions is to represent the process as a reduction. The more positive half-cell reaction is the oxidizing agent (where the anode is) and the less positive half-cell reaction is the reducing agent (where the cathode is). The relationship between the concentration and the electrode potential for a half-cell reaction is represented by Nernst equation where for the half-cell reaction we have: aA + bB + ne  cC + dD E o = x V

4. 4 The Nernst equation can be written as: E = E o – (RT/nF) ln {[C] c [D] d /[A] a [B] b } Where, R is the molar gas constant (R = 8.314 J mol -1 K -1 ), T is the absolute temperature in Kelvin ( T = o C + 273) and F is the Faraday constant ( F = 96485 Coulomb.eq -1 ) and n is the number of electrons per mole. One may write after substitution: E = E o – (0.0592/n) log {[C] c [D] d /[A] a [B] b }

5. 5 Example Calculate the electrode potential for the half-cell below if the solution contains 0.100 M Cu 2+. The half-cell reaction is: Cu 2+ + 2e  Cu (s) E o = 0.521 V Solution E = E o – (0.0592/n) log [Cu(s)]/[Cu 2+ ] E = 0.521 – (0.0592/2) log 1/0.100 E = 0.491 V

6. 6 Example Calculate the electrode potential of a half-cell containing 0.100 M KMnO 4 and 0.0500 M MnCl 2 at pH 1.000. Solution MnO 4 - + 8 H + + 5 e  Mn 2+ + 4 H 2 O E o = 1.51 V E = E o – (0.0592/n) log [Mn 2+ ]/[MnO 4 - ][H + ] 8 E = 1.51 – (0.0592/5) log 0.0500/0.100*(0.100) 8 E = 1.42 V

7. 7 Redox Indicators Redox visual indicators are of two types. The first type is called specific while the other type is called nonspecific. An example of a specific indicator is starch where it forms a blue complex with iodine but not with iodide. When iodine is consumed, the blue complex disappears and the solution turns almost white indicating the end point. Another specific indicator is the permanganate ion where it is also called a self indicator. The purple color of the permanganate ion is converted to the colorless Mn 2+. Nonspecific redox indicators are very weak oxidants or reductants which have different colors of the oxidized and reduced forms. When all the analyte is consumed in a redox reaction, the first drops of excess titrant will react with the indicator, thus changing its color.

8. 8 The redox indicator equilibrium can be represented by the equation below: In OX + n e = In RED E o = x V E = E In o – (0.0592/n In ) log [In RED ]/[In OX ] The color of the oxidized form can be clearly distinguished when 10 [In RED ] = [In OX ] and the color of the reduced form can be clearly distinguished when [In RED ] = 10 [In OX ]. Substituting into the electrode potential equation we get:  E Color Change = E In o + (0.0592/n In ) Therefore, the color change of the indicator occurs around E o In and, in fact, very close to it. For a real titration, E o In should be as close as possible to the electrode potential at the equivalence point, E ep, and should thus be at the potential range of the sharp break of the titration curve

9. 9 Examples of some redox indicators

10. 10 Redox Titrations The idea of redox titrations is that the potential of a redox reaction will change as the concentration changes. Therefore, consider the following redox reaction, Fe 2+ + Ce 4+  Fe 3+ + Ce 3+ If we titrate Fe 2+ with Ce 4+, the concentration of Fe 2+ will decrease upon addition of Ce 4+ and thus the potential of an electrode immersed in a Fe 2+ solution will start to change. This means that the potential of the cell will be a function of Fe 2+ concentration. Thus, if the volume of Ce 4+ added is plotted against the electrode potential, a titration curve will be obtained. See the discussion below.

11. 11 Titration Curves Oxidation-reduction titrations and titration curves can be followed by measuring the potential of an indicator electrode (e.g. platinum) as compared to a reference electrode (has a fixed constant potential, like Saturated Calomel Electrode, SCE). The potential of the cell measured is: E cell = E solution – E reference electrode

12. 12 Voltmeter Indicator electrode Reference electrode Semipermeable barrier

13. 13 E cell = E solution – E reference electrode This means that: E solution = E cell + E reference electrode Assume the process of titration of Fe 2+ with Ce 4+, we have Fe 2+ + Ce 4+  Fe 3+ + Ce 3+ Fe 3+ + e  Fe 2+ E o = 0.771 V Ce 4+ + e  Ce 3+ E o = 1.70 V We have three cases to consider:

14. 14 1. Before equivalence point we have Fe 2+ /Fe 3+ couple and we can use the Nernst equation to calculate the electrode potential where: E Fe = E Fe o – (0.0592/n Fe ) log [Fe 2+ ]/[Fe 3+ ] Therefore, we calculate the concentration of Fe 2+ and Fe 3+ and insert the values in Nernst equation to find the resulting potential of the electrode. We can use the other half reaction to calculate the potential of the cell but it is easier to use the Fe 2+ /Fe 3+ couple since the concentration of Ce 4+ is difficult to calculate at this stage. The Nernst equation for such potential calculation is: E Ce = E Ce o – (0.0592/n Ce ) log [Ce 3+ ]/[Ce 4+ ]

15. 15 2. At the equivalence point, we may combine the two Nernst equations for the two half reactions by adding them up. We get: n Fe E Fe + n Ce E Ce = n Fe E Fe o + n Ce E Ce o – (0.0592) {log [Fe 2+ ]/[Fe 3+ ] + log [Ce 3+ ]/[Ce 4+ ]} n Fe E Fe + n Ce E Ce = n Fe E Fe o + n Ce E Ce o – (0.0592) {log [Fe 2+ ] [Ce 3+ ]/[Fe 3+ ] [Ce 4+ ]} But [Fe 3+ ] = [Ce 3+ ] and [Fe 2+ ] = [Ce 4+ ] at equivalence point.

16. 16 Therefore, the equation simplifies to: n Fe E Fe + n Ce E Ce = n Fe E Fe o + n Ce E Ce o Also, at equivalence point we have E Fe = E Ce = E eq pt, therefore: (n Fe + n Ce ) E eq pt = n Fe E Fe o + n Ce E Ce o The final equation can be written as: E eq pt = (n Fe E Fe o + n Ce E Ce o )/ (n Fe + n Ce )

17. 17 3. After equivalence point, it is easier to calculate the cell potential from the cerium half-cell using the equation: E Ce = E Ce o – (0.0592/n Ce ) log [Ce 3+ ]/[Ce 4+ ] It is easier to calculate the Ce 3+ and Ce 4+ concentrations and it will be difficult to calculate the Fe 2+ concentration since it is totally consumed, except for an equilibrium concentration.