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Electrochemistry

Electrochemistry. Redox in action! Batteries. Electrochemical Cell (Voltaic Cells). Zn + Cu +2  Zn +2 + Cu Anode – Oxidation occurs Cathode – Reduction occurs Electrons travel through the wire from anode to cathode The salt bridge is used to keep the solutions neutral.

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Electrochemistry

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  1. Electrochemistry • Redox in action! • Batteries

  2. Electrochemical Cell (Voltaic Cells) Zn + Cu+2 Zn+2 + Cu • Anode – Oxidation occurs • Cathode – Reduction occurs • Electrons travel through the wire from anode to cathode • The salt bridge is used to keep the solutions neutral

  3. Standard Reduction Potential • SRP = Cell Emf = Voltage of the Cell = E0 • Voltaic Cells – Emf is (+) because it represents a spontaneous reaction • Find the E0 values for each ½ reaction • All E0values represent reduction – switch one to represent oxidation, change the E0values sign • All the E0values to get the overall E0for the reaction

  4. Example The two reduction half reactions in this example represent the half cells of a voltaic cell I2(s) + 2e- 2I-(aq) Fe+2(aq) + 2e- Fe(s) Determine the overall cell reaction and the standard cell potential.

  5. Line Notation Mg(s)|Mg+2(aq)||Al+3(aq)|Al(s) Anode Salt Cathode Bridge

  6. Practice Calculate E0 for the following reactions as written. • 2Ag+ + Cu  Cu+2 + 2Ag • Zn+2 + Ni  Ni+2 + Zn • Fe(s)| Fe+2(aq) || MnO4-(aq), Mn+2(aq) | Pt(s)

  7. The movement of e-s creates a current b/c it is the movement of a charge • Unit for current – Ampere (A) • Unit for charge – Coulomb (C) • 1 mole of e-s = 96,485 C (of charge) Faraday (F) How many grams of copper are deposited on the cathode of a voltaic cell if an electric current of 2.00 A is run through a solution of copper (II) sulfate for a period of 20 mins? 1 C = 1 A*s

  8. How much time would it take to deposit 0.50 g of Ni on a metal object using a current of 3.00 A? 2. What current is needed to deposit 0.50 g chromium metal from a solution of Cr+3 in a period of 1.0 hr?

  9. G = -nFE; at standard conditions G0= -nFE0 4Ag + O2 + 4H+ 4Ag+ + 2H2O • Determine the value of n for this reaction. • Calculate the value of G0for this reaction.

  10. Nernst Equation • As the cell reacts, the concentration of the substances involved changes • When the cell reaches equilibrium, we say the battery is dead and has an E=0 • The Nernst equation allows us to calculate the value of E at different concentrations E = E0 – 0.0592 log Q n @ equilibrium E=0 and Q=K Log K = nE0 0.0592

  11. Practice Calculate the emf at 298 K generated by the cell involving the reaction Cr2O7-2(aq) + 14H+(aq)+ 6I-(aq)  2Cr+3(aq) + 3I2(s) + 7H2O(l) When [Cr2O7-2] =2.0 M; [H+] = 1.0 M; [I-] = 1.0 M; and [Cr+3]= 1.0x10-5M.

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