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Description of a Galvanic Cell.

Description of a Galvanic Cell. Describe completely the galvanic cell based on the following half-reactions under standard conditions: Ag + + e -  Ag E o cell = 0.80 V (1) Fe 3 + + e -  Fe 2+ E o cell = 0.77 V (2) In addition, draw the cell and write the line notation.

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Description of a Galvanic Cell.

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  1. Description of a Galvanic Cell. • Describe completely the galvanic cell based on the following half-reactions under standard conditions: • Ag+ + e-Ag Eocell= 0.80 V (1) • Fe3+ + e-Fe2+Eocell= 0.77 V (2) • In addition, draw the cell and write the line notation.

  2. Cell Potential, Electrical Work • The work that can be accomplished when electrons are transferred (emf) is defined in terms of a potential difference (in volts) betweentwo circuits. • emf= potential difference (V) = work (J) • charge (C) • or 1 V = 1 J/C • Work is viewed from the point of view of the system. • Therefore, E = -w/q or -w = q E

  3. Maximum work • The maximum work is defined as • wmax = -q Emax • *Achieving maximum work is impossible. • In any real, spontaneous process, some energy is always wasted. The actual work realizedis always less than the calculated maximum.

  4. Free Energy • The Faraday (F)is defined as the charge of 1 mole of electrons. • F= 96,485 C/mol e- • The purpose of a voltaic cell is to convert the free energy change of a spontaneous reaction into the KE of electrons moving through an externalcircuit. • wmax= DG

  5. For a galvanic cell • DG = -nFEmaxor DGo= -nFEomax • Where n is the number of moles of electrons transferred in the redox reaction. • Note the units of F is C/mol, and the units of E, volts, is J/C • -nFE = (mol)(C/mol )(J/C) = J • This equation is in Joules, not Kilojoules!!

  6. Spontaneity • If Ecell > 0, then DG< 0 and the process is spontaneous. • If Ecell < 0, then DG> 0 and the process is nonspontaneous. • If Ecell = 0, then DG= 0 and the process is at equilibrium.

  7. Calculating DGo for a Cell Reaction. • Using the Standard Reduction Potential Chart, calculate DGo for the reaction • Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq) • Is the reaction spontaneous? • Predict whether 1 M HNO3will dissolve gold metal to form a 1 M Au3+ solution. • Below is the half reaction of nitric acid acting as an oxidizing agent • 4 H+ + NO3- + 3 e- NO + 2 H2O Eo = .96 V

  8. The Relationship to the Equilibrium Constant • DGo= -R T ln K and DGo= -nFEocell • therefore -R T lnK = -nFEocell • Eocell= RT ln K /n F • Calculating K and DGo from Eocell • When cadmium metal reduces Cu2+ in solution, Cd2+ forms in addition to copper metal. If DGo = -143 kJ, calculate K at 25o C.What would be Eocellin a voltaic cell that used this reaction?

  9. The Effects of Concentration on E • For the cell reaction • 2Al(s) + 3Mn2+(aq) ⇌2Al3+(aq) + 3Mn(s) Eocell= 0.48 V • predict whether Ecellis larger or smaller than Eocellfor the following cases. • a) [Al3+] = 2.0 M, [Mn2+] = 1.0 M • b) [Al3+] = 1.0 M, [Mn2+] = 3.0 M

  10. Nernst Equation • The Nernst equation gives the relationship between the cell potential and the concentrations of cell components. • E = Eo– (.0591/n) log Q • Or at equilibrium • Eo= (.0591/n) log K • Calculations with this equation have been removed from the AP test, however, qualitative understanding of how concentration affects a cell are still on the test.

  11. The Effect of Q • A cell in which the concentrations are not in their standard states will continue to discharge until equilibrium is reached. • Q values are never negative. For values less than 1, the concentration of reactant are higher than product, and the reaction is shifted to the right, increasing the Ecellcompared to Eocell. • When Q < 1, [reactant] > [product], Ecell > Eocell. • For Q values higher than 1, the concentration of product are higher than reactant and the reaction is shifted to the left, decreasing the Ecell. • When Q > 1, [reactant] < [product], Ecell < Eocell. • When Q = 1, [reactant] = [product], Ecell = Eocell.

  12. More K and Q • At equilibrium, K = Q and Ecell = 0. • At equilibrium, the components in the two cell compartments have the same free energy, and DG = 0. • The cell can no longer do work!

  13. Problems • Consider a cell based on the reaction • Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu (s) • If [Cu2+] = 0.30 M, what [Fe2+] is needed to increase Ecellby 0.25 V above Eocellat 25oC?

  14. Describe the cell based on the following half-reactions: • VO2+ + 2H+ + e-VO2+ + H2O • Eo= 1.00 V • Zn2+ + 2e- Zn Eo= -0.76 V • where T = 25oC • [VO2+ ] = 2.0 M • [H+ ] = 0.50 M • [VO2+ ] = 0.01 M • [Zn2+ ] = 0.1 M

  15. Concentration Cells • Concentration cells are constructed with the exact same half-reactions, with the exception of a difference in concentrations. • Voltages are typically small as electrons are transferred from the cell of higher concentration to the cell of lower concentration. • Eocell= 0.00 V, but these never deal with standard conditions because concentrations are not standard (not 1 M).

  16. Calculating the Potential of a Concentration Cell. • A concentration cell is built using two Au/Au3+ half-cells. In half-cell A, [Au3+] = 7.0 x 10-4 M, and in half-cell B, [Au3+] = 2.5 x10-2 M. What is Ecell, and which electrode is the anode?

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