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Electrochemistry - the Science of Oxidation-Reduction Reactions

Electrochemistry - the Science of Oxidation-Reduction Reactions. 1. Constructing electrochemical cells - sketching cells which carry out redox reaction - electrodes and salt bridges 2. Diagramming the cell - cell conventions, anode and cathode Calculating the cell potential and D G°

audrey-brennan
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Electrochemistry - the Science of Oxidation-Reduction Reactions

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  1. Electrochemistry - the Science of Oxidation-Reduction Reactions 1. Constructing electrochemical cells - sketching cells which carry out redox reaction - electrodes and salt bridges 2. Diagramming the cell - cell conventions, anode and cathode Calculating the cell potential and DG° Galvanic and electrolysis cells - Faraday’s Laws

  2. Constructing Electrochemical Cells Overall cell reaction: Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) voltmeter Cu(s) Ag(s) salt bridge Cu2+(aq) Ag+(aq) Ag+(aq) + e- Ag(s) Cu(s) Cu2+(aq) + 2 e- Oxidation half-reaction on the left Reduction half-reaction on the right Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s)

  3. voltmeter Ag(s) (cathode, positive) Cu(s) (anode, negative) salt bridge Ag+(aq) + e- Ag(s) Cu2+(aq) Ag+(aq) The Standard Cell Potential Cu2+(aq) + 2 e- Cu(s) E° = + 0.3442 V E° = + 0.7996 V E° = - (+0.3442) + (+.7996) = + 0.4554 V Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

  4. Standard Reduction Potentials(Appendix E)

  5. Ag+(aq) + e- Ag(s) Relationship of Cell potential and DG Cu2+(aq) + 2 e- Cu(s) E° = + 0.3442 V E° = + 0.7996 V DE° = - (+0.3442) + (+.7996) = 0.4554 V Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) The cell potential directly measures DG for the reaction: DG = - n F DE n = moles of electrons transferred in the cell reaction F = 96,485 C-mol-1 (the Faraday, charge of 1 mol of protons) DG° = - (2) (96 485) (0.4554 V) = - 87880 kJ (We have assumed standard conditions, i.e., unit activities.)

  6. voltmeter Cu(s) Ag(s) salt bridge Cu2+(aq, 0.05 M) Ag+(aq, 0.15 M) Cu(s) | Cu2+(aq, 0.05 M) || Ag+(0.15M) | Ag(s) metal electrode (anode) phase boundary salt bridge phase boundary metal electrode (cathode) Diagramming the Cell Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

  7. Answer: Pt(s) O2(g) salt bridge H3O+(aq) Draw a Cell which Carries Out the Following Reaction Overall cell reaction: 4 Fe3+(aq) + 6 H2O(l) O2(g) + 4 H3O+(aq) + 4 Fe2+(aq) voltmeter Pt(s) Fe3+(aq) Fe2+(aq) O2(g) + 4 H3O+(aq) + 4 e- 6 H2O (l) Fe3+(aq) + e- Fe2+(aq) Pt(s) | O2(g, PO2) | H3O+(x M) || Fe2+(y M), Fe3+(z M) | Pt(s) If the cell is not a standard cell, you must specify partial pressures and concentrations.

  8. Further Description of the Cell 1. Calculate the standard cell potential: M(-1) O2(g) + 4 H3O+(aq) + 4 e- 6 H2O E° = +1.229 V M(+1) Fe3+(aq) + e- Fe2+(aq) E° = +0.770 V 4 Fe2+(aq) + 6 H2O(l) O2(g) + 4 H3O+(aq) + 4 Fe3+(aq) E° = - 0.459 V 2. Diagram the cell: Pt(s) | Fe2+(aq), Fe3+(aq) || H3O+(aq) | O2(g, 1 atm) | Pt(s) Calculate DG° for the cell reaction: Is the reaction spontaneous as written? no Which electrode is the anode? the cathode? DG° = +177 kJ 4. Which electrode has a positive potential in the galvanic cell? In a galvanic cell, the anode has a negative potential, the cathode a positive potential In an electrolytic cell, the anode is positive, the cathode negative..

  9. Analyzing Cells: a Step-by-Step Method Write down the cell reaction. Note that the reaction may be spontaneous or non-spontaneous as written. Calculate DE°cell. Write both half-reactions as reduction half-reactions (even if the half-reaction occurs as an oxidation in the cell reaction). Find the standard reduction potentials for both reduction half-reactions. Add the half-reactions to give the overall cell reaction. Notice how this is done in the example. If DE°cell is positive, the reaction is spontaneous as written under standard conditions. If not, the reverse reaction is spontaneous under standard conditions. Cu2+(aq) + 2 e- Cu(s) M(-1) E° = + 0.3442 V M(+2) Ag+(aq) + e- Ag(s) E° = + 0.7996 V DE° = - (+0.3442) + (+.7996) = + 0.4554 V Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) The multiplicative factor is negative, so E° is multiplied by (-1) The muliplicative of (+2) is NOT applied to E°. Cell potentials are an intrinsic, not extrinsic, property.

  10. Ag(s) Cu(s) Cu2+(aq) salt bridge Ag+(aq) Galvanic Cells and Electrolysis Cells Galvanic cell: cell reaction runs in the spontaneous direction Electroytic cell: cell reaction runs in the non-spontaneous direction, driven by an applied electrical potential (a voltage) that is greater than the spontaneous potential negative positive anode if galvanic cathode if electrolytic cathode if galvanic anode if electrolytic galvanic: E°= 0.+4554 V Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) electrolytic: E (applied) > E°

  11. Cathode and Anode in Galvanic Cells and Electrolysis Cells By definition, the cathode is where reduction occurs and the anode is where oxidation occurs. In the galvanic cell, the copper electrode is the cathode, silver is the anode. In the electroytic cell, the silver electrode is the cathode, copper is the anode. Ag(s) (positive) Cu(s) (negative) Cu2+(aq) salt bridge Ag+(aq) Note that the sign of the potential doesn't change - copper is always negative, silver is always positive. But the designations "cathode" and "anode" change between galvanic and electroytic cells.

  12. voltmeter Cu(s) Ag(s) Cu2+(aq) salt bridge Ag+(aq) Analyzing Cells: a Step-by-Step Method 4. Draw a sketch of the cell. The anode (where the oxidation occurs) is the left hand compartment, the cathode is the right hand compartment. Note that if DE°cell is positive, the reaction is spontaneous as written under standard conditions. If not, the reverse reaction is spontaneous under standard conditions. sketch of the cell Cu(s) Cu2+(aq) + 2 e- Ag+(aq) + e- Ag(s) Oxidation half-reaction on the left Reduction half-reaction on the right

  13. voltmeter Cu(s) Ag(s) salt bridge Cu2+(aq) Ag+(aq) Cu(s) | Cu2+(aq, 0.05 M) || Ag+(0.15M) | Ag(s) metal electrode (anode) phase boundary salt bridge phase boundary metal electrode (cathode) Analyzing Cells: a Step-by-Step Method 5. Diagram the cell. Note that diagramming the cell is a formal procedure following well-defined rules. It is not the same as “sketching the cell”. sketch cell diagram

  14. Analyzing Cells: a Step-by-Step Method 6. Calculate DG°298 for the cell reaction using the general relationship: DG° = - n F DE°cell. If the cell potential is positive, DG° is negative and the reaction is spontaneous as written. If the cell potential is negative, the reaction as written is non-spontaneous. Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) DE° = + 0.4554 V DG°298 = - n F DE°cell = - 2 * 96485 * DE°298 = - 87.9 kJ DG = - nF DE positive DE corresponds to a negative DG

  15. Analyzing Cells: a Step-by-Step Method 7. Electroytic and Galvanic Cells Which electrode is the cathode (anode)? which is positive (negative)? By definition,the reduction occurs at the cathode, oxidation occurs at the anode. This is true in both galvanic and electrolytic cells. For a galvanic cell: The cell reaction occurs in the spontaneous direction. By convention, the sketch of the cell places the reduction half-reaction in the right-hand half-cell, which is therefore the cathode. The left- hand half-cell is the anode. For an electrolysis cell: The reaction runs in the non-spontaneous direction driven by an applied external voltage. The signs of the potentials of the two electrodes are unchanged, but the anode of the galvanic cell becomes the cathode of the electrolytic cell, and vice versa.

  16. 8. Standard and non-standard cells. In a standard cell, all activities are unity. When activities are not all unity, the cell is not a standard cell. In either case, DG is directly related to DE: DG° = - n F DE°cell (standard)or DG = - n F DE (non-standard) • The Nernst Equation describes the dependence of the cell potential on activities. since DG = DG° + RT lne Q and DG = - n F E, it follows that DE = DE° - (RT/nF) lne Q At T=298 K, this can be written in the form DE = DE° - (0.0592/n) log10 Q (the Nernst Equation) Note that the Nernst Equation contains a base 10 logarithm, rather than base e. Thus DG and the cell potential contain precisely the same physical information. One can be calculated from the other. Also, since DG° = - RT lne K, it follows that DE°cell = + (RT/nF) lne K Analyzing Cells: a Step-by-Step Method

  17. Faraday's Laws of Electrolysis Mass is proportional to electric charge passed through the cell. Equivalent masses of different substances require equal amounts of electric charge passed through the cell. An electrical current of 1 ampere equals one coulomb per second - Q (coulombs) = I (amperes) * t (time) 1 mole of electrons has a charge of -96,485 C (the Faraday) moles of e- = (coulombs passed through cell) / 96485 = (I*t ) / 96,485 3 Cu(s) + 2 Au3+(aq) 3 Cu2+(aq) + 2 Au(s) How much gold is deposited if 100 A is passed through this electrolysis cell for an hour? (Ans. 1.24 moles or 245 g)

  18. For home practice: Constructing Electrochemical Cells (1) Construct a cell which burns hydrogen and oxygen to water: (-2)* [2 H3O+(aq) + 2 e- H2(g) + 2 H2O(l) ] E°H+/H2 = 0.000 v (+1)* [O2(g) + 4 H3O+(aq) + 4 e- 6 H2O(l)]E°O2/H2O = +1.229 v 2 H2(g) + O2(g) 2 H2O(l) DE°cell = +1.229 - (+0.000) = +1.229 volts Diagram the cell: Pt | H2(g) | H+(aq) || H+(aq) | O2(g) | Pt Draw the cell that carries out this reaction. In the galvanic cell, what chemical reaction takes place at the anode? at the cathode? Which electrode is positive? How does this change in an electrolytic cell? Calculate Calculate K Is the cell reaction spontaneous as written?

  19. For home practice: Constructing Electrochemical Cells (2) Construct a cell to carry out the following redox reaction: (+1)* [PbO2(s) + SO42- + 4 H3O+ + 2 e- PbSO4(s) + 6 H2O ] E° = +1.685 v (-1)* [ I2(s) + 2 e- 2 I-(aq)]E° = +0.535 v 2 I-(aq) + PbO2(s) + SO42- + 4 H3 O+I2(s) + PbSO4(s) + 6 H2O DE°cell = +1.685 - (+0.535) = +1.150 volts Diagram the cell: Pt | I2(s) | I-(aq) || SO42-(aq), H3O+ | PbO2(s) | PbSO4(s) | Pt In a galvanic cell, what chemical reaction takes place at the anode? at the cathode? Which electrode is positive? How does this change in an electrolytic cell? Draw the cell that carries out this reaction. Calculate Is the cell reaction spontaneous as written?

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