CHEM Pharmacy Week 9: Nernst Equation. Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone:

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1 CHEM Pharmacy Week 9: Nernst Equation Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: siegbert.schmid@sydney.edu.au

2 Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd ISBN:

3 Electrochemistry Blackman, Bottle, Schmid, Mocerino & Wille: Chapter 12, Sections 4.8 and 4.9 Key chemical concepts: Redox and half reactions Cell potential Voltaic and electrolytic cells Concentration cells Key Calculations: Calculating cell potential Calculating amount of product for given current Using the Nernst equation for concentration cells Lecture 25-3

4 Recap: Standard cell potential The measured voltage across the cell under standard conditions is the standard cell potential E 0 cell (also called emf). Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Zn (s) + Cu 2+ (aq) à Zn2+ (aq) + Cu (s) Lecture 26-4

5 Tricks to memorise anode/cathode 1. Anode and Oxidation begin with a vowels, Cathode and Reduction with consonants. 2. Alphabetically, the A in anode comes before the C in cathode, as the O in oxidation comes before the R in reduction. 3. Think of this picture: AN OX and a RED CAT (anode oxidation reduction cathode) Lecture 26-5

6 Standard cell potential and free energy For a spontaneous reaction, E 0 cell and also ΔG0 For a non-spontaneous reaction, E 0 cell and also ΔG 0 So there is a proportionality between E 0 and -ΔG 0. You also know that the maximum work done on the surroundings is -w max = ΔG Electrical work done by the cell is w = E cell charge Lecture 26-6

7 Standard cell potential and free energy The emf E 0 cell is related to the change in free energy of a reaction: G 0 = nfe 0 cell G 0 = Standard change in free energy n = number of electrons exchanged Also, away from standard conditions: F = C/mol e - (Faraday constant) G = nfe cell v But what is E cell? Lecture 26-7

8 Example Calculate G 0 for a cell reaction: Cu 2+ (aq) + Fe (s) D Cu (s) + Fe2+ (aq) Is this a spontaneous reaction? Cu e D Cu Fe e D Fe E 0 cell = E 0 = 0.34 V E 0 = 0.44 V G 0 = nfe 0 cell G 0 = This process is spontaneous as indicated by the negative sign of ΔG 0 and the positive sign for E 0 cell. Lecture 26-8

9 Example: a Dental Galvanic Cell Al(s) Al 3+ (aq) O 2 (aq), H + (aq), H 2 O(aq) Ag,Sn,Hg Al is very easily oxidised, Al e à Al E 0 = -1.66V. The filling is an inactive cathode for the reduction of oxygen, O 2 + 4H + + 4e à 2H 2 O and saliva is an electrolyte. Put the three together (biting on a piece of foil) results in generation of a current and possible pain. Lecture 26-9

10 Example: a Dental Galvanic Cell Al(s) Al 3+ (aq) O 2 (aq), H + (aq), H 2 O(aq) Ag,Sn,Hg O 2 + 4H + + 4e à 2H 2 O E 0 = 1.23 V Al e à Al E 0 = 1.66 V 12H + + 3O 2 + 4Al à 6H 2 O + 4Al 3+ E 0 cell = G 0 = nfe 0 cell = Lecture 26-10

11 Nernst Equation Recall ΔG = ΔG 0 + RT ln(q) (1) Since ΔG 0 = -nfe 0 cell and ΔG = -nfe Image fromnobelprize.org Equation (1) becomes Walther Nernst Nobel Prize nfe = -nfe 0 cell + RTln(Q) Q = [products] / [reactants] dividing both sides by nf gives: E = E 0 RT ln(q) nf Lecture 26-11

12 Nernst Equation E cell = E 0 RT ln(q) nf Since ln (x) = log (x) E = E RT log(q) nf E = cell potential E 0 = Standard cell potential R = Real Gas Constant = JK -1 mol -1 T =Temperature (K) n = no. of e - transferred F = Faraday constant = C mol -1 Q = Reaction quotient (Q = K at equilibrium) At 25 C, (2.303 R 298)/96485 = E cell = E n log( Q) Nernst equation more commonly written like this (note: only at 25 C) Lecture 26-12

13 Example calculation 1 Calculate the expected potential for the following cell: Zn(s) + Cu 2+ (aq) D Zn 2+ (aq) + Cu(s) E 0 = 1.1 V i) [Cu 2+ ] = 1.0 M; [Zn 2+ ] = 10-5 M ii) [Cu 2+ ] = 10-5 M; [Zn 2+ ] = 1.0 M E cell = E log( Q) n Firstly, work out the value of n : Cu e - D Cu Zn D Zn e - Zn(s) + Cu 2+ (aq) D Zn 2+ (aq) + Cu(s) Lecture 26-13

14 Example calculation Zn(s) + Cu 2+ (aq) D Zn 2+ (aq) + Cu(s) E cell = E log( Q) n (n=2) [ Zn Q = [ Cu ] ] [Zn 2+ ] (M) [Cu 2+ ] (M) Q log(q) E cell (V) Lecture 26-14

15 Demo: The effect of concentration 0.00 V Cu Cu 2+ Cu 2+ Cu Both compartments of the voltaic cell are identical. E 0 cell = E0 copper E0 copper = 0 (in standard conditions, 1M concentrations) What happens when the concentration of one cell is changed? Lecture 26-15

16 Demo: Cu Concentration Cell Low [Cu 2+ ] Cu D Cu e- High [Cu 2+ ] Cu e- D Cu Cu Cu 2+ Cu 2+ Cu E 0 same for both half-reactions, so E 0 cell = 0. However, we have reduced the concentration of Cu 2+ in one cell = nonstandard conditions. Electrical energy is generated until the concentrations in each half-cell become equal (equilibrium is attained). Add Na 2 S precipitate forms. Can we explain this? Lecture 26-16

17 Cu Concentration Cell Cu Cu 2+ Cu 2+ Cu Low [Cu 2+ ] High [Cu 2+ ] Cu D Cu e - Cu e - D Cu E 0 cell is same on both sides, but the Cu concentrations are different. More charge carriers in one half-cell. If we poured the two solutions together, we would expect spontaneous mixing of two solutions of different concentrations to give one of equal concentration. The electrical connection allows electrons to pour from one half-cell to the other. Lecture 26-17

18 Concentration cells The measured cell potential in our experiment was Voltage. Let s work out what the voltage should be: Voltage Cu à Cu 2+ (0.01 M) + 2e - Cu 2+ (0.1M) + 2e - à Cu Cu 2+ (0.1M) à Cu 2+ (0.01 M) E cell = Voltage E cell y = E log( Q) = log n 0.1 = Voltage Solve for Voltage : Voltage = V Lecture 26-18

19 Concentration cells The cell potential depends on the concentration of reactants. Corollary: It is useful to define a standard concentration, which is 1 M. Implication: We need to specify concentration when referring to the cell potential. The overall potential for the Cu/Cu 2+ concentration cell is: E = E 0 cell /2 log [Cu2+ ] dil / [Cu 2+ ] conc Lecture 26-19

20 Reference Electrodes 1. Standard Hydrogen Electrode (SHE) 2. Metal-Insoluble Salt Electrode: Standard Calomel Electrode (SCE) and Silver Electrode 3. Ion-Specific: ph electrode Lecture 26-20

21 Standard Hydrogen Electrode (SHE) Finely divided surface Pt electrode. HCl solution with [H + ] =1, H 2 p= 1 atm bubbling over the electrode. H 2 absorbs on the Pt, forming the equivalent of a 'solid hydrogen electrode in equilibrium with H +. Platinum gas electrode H 2 electrode H 2 D 2H e - E o = 0.00 V Metal Metal ion Electrode Cu e - D Cu E o = 0.34 V Pt H 2 (g) H + (aq) Cu 2+ (aq) Cu(s) E o cell = = 0.34 V Lecture 26-21

22 Metal-insoluble salt electrodes The Standard Hydrogen Electrode (SHE) is not convenient to use in practice (can be contaminated easily by O 2 or organic substances). There are more practical choices, like metal - insoluble salt electrodes. The potential of these electrodes depends on the concentration of the anion X - in solution. In practice 2 interfaces: 1. M / MX insoluble salt: MX (s) + e - à M (s) + X - (s) C + X - X - C + MX 2. coating/solution: X - (s) à X - (aq) Overall: MX (s) + e - (metal) à M (s) + X - (aq) X - C + M X - C + Lecture 26-22

23 Metal-insoluble salt electrodes The concentration of anions in solution is controlled by the salt's solubility: MX C + X - X - M X - C + C + [Ag + ] [X - ] = K sp C + X - Normal calomel electrode, Pt Hg Hg 2 Cl 2 KCl (1M) E 0 = 0.28 V Saturated calomel electrode Pt Hg Hg 2 Cl 2 KCl(sat.) E 0 = 0.24 V Silver/Silver chloride, Ag AgCl Cl - (1M) E 0 = 0.22 V (used in ph meters) Lecture 26-23

24 Saturated Calomel Electrode The saturated calomel electrode (SCE) features the reduction halfreaction: Hg + + e D Hg Hg 2 Cl 2 D 2Hg + + 2Cl Overall: Hg 2 Cl 2 (s) + 2e à 2 Hg (s) + 2Cl (sat) Pt Hg Hg 2 Cl 2 KCl 5 M Standard cell potential of E 0 = 0.24 V. 2 Lecture 26-24

25 Cell Potentials 1 Q: The standard reduction potential of Zn 2+ /Zn is V. What would be the observed cell potential for the Zn/Zn 2+ couple when measured using the SCE as a reference? Calomel: Hg 2 Cl 2 (s) + 2e - à 2Hg(l) + 2Cl - (aq) E 0 = 0.24V Zn à Zn e - E 0 = V (reversed because it is written as an oxidation) 0.24 Hg + /Hg 0.0V H 2 /H + Ans: The Zn will be oxidised (lower reduction potential), so E (cell) = = 1.00 V respect to the SCE So to get the oxidation half-reaction E 0 using the SCE as cathode, subtract 0.24 V from the volt meter reading Zn 2+ /Zn Lecture 26-25

26 Summary CONCEPTS q q Concentration cells Nernst equation CALCULATIONS q q Work out cell potential from reduction potentials; Work out cell potential for any concentration (Nernst equation) Lecture 27-26

Chapter 18 Electrochemistry. Electrochemical Cells

Chapter 18 Electrochemistry. Electrochemical Cells Chapter 18 Electrochemistry Chapter 18 1 Electrochemical Cells Electrochemical Cells are of two basic types: Galvanic Cells a spontaneous chemical reaction generates an electric current Electrolytic Cells