HW #3 - Chemistry 132, Winter 2008 Solutions to Homework...

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Chemistry 132, Winter 2008 Solutions to Homework No. 3 Problem 1. See Levine problem 14.12. (a) True. Taking the derivative of E (Levine eq. 14.41) with respect to the activity a j of substance j gives E /∂a j = - ν j RT/nFa j , which is negative when ν j > 0, i.e., when substance j is a product. Therefore, the emf decreases when the activity of a product is increased. (b) True. The cell’s charge number n is the stoichiometric coefficient of the electrons on the left terminal (Levine eq. 14.37). These electrons are considered to be a product of the electrochemical reaction. Stoichiometric coefficients have no units (Levine p. 132) and the stoichiometric coefficients of products are positive (Levine p. 132). Therefore, n is positive and has no units. (c) True. Doubling all the stoichiometric coefficients ν i in a cell’s reaction causes twice as many electrons to be trans- ferred, thus causing n to also double. For a reversible cell, E = - 0 i ν i μ i /nF (Levine eq. 14.39), thus doubling both the ν i ’s and n leaves E unchanged. (d) False. The standard emf of the cell E is the limiting value of the left-hand side of Levine eq. 14.49 as all molalities approach zero. The left-hand side of that equation is not the same as E . Problem 4. See Levine problem 14.26. The half-reactions for this cell are shown in the text (Levine eq. 14.33). From these half- reactions and from Levine table 14.1, the standard emf for the cell is found to be E = E R - E L = (0 . 2222 V - 0 V) = 0 . 2222 V. The activity quotient is given by Q = exp[ nF ( E - E ) /RT ] (Levine eq. 14.43) and for this cell n = 2. (a) E = - 1 . 00 V E - E = (0 . 2222 V) - ( - 1 . 00 V) = 1 . 2222 V , Q = exp ± 2(96485 C / mol)(1 . 2222 V) (8 . 3145 J / K · mol)(298 . 15 K) ² = 2 . 08 × 10 41 . (b) E = 1 . 00 V. The calculations are the same as in part (a), but in this case E - E = (0 . 2222 V) - (1 . 00 V) = 0 . 7778 V, so that Q = 5 . 07 × 10 - 27 . Problem 7. See Levine problem 14.31. The given half reactions and their combination are as follows, where E 3 is to be determined: Cr 3+ ( aq ) + e - Cr 2+ ( aq ) E 1 = - 0 . 424 V n 1 = 1 Cr 2+ ( aq ) + 2 e - Cr( s ) E 2 = - 0 . 90 V n 2 = 2 Cr 3+ ( aq ) + 3 e - Cr( s ) E 3 =? n 3 = 3 When each of the above half-reactions takes place on the right electrode of a cell that has a hydrogen electrode as the left electrode, the standard molar Gibbs energy change of each overall reaction is Δ G j = - n j F E j (Levine eq. 14.42), where j = 1 , 2 , 3. Since the third half-reaction is the sum of the first and second half-reactions Δ G 3 = Δ G 1 + Δ G 2 - n 3 F E 3 = - n 1 F E 1 - n 2 F E 2 E 3 = n 1 E 1 + n 2 E 2 n 3 = ( - 0 . 424 V) + 2( - 0 . 90 V) 3 = - 0 . 7413 V . Problem 10.
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HW #3 - Chemistry 132, Winter 2008 Solutions to Homework...

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