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CHEM1612 - Pharmacy Week 9: Nernst Equation

CHEM1612 - Pharmacy Week 9: Nernst Equation. Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: siegbert.schmid@sydney.edu.au. Unless otherwise stated, all images in this file have been reproduced from:

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CHEM1612 - Pharmacy Week 9: Nernst Equation

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  1. CHEM1612 - PharmacyWeek 9: Nernst Equation Dr. SiegbertSchmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: 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. 2008      ISBN: 9 78047081 0866

  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

  4. Recap: Standard cell potential • The measured voltage across the cell under standard conditions is the standard cell potential E0cell (also called emf). Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) Zn(s) + Cu2+(aq)Zn2+(aq) + Cu(s) E0cell = E0cathode – E0anode = 0.34 - ( - 0.76)= 0.34 +0.76 = 1.10 V

  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)

  6. Standard cell potential and free energy • For a spontaneous reaction, E0cell > 0 and also ΔG0 < 0 • For a non-spontaneous reaction, E0cell <0 and also ΔG0 > 0 • So there is a proportionality between E0 and -ΔG0. • You also know that the maximum work done on the surroundings is -wmax = ΔG • Electrical work done by the cell is w = Ecell× charge

  7. ∆G0 = –nFE0cell ∆G = –nFEcell Also, away from standard conditions: Standard cell potential and free energy • The emfE0cell is related to the change in free energy of a reaction: ∆G0 = Standard change in free energy n = number of electrons exchanged F = 96485 C/mol e- (Faraday constant) • But what is Ecell?

  8. Example Calculate ∆G0 for a cell reaction: Cu2+(aq) + Fe (s) Cu(s) + Fe2+(aq) Is this a spontaneous reaction? Cu2+ + 2e– Cu E0 = 0.34 V Fe2+ + 2e– Fe E0 = –0.44 V E0cell = 0.34 - (-0.44) = 0.78 V ∆G0 = –nFE0cell ∆G0 = – 2 · 96485 · 0.78 = – 1.5 · 105 J This process is spontaneous as indicated by the negative sign of G0 and the positive sign for E0cell.

  9. Example: a Dental Galvanic Cell Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg Al is very easily oxidised, Al3+ + 3e−Al Eo = -1.66V. The filling is an inactive cathode for the reduction of oxygen, O2 + 4H+ + 4e−2H2O and saliva is an electrolyte. Put the three together (biting on a piece of foil) results in generation of a current and possible pain.

  10. Example: a Dental Galvanic Cell Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg O2 + 4H+ + 4e– 2H2OE0= 1.23 V Al3+ + 3e– AlE0= –1.66 V 12H+ + 3O2 + 4Al 6H2O + 4Al3+ E0cell= 2.89 V ∆G0=–nFE0cell= –12 · 96485 · 2.89 = –3346 kJmol–1

  11. Nernst Equation • But what is Ecell? Image fromnobelprize.org Recall ΔG = ΔG0 + RT ln(Q) (1) Since ΔG0 = -nFE0cell and ΔG = -nFE Equation (1) becomes -nFE = -nFE0cell + RTln(Q) Q = [products] / [reactants] dividing both sides by –nF gives: E = E0 – RTln(Q) nF Walther Nernst Nobel Prize 1920

  12. Nernst Equation E = cell potential E0 = Standard cell potential R = Real Gas Constant = 8.314 JK-1mol-1 T =Temperature (K) n = no. of e- transferred F = Faraday constant = 96485 C mol-1 Q = Reaction quotient (Q = K at equilibrium) Ecell = E0 – RT ln(Q) nF Since ln (x) = 2.303 log (x) E = E0 – 2.303 ·RT log(Q) nF At 25°C, (2.303·R·298)/96485 = 0.0592 Nernst equation more commonly written like this (note: only at 25°C)

  13. Cu2+ + 2e-  Cu Zn  Zn2+ + 2e- 2 mol e- transferred per mole of reaction: n = 2 Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Example calculation 1 Calculate the expected potential for the following cell: i) [Cu2+] = 1.0 M; [Zn2+] = 10-5M ii) [Cu2+] = 10-5M; [Zn2+] = 1.0 M Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) E0 = 1.1 V Firstly, work out the value of n :

  14. Example calculation Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) (n=2) 1.0 10-5 105 0.0 -5.0 5.0 1.10 1.25 0.95

  15. 0.00 V Demo: The effect of concentration Cu|Cu2+||Cu2+|Cu Both compartments of the voltaic cell are identical. E0cell = E0copper – E0copper = 0 (in standard conditions, 1M concentrations) What happens when the concentration of one cell is changed?

  16. High [Cu2+] Low [Cu2+] Cu2+ + 2e- Cu Cu  Cu2++ 2e- Add Na2S –precipitate forms. Demo: Cu Concentration Cell Cu |Cu2+||Cu2+|Cu E0same for both half-reactions, so E0cell= 0. However, we have reduced the concentration of Cu2+ in one cell = non-standard conditions. Electrical energy is generated until the concentrations in each half-cell become equal (equilibrium is attained). Can we explain this?…

  17. Cu Concentration Cell Cu|Cu2+||Cu2+|Cu Low [Cu2+] High [Cu2+] Cu  Cu2++ 2e- Cu2+ + 2e-Cu • E0cellissame 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.

  18. Cu Cu2+ (0.01 M) + 2e- Cu2+ (0.1M) + 2e- Cu Cu2+ (0.1M) Cu2+ (0.01 M) Ecell = Concentration cells “Voltage” • The measured cell potential in our experiment was “Voltage”. Let’s work out what the voltage should be: “Voltage” 0.01 = “Voltage” Solve for “Voltage”: “Voltage” = 0.0296 V

  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/Cu2+ concentration cell is: E = E0cell – 0.0592/2 · log [Cu2+]dil / [Cu2+]conc

  20. Reference Electrodes • Standard Hydrogen Electrode (SHE) • Metal-Insoluble Salt Electrode: Standard Calomel Electrode (SCE) and Silver Electrode • Ion-Specific: pH electrode

  21. Standard Hydrogen Electrode (SHE) Finely divided surface Pt electrode. HCl solution with [H+] =1, H2 p= 1 atm bubbling over the electrode. H2 absorbs on the Pt, forming the equivalent of a 'solid hydrogen‘ electrode in equilibrium with H+. Platinum – gas electrode H2 electrode H2 2H+ + 2 e- Eo = 0.00 V Metal – Metal ion Electrode Cu2+ + 2e- Cu Eo= 0.34 V Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s) Eocell= 0.34 – 0 = 0.34 V anode cathode

  22. X- C+ MX C+ X- M C+ X- C+ X- Metal-insoluble salt electrodes • The Standard Hydrogen Electrode (SHE) isn't convenient to use in practice (can be contaminated easily by O2 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-  M (s) + X-(s) 2. coating/solution: X- (s) X- (aq) Overall: MX (s) + e-(metal) M (s) + X- (aq)

  23. X- X- C+ MX C+ X- M C+ X- C+ Metal-insoluble salt electrodes • The concentration of anions in solution is controlled by the salt's solubility: [Ag+] [X-] = Ksp • Normal calomel electrode, Pt | Hg | Hg2Cl2 | KCl (1M) E 0 = 0.28 V • Saturated calomel electrode Pt | Hg | Hg2Cl2 | KCl(sat.) E0 = 0.24 V • Silver/Silver chloride, Ag | AgCl | Cl- (1M) E 0 = 0.22 V (used in pH meters)

  24. 2 Saturated Calomel Electrode • The ‘saturated calomel electrode’ (SCE) features the reduction half-reaction: Hg+ + e–  Hg Hg2Cl2 2Hg+ + 2Cl– • Overall: Hg2Cl2 (s) + 2e– 2 Hg (s) + 2Cl– (sat) Pt | Hg | Hg2Cl2 | KCl || • Standard cell potential of E0 = 0.24 V. 5 M

  25. 0.24 Hg+/Hg 0.0V H2/H+ -0.76 Zn2+/Zn Cell Potentials 1 Q: The standard reduction potential of Zn2+/Zn is - 0.76 V. What would be the observed cell potential for the Zn/Zn2+ couple when measured using the SCE as a reference? Calomel: Hg2Cl2(s) + 2e-2Hg(l) + 2Cl-(aq)E0 = 0.24V Zn Zn2+ + 2e-E0 = + 0.76V (reversed because it is written as an oxidation) Ans: The Zn will be oxidised (lower reduction potential), so E(cell) = 0.76 + 0.24 = 1.00 V respect to the SCE So to get the oxidation half-reaction E0 using the SCE as cathode, subtract 0.24 V from the volt meter reading.

  26. Summary CONCEPTS • Concentration cells • Nernst equation CALCULATIONS • Work out cell potential from reduction potentials; • Work out cell potential for any concentration (Nernst equation)

  27. Silver electrode Ag+ + e– Ag AgCl Ag+ + Cl– Overall: AgCl + e– Ag + Cl– E0 = 0.22 V AgCl (s) + e–Ag (s) + Cl– (sat) A thin coating of AgCl is deposited on the pure metal surface. Ag|AgCl|Cl–

  28. ? 0.22 Ag+/Ag 0.0V H2/H+ Cell Potentials 2 • A Fe3+/Fe2+ cellwith [Fe3+]=[Fe2+] =1 M has a potential of 0.55V respect to the Ag/AgCl electrode (E0= 0.22 V). What is the potential of this electrode with respect to the SHE? Answer. The reactions that occur in the two half-cells are: Fe3+ + e- → Fe2+ at the cathode E = 0.55 V; E0 = ? Ag + Cl- +e-→ AgClat the anode E0= 0.22 V The potential of this electrode with respect to the SHE is the difference of the two electrode potentials: E = E0 - 0.22 E0Fe3+/Fe2+ = 0.55 + 0.22 = 0.77 V

  29. Measurement of pH We could construct a concentration cell, using the standard hydrogen electrode (SHE) and a hydrogen electrode: H2(g, 1 atm) 2H+(aq, unknown) + 2e- anode 2H+ (aq, 1M) + 2e-H2(g, 1 atm) cathode 2H+ (1M) 2H+ (unknown) Ecell = ? Using Nernst equation: i.e. the measurement of the cell potential provides pH directly! unknown at 25°C,

  30. pH electrode The pH electrode potential is typically measured versus a fixed reference calomel electrode. E’ is the sum of the constant offset potentials of the inner glass surface/solution, the Ag/AgCl electrode, and the calomel electrode. Eglass electrode= E’ + RT/2.303F log [H+] • Based on a thin glass membrane: a modified glass enriched in H+ and resulting in a hydration layer a few micrometers thick. • Inside the membrane is a 'reference solution' of known [H+] (1M HCl). • The potential difference relevant to pH measurement builds up across the outside glass/solution interface marked ||.

  31. Nernst Equation The Nernst equation describes the effect of concentration on cell potential. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) When Q < 1, [reactants] >[products] and the cell can do more work. When Q = 1, Ecell = E0cell (standard conditions [x] = 1 M). When Q > 1 , [products] > [reactants] and Ecell is lower.

  32. Difference between Q and K Figure from Silberberg, “Chemistry”, McGraw Hill, 2006. • Breakdown of N2O4 to NO2: N2O4 (g, colourless) → 2 NO2 (g, brown) Q =K

  33. - K 0 V ~1037 Ecell decreases as the reaction proceeds, until at equilibrium Ecell =0 and . Potential of an electrochemical cell Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Ecell (Volts)

  34. Recap: Examining Q To K Ratios If Q/K < 1 • Ecell is positive for the reaction as written. The smaller the Q/K ratio, the greater the value of Ecell and the more electrical work the cell can do. If Q/K = 1, • Ecell = 0. The cell is at equilibrium and can no longer do work. If Q/K > 1, • Ecell is negative for the reaction as written. The cell will operate in reverse – the reverse reaction will take place and do work until Q/K = 1 at equilibrium.

  35. At equilibrium Q=K Link between E 0 and K Q: What happens if the reaction proceeds until equilibrium is reached? A: The reaction stops, therefore the voltage, or electrical potential, is zero (the battery is flat). In mathematical terms: So the equilibrium constant determines the cell potential. Large K products favoured  large standard cell potential, E0

  36. Redox reactions are special Figure from Silberberg, “Chemistry”, McGraw Hill, 2006. • For redox reactions there is a direct experimental method to measure K and ΔG°.

  37. Relation between E 0 and K Figure from Silberberg, “Chemistry”, McGraw Hill, 2006. K is plotted on a logarithmic scale to give a straight line.

  38. Example question 2 Q: A voltaic cell consisting of a Ni/Ni2+ half-cell and Co/Co2+ half-cell is constructed with the following initial concentrations: [Ni2+] = 0.80 M; [Co2+]=0.2 M. a) What is the initial Ecell? b) What is the [Ni2+] when the voltage reaches 0.025 V? c) What are the equilibrium concentrations of the ions? Given: E 0Ni2+/Ni = -0.25 V; E 0Co2+/Co = -0.28 V Ni2+ + 2e- →NiE 0 = -0.25V Co →Co2+ + 2e- E 0 = +0.28V Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s) E 0 = 0.03 V

  39. Example question 2a a) What is the initial Ecell? Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s) E 0 = 0.03 V

  40. Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s) 0.80 - x 0.20+x x = 0.40 Example question 2b b) What is the [Ni2+] when the voltage reaches 0.025V? Q = 1.47 So when Ecell = 0.025 V [Co2+] = 0.60 M [Ni2+] = 0.40 M

  41. Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s) K = 10.24 0.80-x 0.20+x x = 0.71 Example question 2c c) What are the equilibrium concentrations of the ions? So at equilibrium, [Co2+] = 0.91 M [Ni2+] = 0.09 M

  42. Concentration Cells in Nature Concentration cells are present all around us, e.g. • nerve signalling: concentration gradients produce electrical current • ion pumps across cell membranes: Na+ / K+ pump, Ca2+ pump • energy production and storage in cells: ATP

  43. Nerve cells Energy from ATP hydrolysis is used by ion pumps, so that across the nerve cell membrane concentration gradients are maintained: Ion Concentration Gradient: Inside Outside K+High Low Na+Low High The membrane potential is more positive outside than inside the cell. • On nerve stimulation, Na+ enters cell, the inside cell membrane becomes > +ive, then K+ ions leave cell to re-equilibrate the outside. • These rapid (ms) changes in charge across the membrane stimulate the neighbouring region and the electrical impulse moves down the length of the cell. Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

  44. Consider K+: [K+]outside = 3 mM, [K+]inside = 135 mM mV mV Nerve cells • Nernst Equation gives the membrane potential generated by the differing extracellular versus intracellular concentrations of each ion: (Eo = 0) Substituting n = 1, T = 37oC: ( in mV) 1 mV = 10-3 V

  45. Cellular Electrochemistry • Biological cells apply the principles of electrochemical cells to generate energy in a complex multistep process. • Bond energy in food is used to generate an electrochemical potential. • The potential is used to create the bond energy of the high-energy molecule adenosine triphosphate (ATP) (energy currency for the cell). ATP4- + H2O→ADP3- + HPO42- + H+ ΔG °’ = -30.5 kJ mol-1 ΔGo’ (solution at pH 7 and at human body temperature 37oC.)

  46. Inside mitochondria, redox reactions are performed by a series of proteins that form the electron-transport chain (ETC) which contain redox couples, such as Fe2+/Fe3+. Large potential differences provide enough energy to convert ADP into ATP. Bond Energy to Electrochemical Potential Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

  47. Bond Energy to Electrochemical Potential In nature, the most important reducing agent is a complex molecule named nicotinamide adenine dinucleotide, abbreviated NADH, which functions as a hydride donor (H-). = NADH = biological reducing agent NAD+ = biological oxidising agent

  48. Electron Transport Chain (ETC) ETC consists of three main steps, each of which has a high enough potential difference to produce one ATP molecule. The reaction that ultimately powers ETC is the reduction of oxygen in the presence of NADH: 2H+ + 2e- + ½ O2→H2O Eo’ = +0.82 NADH + H+→NAD+ + 2H+ + 2e-Eo’ = -0.32 NADH(aq) + H+(aq) + ½O2(aq)→NAD+(aq) + H2O(l) Eo’overall = 1.14 V Substantial energy release! ΔGo’ = -nFEo’ = -2 · 96485 C mol-1· 1.14 V = - 220 kJ mol-1

  49. ATP Synthesis • In short: e- are transported along the chain, while protons are forced into the intermembrane space. • This creates a H+concentration cell across the membrane. • In this step, the cell acts as an electrolytic cell, i.e. uses a spontaneous process to drive a non-spontaneous process. • When [H+]intermembrane/[H+]matrix ~ 2.5 a trigger allows protons to flow back across membrane, and ATP is formed. Figure from Silberberg, “Chemistry”, McGraw Hill, 2006. • It’s not simple: Noble prize in 1997 to Boyer and Walker for elucidating this.

  50. Summary CONCEPTS • Concentration cells • Nernst equation • E 0 and K • Link between E , Q and K • Applications of concentration cells CALCULATIONS • Work out cell potential from reduction potentials; • Work out cell potential for any concentration (Nernst equation) • Work out K from E 0 • Work out pH from concentration cell

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