Lecture 8

# Lecture 8 - Summary • Construc.on of a Voltaic...

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Unformatted text preview: Summary • Construc.on of a Voltaic cell • Nota.on of a Voltaic Cell • Cell Poten.al • Standard Cell Poten.al • Determina.on of standard poten.al of a half cell • Spontaneous redox reac.ons Writing Spontaneous Redox Reactions • By convention, electrode potentials are written as reductions. • When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. • The reduction half-cell potential and the oxidation half-cell potential are added to obtain the Eocell. • When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Example: Zn(s) + stronger reducing agent Cu2+(aq) Zn2+(aq) stronger weaker oxidizing agent oxidizing agent + Cu(s) weaker reducing agent Sample Problem 21.4 Writing Spontaneous Redox Reactions PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate Eocell for each. (1) NO3-(aq) + 4H+(aq) + 3e(2) N2(g) + 5H+(aq) + 4e(3) MnO2(s) + 4H+(aq) + 2e- NO(g) + 2H2O(l) N2H5+(aq) Mn2+(aq) + 2H2O(l) Eo = 0.96 V Eo = -0.23 V Eo = 1.23 V Sample Problem 21.4 Writing Spontaneous Redox Reactions continued (2 of 3) SOLUTION: (a) (1) NO3-(aq) + 4H+(aq) + 3e- Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (1) NO3-(aq) + 4H+(aq) + 3e(2) N2H5+(aq) (A) x4 4NO(g) + 3N2(g) + 8H2O(l) (B) 2NO(g) + 3MnO2(s) + 4H+(aq) Eo = -0.96 V Mn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- (3) MnO2(s) + 4H+(aq) + 2e- Eocell = 1.19 V x3 NO3-(aq) + 4H+(aq) + 3e- (3) MnO2(s) + 4H+(aq) + 2e(1) NO(g) + 2H2O(l) Eo = +0.23 V NO(g) + 2H2O(l) N2(g) + 5H+(aq) + 4e- 4NO3-(aq) + 3N2H5+(aq) + H+(aq) Rev (1) NO(g) + 2H2O(l) NO(g) + 2H2O(l) Eo = 0.96 V x2 Eo = 1.23 V Eocell = 0.27 V Mn2+(aq) + 2H2O(l) x 3 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) Writing Spontaneous Redox Reactions Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) + 4H+(aq) + 2e(2) N2H5+(aq) Mn2+(aq) + 2H2O(l) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) + 4H+(aq) + 2e(C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) E0 = +0.23 V E0 = 1.23 V E0cell = 1.46 V Mn2+(aq) + 2H2O(l) x 2 N2(g) + 2Mn2+(aq) + 4H2O(l) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of Oxidizing agents: MnO2 > NO3- > N2 Reducing agents: N2H5+ > NO > Mn2+ Oxidizing and Reducing Agents •  The strongest oxidizers have the most posi.ve reduc.on poten.als. •  The strongest reducers have the most nega.ve reduc.on poten.als. Oxidizing and Reducing Agents The greater the diﬀerence between the two, the greater the voltage of the cell. Figure 21.9 The reaction of calcium in water. Relative Reactivities (Activities) of Metals 1. Metals that can displace H2 from acid. 2. Metals that cannot displace H2 from acid. 3. Metals that can displace H2 from water. 4. Metals that can displace other metals from solution. Free Energy and Electrical Work ΔG α -Ecell -Ecell = -wmax charge charge = nF n = mol eF = Faraday constant F= 96,485 C mol e- 1V=1 F= J C 9.65 x 104 J V•mol e- ΔG = wmax = charge x (-Ecell) ΔG = -nFEcell In the standard state ΔGo = -nFEocell ΔGo = - RT ln K RT Eocell = - ( n F )ln K Figure 21.10 The interrelationship of ΔG0, E0cell, and K. <0 >1 >0 0 1 0 >0 <1 <0 ΔGo = -RT lnK Sample Problem 21.5 PROBLEM: Calculating K and ΔGo from Eocell Lead can displace silver from solution: Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s) As a consequence, silver is a valuable byproduct in the industrial extraction of lead from its ore. Calculate K and ΔGo at 298.15 K for this reaction. PLAN: Break the reaction into half-reactions, find the Eo for each half-reaction and then the Eocell. Substitute into the equations from previous slide. Nernst Equa.on •  We learned that ΔG = ΔG° + RT ln Q •  This means −nFE = −nFE° + RT ln Q Nernst Equa.on Dividing both sides by −nF, we get the Nernst equa.on: RT ln Q E = E° − nF or, using base ­10 logarithms, 2.303 RT ln Q E = E° − nF Nernst Equa.on At room temperature (298 K), 2.303 RT = 0.0592 V F The equa.on becomes E = E° − 0.0592 ln Q n The Effect of Concentration on Cell Potential Ecell = Eo cell - RT nF 0.0592 Ecell = Eocell - ln Q log Q n • When Q < 1 and thus [reactant] > [product], ln Q < 0, so Ecell > Eocell. • When Q = 1 and thus [reactant] = [product], ln Q = 0, so Ecell = Eocell. • When Q >1 and thus [reactant] < [product], ln Q > 0, so Ecell < Eocell. Sample Problem 21.6 Using the Nernst Equation to Calculate Ecell PROBLEM: a test of a new reference electrode, a chemist constructs a voltaic In cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+] = 0.010 M [H+] = 2.5 M PH = 0.30 atm Calculate Ecell at 298.15 K. 2 PLAN: Find Eocell and Q in order to use the Nernst equation. SOLUTION: Determining Eocell : 2H+(aq) + 2e- H2(g) Eo = 0.00 V Zn2+(aq) + 2e- Zn(s) Eo = -0.76 V Zn(s) Ecell = Eocell Ecell = 0.76 – Zn2+(aq) + 2e0.0592 V n 0.0592 2 Eo = +0.76 V log Q log(4.8 x 10-4) = 0.86 V Q= P x [Zn2+] H2 [H+]2 Q= (0.30)(0.010) (2.5)2 Q = 4.8 x 10-4 The relation between Ecell and log Q for the zinc-copper cell. ...
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## This note was uploaded on 02/18/2012 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.

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