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Applications of standard potentials

Applications of standard potentials. 10.6 The electrochemical series. For two redox couples Ox 1 /Red 1 and Ox 2 /Red 2 , Red 1 , Ox 1 || Red 2 , Ox 2 E θ = E θ 2 – E θ 1 The cell reaction: Red 1 + Ox 2 → Ox 1 + Red 2

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Applications of standard potentials

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  1. Applications of standard potentials

  2. 10.6 The electrochemical series • For two redox couples Ox1/Red1 and Ox2/Red2, Red1, Ox1 || Red2, Ox2 Eθ = Eθ2 – Eθ1 The cell reaction: Red1 + Ox2→ Ox1 + Red2 If Eθ > 0 then ΔrGθ< 0 (Nernst equation), the reaction will take place spontaneously. In other words, if Eθ2 > Eθ1, the Ox2 has the thermodynamic tendency to oxidize Red1.

  3. 10.7 Measuring pH or pKa • The half reaction at a hydrogen electrode: H+(aq) + e- → ½ H2(g) Here v= 1, Q = The potential E(H+/H2) = Eө - E(H+/H2) = - Assuming that the pressure of H2 gas equals 1 bar E(H+/H2) = - = = E(H+/H2) = -

  4. The potential of a single electrode cannot be measured separately. Another electrode is required to build up an electrochemical cell. This is also why we need a reference electrode when measuring the pH of a solution. • A regular reference electrode is calomel (Hg2Cl2(s)). • the hydrogen electrode is used as the right hand electrode, i.e., cathode. • E(cell) = E(H+/H2) - E(ref.) • E(cell) + E(ref) = - • Why do we need to calibrate the pH electrode before its usage?

  5. 10.8 Thermodynamic function • Nernst equation is the bridge to the connect the thermodynamic quantify, Gibbs energy, and the electromotive force. • Consider: Pt(s)|H2(g)|H+(aq)|Ag+(aq)|Ag(s) Eө = 0.80V. Calculate the ΔfGө(Ag+(aq)). Solution: First, write down the cell reaction. To do that, write down the two half reactions first and then do a simple subtraction (R-L) to get the cell reaction. R: Ag+(aq) + e- → Ag(s) L: H+(aq) + e- → 1/2 H2(g) Cell : Ag+(aq) + 1/2 H2(g) → Ag(s) + H+(aq)

  6. Continued ΔrGө = ΔfGө(Ag(s)) + ΔfGө (H+) - ΔfGө (Ag+) - (1/2)ΔrGө(H2(g)) ΔrGө = 0 + 0 - ΔfGө (Ag+) - 0 Since ΔrGө = -νFEө ΔfGө(Ag+) = vFEө = 9.648 x 104 C mol-1 * 0.80V = 7.715 x 104 CV mol-1 = 7.715 x 104 J mol-1 ( 1CV = 1J)

  7. Temperature dependence of emf • ΔrGө = -νFEө • take the derivate of temperature for both sides • is called temperature coefficient of standard cell emf. • Because one gets therefore, one can use electrochemical method to obtain reaction entropy and relate them to entropies of ions in solution.

  8. noncalorimetric method of measuring ΔrHө ΔrHө = ΔrGө + TΔrSө = -vFEө + T(vF ) = -vF(Eө - T ) • Example:The standard electromotive force of the cell Pt(s)|H2(g)|H+(aq)||Ag+(aq)|Ag(s) was measured over a broad range of temperatures, and the data were fitted to the following polynomial: Eө/V = 0.07 – 4.11x10-4(T/K – 298) – 3.2x10-6(T/K -298)2 Evaluate the standard reaction Gibbs energy, enthalpy, and entropy at 298K. Solution: The standard reaction Gibbs energy can be calculated once we know the standard emf of the above cell: At 298K, Eө/V = 0.07- 4.11x10-4(298K/K – 298) – 3.2x10-6(298K/K -298)2 Eө/V = 0.07 Eө = 0.07 V

  9. to identify the value of v, we need to write down the cell reaction Ag+ + 1/2H2(g) → H+(aq) + Ag(s) ΔrGө = -vFEө = - (1) x 9.6485 x104 C mol-1 x (0.07V) = - 6.754 x 103 CVmol-1 = - 6.754 x 103 J mol-1 Calculate the temperature coefficient of the cell electromotive force = – 4.11x10-4 K-1V – 3.2x10-6x2x(T/K -298) K-1V = - 4.11x10-4 K-1V (at 298K) then ΔrSө = vF = 1 x 9.6485 x 104 C mol-1 x (- 4.11x10-4 K-1V ) = - 39.66 CV K-1 mol-1 = - 39.66 J K-1 mol-1 to calculate the standard reaction enthalpy: ΔrHө = ΔrGө + TΔrSө = -6.754kJ mol-1 + 298K (- 39.66 J K-1 mol-1) = -18.572 kJ mol-1

  10. Example: The standard cell potential of Pt(s)|H2(g)|HCl(aq)| Hg2Cl2(s)|Hg(l) Was found to be +0.269 V at 293 K and + 0.266 V at 303K. Evaluate the standard reaction Gibbs function, enthalpy, and entropy at 298K. Solution: Write the cell reaction Hg2Cl2(s) + H2(g) → 2Hg(l) + 2HCl(aq) So v = 2, To find the ΔrGө at 298 K, one needs to know the standard emf at 298K, which can be obtained by linear interpolation between the two temperatures. Eө = 0.2675 V ΔrGө = -2F Eө = -51.8 kJ mol-1 The standard reaction entropy can then be calculated from = (0.266V- 0.269V)/10K = -3.0x10-4 VK-1 ΔrSө = 2F = - 58 JK-1 mol-1 then ΔrHө = ΔrGө + TΔrSө = -69 kJ mol-1

  11. Evaluate the reaction potential from two others • Example Calculate the standard potential of the Fe3+/Fe from the values for the Fe3+/Fe2+ (+0.77V) and Fe2+/Fe( -0.44V). • Solution: : first write down the half reactions for these three couples: 1) Fe3+ + e- → Fe2+ 2) Fe2+ + 2e- → Fe 3) Fe3+ + 3e- → Fe Reaction 3 is the sum of 1 and 2, yet one cannot use E3 = E1+ E2 ΔrGө(1) = - 1x F x 0.77V ΔrGө(2) = - 2x F x (-0.44)V ΔrGө(3) = ΔrGө(1) + ΔrGө(2) = 0.11F V ΔrGө(3) = - 3*F*E3 = 0.11 F V E3 = -0.033V

  12. Example: Given that the standard potentials of the Cu2+/Cu and Cu+/Cu couples are +0.340V and + 0.522V, respectively, evaluate Eө(Cu2+, Cu+). Solution: Again, we should go through the standard Gibbs energy to calculate it. First write the half-reactions: (1) Cu2+(aq) + 2e- → Cu(s) (2) Cu+(aq) + e- → Cu(s) (3) Cu2+(aq) + e- → Cu+ it can be identified easily that reaction (3) equals (1) - (2) thus: ΔrGө(3) = ΔrGө(1) - ΔrGө(2) = (-2*F*0.340V) – (-1*F*0.522V) ΔrGө(3) = -0.158F V -1*F*Eө(Cu2+/Cu+) = -0.158F V Eө(Cu2+/Cu+) = 0.158 V

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