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20 Electrochemistry Solutions to Exercises 20.102 (a) Ag(s)] Ni(s) (-0.28) = 1.08 V (b) As the reaction proceeds, Ni²⁺(aq) is produced, so [Ni²⁺] increases as the cell operates. (c) 0.04(2) 0.0592 0.0592 = -2.000 + 1.351 = -0.649; = 0.255 M; = 0.474 = 0.5 M (Strictly speaking, [E - E°] having only one sig fig leads (after several steps) to the answer having only one sig fig. This is not a very precise or useful result.) 20.103 (a) I₂(s)+2e⁻ (aq) 2[Cu(s) I₂(s)+2Cu(s) n E = +0.015 0.0592 2 = + 0.0034 = 0.018 V (b) Since the cell potential is positive at these concentration conditions, the reaction as written in part (a) is spontaneous in the forward direction. Cu is oxidized and Cu(s) is the anode. (c) Yes. E° is positive, so Cu is oxidized and Cu(s) is the anode at standard conditions. (d) 2(0.015) 0.0592 = log +2 log [I⁻]; 20.104 Use the relationship developed in Solution 20.101 to calculate K from E°. Use data from Appendix E to calculate E° for the disproportionation. Cu(s) = 0.153 V (aq) E° = 0.521 V 0.153 V = 0.368 V E° n 20.105 (a) In discharge: Cd(s) + 2NiO(OH)(s) + Cd(OH)₂(s) + 2Ni(OH)₂(s) In charging, the reverse reaction occurs. 641