Chap 17 Solns-6E - CHAPTER 17 CORROSION AND DEGRADATION OF...

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CHAPTER 17 CORROSION AND DEGRADATION OF MATERIALS PROBLEM SOLUTIONS 17.1 (a) Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or electrons) and becomes an anion. (b) Oxidation occurs at the anode; reduction at the cathode. 17.2 (a) This problem asks that we write possible oxidation and reduction half-reactions for magnesium in various solutions. (i) In HCl Mg Mg 2+ + 2e - (oxidation) 2H + - H 2 (reduction) (ii) In an HCl solution containing dissolved oxygen Mg Mg 2+ - (oxidation) 4H + + O 2 + 4e - 2H 2 O (reduction) (iii) In an HCl solution containing dissolved oxygen and Fe 2+ ions Mg Mg 2+ - (oxidation) 4H + 2 - 2H 2 O (reduction) Fe 2+ - Fe (reduction) 1
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(b) The magnesium would probably oxidize most rapidly in the HCl solution containing dissolved oxygen and Fe 2+ ions because there are two reduction reactions that will consume electrons from the oxidation of magnesium. 17.3 Iron would not corrode in water of high purity because all of the reduction reactions, Equations (17.3) through (17.7), depend on the presence of some impurity substance such as H + or M n+ ions or dissolved oxygen. 17.4 (a) The Faraday constant is just the product of the charge per electron and Avogadro's number; that is F = eN A = 1.602 x 10 -19 C/electron () 6.023 x 10 23 electrons/mol ( ) = 96,488 C/mol (b) At 25 ° C (298 K), RT nF ln(x) = (8.31 J/mol - K)(298 K) (n)(96,500 C /mol) (2.303) log(x) = 0.0592 n log (x) This gives units in volts since a volt is a J/C. 17.5 (a) We are asked to compute the voltage of a nonstandard Cd-Fe electrochemical cell. Since iron is lower in the emf series (Table 17.1), we begin by assuming that iron is oxidized and cadmium is reduced, as Fe + Cd 2+ Fe 2+ +Cd and V = (V Cd o V Fe o ) 0.0592 2 log [Fe 2 + ] [Cd 2 + ] 0.403 V ( 0.440 V) [] 0.0592 2 log 0.40 2x10 3 2
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= -0.031 V (b) Since this V is negative, the spontaneous cell direction is just the reverse of that above, or Fe 2+ + Cd Fe + Cd 2+ 17.6 This problem calls for us to determine whether or not a voltage is generated in a Zn/Zn 2+ concentration cell, and, if so, its magnitude. Let us label the Zn cell having a 1.0 M Zn 2+ solution as cell 1, and the other as cell 2. Furthermore, assume that oxidation occurs within cell 2, wherein = 10 -2 M. Hence, Zn 2 2 + Zn 2 + Zn 1 2+ Zn 2 2+ +Zn 1 and V = 0.0592 2 log Zn 2 2 + [ ] Zn 1 2 + [ ] = 0.0592 2 log 10 2 1.0 = + 0.0592 V Therefore, a voltage of 0.0592 V is generated when oxidation occurs in the cell having the Zn 2+ concentration of 10 -2 M. 17.7 We are asked to calculate the concentration of Pb 2+ ions in a copper-lead electrochemical cell. The electrochemical reaction that occurs within this cell is just Pb + Cu 2+ Pb 2+ +Cu while V = 0.507 V and [Cu 2+ ] = 0.6 M. Thus, Equation (17.20) is written in the form V = V Cu o V Pb o () 0.0592 2 log [Pb 2 + ] [Cu 2 + ] Solving this expression for [Pb 2+ ] gives 3
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[Pb 2+ ] = [Cu 2+ ] exp (2.303) V V Cu o V Pb o ( ) 0.0296 The standard potentials from Table 17.1 are = +0.340 V and = -0.126 V. Therefore, V Cu o V Pb o [Pb 2+ ] = (0.6 M) exp (2.303) 0.507 V (0.340 V + 0.126 V) 0.0296
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Chap 17 Solns-6E - CHAPTER 17 CORROSION AND DEGRADATION OF...

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