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Chapter 19 Electrochemistry 441 mol H₂ mol e C S min hr 2 mol 96,485 C 1s 1 min 1 hr 1 mol H₂ 1 mol e⁻ 7.8 60 60 min Solution: T = + 273.15 = 298 K, then PV = nRT Rearrange to solve for n. PV 25.0 atm X RT = 25.5583 mol H₂. The hydrolysis of water reaction is as follows: 0.08206 + + and + + 2 Check: The units (hr) are correct. Because we have 25L of gas at 25 atm and 25 °C, we expect ~25 moles of gas (remem- ber that 1 mole of gas at STP = A very long time is expected because we have many moles of gas to generate. 19.119 Given: Cu(s) CuI(s) (aq, = 1.1 X 10⁻¹² Find: Conceptual Plan: Write half-reactions from line notation. Because this is a concentration cell, = 0.00 V. Then [I⁻] then Solution: The half-reactions are Cu(s) Cu⁺(aq) + and Cu(s). Because this is a concentra- tion cell, = 0.00 V and n = 1. Because = rearrange to solve for [Cu+ (ox). = = 1.1 1.0 = 1.1 X then Q = = 1.1 1.0 = 1.1 X 10⁻¹² then = Q = 0.00 0.0592 V 1 log (1.1 X 10⁻¹²) = 0.71 V Check: The units (V) are correct. Because [Cu⁺] is so low and red is high, the Q is very small; the voltage increase compared to the standard value is significant. 19.121 Given: (a) disproportionation of to Mn(s) and MnO₂(s) and (b) disproportionation of MnO₂(s) to Mn²⁺ (aq) and in acidic solution Find: and K Conceptual Plan: Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. Balance each half-reaction with respect to mass in the following order: (1) Balance all elements other than H and (2) balance by adding H₂O, and (3) balance H by adding Balance each half-reaction with respect to charge by adding electrons. (The sum of the charges on both sides of the equation should be made equal by adding electrons as necessary.) Make the number of electrons in both half-reactions equal by mul- tiplying one or both half-reactions by a small whole number. Add the two half-reactions, canceling electrons and other species as necessary. Verify that the reaction is balanced with respect to both mass and charge. Look up half-reactions in Table 19.1. Calculate the standard cell potential by subtracting the electrode potential of the anode from the electrode potential of the cathode: E°cell = Then calculate using = Finally, °C K. Solution: (a) Separate: (aq) MnO₂(s) and Mn(s) Balance non 0 elements: (aq) MnO₂(s) and Mn(s) Balance with H₂O: (aq) + MnO₂(s) and Mn(s) Balance H with Mn²⁺ (aq) + MnO₂(s) + (aq) Mn(s) Equalize electrons: (aq) + MnO₂(s) + + and (aq) + Mn(s) Mn(s) Add electrons: Mn²⁺ (aq) + MnO₂(s) + + and (aq) + Add half-reactions: (aq) + + (aq) + 20 MnO₂(s) + + 20 + Mn(s) Cancel electrons: 2 (aq) + MnO₂(s) + + Mn(s) Look up cell potentials. Mn is oxidized in the first half-cell reaction, so = = +1.21 V. Mn is reduced in the second half-cell reaction, so = = 1.18 V. Then = Ecathode = 1.21 = -2.39 V and n = 2 so Copyright © 2017 Pearson Education, Inc.