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CHEM1612 - Pharmacy Week 9: Galvanic Cells

CHEM1612 - Pharmacy Week 9: Galvanic Cells. 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: Galvanic Cells

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  1. CHEM1612 - PharmacyWeek 9: Galvanic Cells 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. Electromotive Force • The measured voltage across the cell is called cell potential (Ecell). • This driving force for the reaction is also called ELECTROMOTIVE FORCE (emf)of the cell. • Every galvanic cell has a different cell potential. • How can we measure the cell potential relative to each species? • How can we tabulate cell potentials? Zn(s) + Cu2+(aq)Zn2+(aq) + Cu(s)

  5. Indicates standard conditions Standard Reduction Potential • The electromotive force cannot be measured absolutely for one element (we cannot measure EZn and ECu in isolation). • Only differences in potential have meaning (ΔE). • We assign a “half-cell” potential to each “half-reaction”, and then add up two half-reactions to get the full reaction and full potential. • We choose as a standard the reaction of hydrogen : 2H+ (aq) + 2e– H2 (g) E0 = 0.00 V (1 atm H2, [H+] = 1 M, at all temperatures)

  6. Determining E0cell • We measure half-cell potentials relative to this reference hydrogen (H+/H2) electrode, which has E0ref = 0.00 V. • We build voltaic cells that have this reference half-cell and another half-cell whose potential is unknown, E0unknown. We define: E0cell = E0cathode – E0anode • When H2 is oxidised, the reference electrode is the anode, so: E0cell= E0cathode – E0ref = E0unknown – 0.00 = E0unknown • When H+ is reduced, the reference electrode is the cathode, so: E0cell= E0ref – E0anode = 0.00 - E0unknown = -E0unknown

  7. Potential(V) 0.76 V Demo: Potential of cell Zn/H2 Experiment 1: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Zn(s)|Zn2+||H+(aq)|H2(g)|Pt We measure E0cell = 0.76 V; Since E0cell = 0 - E0unknown Then E0Zn = -0.76 V Standard Hydrogen Electrode all reagents at standard concentration of 1.0 M

  8. Potential(V) +0.34 V Demo: Potential of cell Cu/H2 Cu2+(s) + H2(g) 2H+(aq) + Cu(s) Experiment 2: H2(g)| H+(aq)|| Cu2+(s)|Cu(s)| Pt We measure E0cell =0.34 V Since E0cell = E0unknown Then E0Cu = 0.34 V all reagents at standard concentration of 1.0 M

  9. Potential(V) E0Zn -0.76 V E0Cu0.34 V 1.10 V Experiment 3: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Demo: Potential of Zn/Cu cell Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) We measure a potential E0cell = 1.1 V. E0cell = E0cathode – E0anode = 0.34 – (–0.76)= 0.34 + 0.76 = 1.10 V

  10. Potential(V) 0.34 V Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) - Experiment 1: Cu2+(aq) + H2(g)Cu(s) + 2H+(aq) Experiment 2: -(-0.76 V) = 1.10 V Cell Potential Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Experiment 3: Conclusion 1: Reverse the reaction – reverse the sign of the potential. Conclusion 2: You can add cell potentials as you add chemical reactions. all reagents at standard concentration of 1.0 M

  11. Electromotive Force ΔE • We measured for the reaction Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) • A cell potential Ecell = ΔE = 1.1 V • Measured in volts, V = J·C–1 • Moving one Coulomb of charge from Zn to Cu2+ releases 1.1 J of energy

  12. Standard Reduction Potentials • We define the standard reduction potentialE0 as the potential of each redox couple when all reagents are in the standard state, i.e. 1 M concentration, 298 K, and gases at 1 atm pressure. • The standard reduction potentials are tabulated. • In data tables all half reactions are written as reductions. Fe3+ + e– Fe2+E0= 0.77 V Cu2+ + 2e– Cu E0= 0.33 V 2H+ + 2e– H2 (g) E0= 0.00 V Zn2+ + 2e– Zn E0= –0.76 V • The higher E (more positive), the greater the tendency to acquire electrons (be reduced).

  13. Weak reducing agent Strong oxidising agent Strong reducing agent Weak oxidising agent Reduction potential table No/slow oxidation by H+ due to an over-potential * by definition

  14. Using the redox potential tables • Write down the two half-reactions. • Work out which is the oxidation and which is the reduction half-reaction. • Balance the electrons. • Add up the half-reactions to get full reaction. • Add up half-cell potentials to get E 0. • A spontaneous (working) voltaic cell, ALWAYS has a positive cell potential, E 0.

  15. Approach Calculate the standard cell potential for a galvanic cell formed by Mg2+(aq) | Mg(s) in one half-cell and Sn2+(aq) | Sn(s) in the other. Sn2+ + 2e- SnE0= -0.14V Mg2+ + 2e- Mg E0= -2.37V Which reaction is the oxidation and which is the reduction? Which half reaction is turned around? In general, you reverse the reaction that appears lower in the table of standard reduction potentials.

  16. Example 1 Calculate the standard cell potential for a galvanic cell formed by Mg2+(aq) | Mg(s) in one half-cell and Sn2+(aq) | Sn(s) in the other. Sn2+ + 2e- SnE0= -0.14V Mg2+ + 2e- Mg E0= -2.37V We saw that Mg is a stronger reducing agent than Sn, i.e. it likes to be oxidised. Turn the Mg reaction around to get an oxidation reaction. Sn2+ + 2e-  SnE 0 = -0.14V Mg  Mg2+ + 2e- E 0 = +2.37V Mg(s) + Sn2+(aq)  Mg2+(aq) + Sn(s) E 0 = +2. 23V

  17. Note! Example 2 Calculate the standard cell potential for a galvanic cell with Ag+(aq) | Ag(s) in one half-cell and Cr3+(aq) | Cr(s) in the other. E0 (Ag+(aq)|Ag(s)) = +0.80 V; E0(Cr3+(aq)|Cr(s)) = -0.74V Cr half reaction is lower (more negative), turn it around… Ag+ + e- AgE 0 = +0.80V Cr Cr3+ + 3e- E 0 = +0.74V Balance the electrons… 3Ag+ + 3e-  3AgE 0 = +0.80V Cr  Cr3+ + 3e- E 0 = +0.74V Cr(s) + 3Ag+(aq)  Cr3+(aq) + 3Ag(s) E 0 = +1.54V Note: E0does not depend on stoichiometry!

  18. Balancing Redox Equations – Example 1 Al + Fe2+Al3+ + Fe ONs: Al Al3+ Fe2+Fe Oxidation half cell: Al Al3+ + 3e- Reduction half cell: 2e- + Fe2+Fe Combine: 2 x (Al Al3+ + 3e-) 3 x (2e- + Fe2+Fe) 2Al + 3Fe2+2Al3+ + 3Fe (oxidation) (reduction) balance charge with e- balance charge with e-

  19. Balancing Redox Equations – Example 2 Cr2O72- + I-Cr3+ + I2 (in acidic solution) O.N.s: Cr2O72- (+6) Cr3+ (+3) I- (-1) I2 (0) Oxidation half cell: 2I-I2 + 2e- Reduction half cell: 6e- + 14H+ + Cr2O72-2Cr3+ + 7H2O Combine: 3 x (2I-I2 + 2e-) 6e- + 14H+ + Cr2O72-2Cr3+ + 7H2O 6 I- + 14H+ + Cr2O72-3I2 + 2Cr3+ + 7H2O (reduction) (oxidation)

  20. Balancing Redox Equations – Example 3 HgO + Zn Hg + Zn(OH)2 (in basic solution) • ONs: HgO (+2) Hg (0) Zn (0) Zn2+ (+2) • Oxidation half cell: Zn Zn2+ + 2e- • Reduction half cell: 2e- + H2O + HgOHg + 2OH- (Add OH- and water to neutralise the charge and balance O and H) • Combine: Zn Zn2+ + 2e- 2e- + H2O + HgOHg + 2OH- Zn + H2O + HgOZn2+ + Hg + 2OH- (reduction) (oxidation)

  21. Balancing Redox Equations – oxidation number method Step 1 Assign O.N. to all elements. Step 2 Identify the species being reduced/oxidised. Step 3 Calculate and balance the gain/loss electrons. Step 4 Balance the number of all the atoms other than O and H. Step 5 Balance O by adding H2O to either side as required. Balance H and charges by adding H+ to either side as required. Step 6 If basic conditions are specified, add OH– as required. (H+ and OH- cannot be present at the same time, they will convert into H2O). Finally, check that the whole reaction is balanced, i.e that the charges and the moles of reactants and products are balanced.

  22. Example 1 Balance the equation for the oxidation of HCl by MnO4 – MnO4–+ HClMn2+ + Cl2 Answer: 2 MnO4–(aq)+ 10 HCl(aq)+ 6 H+(aq) 2 Mn2+(aq)+ 5 Cl2(aq)+ 8 H2O(l)

  23. Example 2 • Balance the reaction of oxidation of H2SO3 by Cr2O72-: Cr2O72- + H2SO3Cr3+ + SO42- Answer: Cr2O72- + 3 H2SO3 + 2 H+ → 2 Cr3+ + 3 SO42- + 4H2O

  24. Example3 • Balance the reaction between NaMnO4 and Na2C2O4: MnO4- + C2O42- MnO2 + CO32- Answer: 2 MnO4- + 3 C2O42-+ 2 H2O→ 2 MnO2 +6 CO32- + 4 H+

  25. Summary CONCEPTS • Galvanic/Voltaic cells • Cell notation • Standard Reduction Potentials • Balancing Redox Equations CALCULATIONS • Work out cell potential from reduction potentials

  26. Half-cell standard reduction potentials

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