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2-5-2011

2-5-2011. Galvanic Cell: Electrochemical cell in which chemical reactions are used to create spontaneous current (electron) flow. Galvanic Cells. anode oxidation. cathode reduction. spontaneous redox reaction. Zn ( s ) + Cu 2+ ( aq ) Cu ( s ) + Zn 2+ ( aq ). Galvanic Cells.

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2-5-2011

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  1. 2-5-2011

  2. Galvanic Cell: Electrochemical cell in which chemical reactions are used to create spontaneous current (electron) flow

  3. Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction

  4. Zn (s) + Cu2+(aq) Cu (s) + Zn2+(aq) Galvanic Cells • The difference in electrical potential between the anode and cathode is called: • cell voltage • electromotive force (emf) • cell potential Cell Diagram [Cu2+] = 1 M & [Zn2+] = 1 M Cell Representation Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) anode cathode

  5. For example, look at the following cell representation: Cu(s) | CuSO4 (0.100 M) || ZnCl2 (0.200 M) | Zn(s) This cell is read as follows: a copper electrode is immersed in a 0.100 M CuSO4 solution (this is the first half-cell, anode), 0.200 M ZnCl2 solution in which a Zn(s) electrode is immersed (this is the second half-cell, cathode). If the Ecell is a positive value, the reaction is spontaneous (galvanic cell) and if the value is negative, the reaction is nonspontaneous (electrolytic cell) in the direction written and will be spontaneous in the reverse direction.

  6. 2e- + 2H+ (1 M) H2 (1 atm) Standard Reduction Potentials and the Standard hydrogen electrode (SHE) Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm (standard state). Reduction Reaction E0 = 0.00 V

  7. 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: Ecell = Ecathode - Eanode Sometimes, this relation is written as: Ecell = Eright – Eleft Ecell = Ereduction half cell - Eoxidation half cell

  8. 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) Zn2+ (1 M) + 2e- Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm) Standard Reduction Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Cathode (reduction):

  9. E0 = 0.76 V E0 = Ecathode - Eanode E0 = EH /H - EZn /Zn cell cell cell Standard emf (E0 ) cell 0 0 0 0 2+ + 2 0.76 V = 0 - EZn /Zn 0 2+ EZn /Zn = -0.76 V 0 2+ Zn2+ (1 M) + 2e- ZnE0 = -0.76 V Standard Reduction Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

  10. 0 0 0 Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 0 2+ 0 ECu /Cu = 0.34 V 2+ E0 = Ecathode - Eanode E0 = 0.34 V cell cell 0 0 H2 (1 atm) 2H+ (1 M) + 2e- 2e- + Cu2+ (1 M) Cu (s) H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M) Standard Reduction Potentials Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): Cathode (reduction):

  11. Combining the two half reactions: These electrodes are Active (they take part in the reaction)

  12. Electrodes are passive (not involved in the reaction)

  13. Standard Electrode Potential

  14. E0 is for the reaction as written • The more positive E0 the greater the tendency for the substance to be reduced • The half-cell reactions are reversible • The sign of E0changes when the reaction is reversed • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

  15. Diagonal Rule It is possible to determine whether an electrochemical reaction would occur spontaneously or not by considering the values of the standard electrode potentials of the two half reactions. Steps: • Write down the two half reactions placing the one with the larger Eo up. • Draw a diagonl from up left pointing to down right. • The species in the up left will react with the species in the products of the down right.

  16. Predict what will happen if Br2 is added to a solution containing NaCl and NaI. Assume all species are in their standard states. Solution: From Eo Table we have: Cl2 + 2e g 2Cl- Eo = 1.36V Br2(l) + 2e g 2Br- Eo = 1.07V I2(s) + 2e g2I- Eo = 0.53V Applying the diagonal rule suggests that Br2 will not be able to react with Cl-, but will react with I-.

  17. Br2(l) + 2e g 2Br- Eo = 1.07V 2I-g I2(s) + 2e Eo = 0.53V Br2(l) + 2I-g I2(s) + 2Br- Eorxn = Eocathode - Eoanode Eorxn = 1.07 – 0.53 Eorxn = 0.54V

  18. Can Sn(s) reduce Zn2+ under standard state conditions? Sn(s) + Zn2+g Sn2+ + Zn(s) ? To answer this question, we should write the two half reactions starting with the one with larger Eo Sn2+ + 2e g Sn(s) Eo = -0.14V Zn2+ + 2e g Zn(s) Eo = -0.76V Applying the diagonal rule, we conclude that Sn(s) will not be able to reduce Zn2+. The spontaneous reaction will be: Sn2+ + Zn(s) g Zn2+ + Sn(s) This means that Zn(s) will reduce Sn2+.

  19. Calculation of the standard emf of an electrochemical cell The procedure is simple: • Arrange the two half reactions placing the one with the larger Eo up (the cathode). • The half reaction with the lower Eo is placed down (the anode). • Eocell = Eocathode - Eoanode

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