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Consider the following stack at 25 C: Sn(s) Sn*(0,150 mol L) | Pb+ (0,550 mol L) Pb(s) ...
Consider the following stack at 25 °C: Sn(s) Sn²*(0,150 mol L¹) | Pb²+ (0,550 mol L´¹) Pb(s) a. Calculate the initial change in potential, AE, of the battery. b. What should happen to the value of AE of the battery as a function of its operating time? Explain. c. Calculate the value of AE of the battery when [Pb²+] decays to 0.500 mol L-¹ d. Calculate the concentrations of Sn²+ and Pb²+, in mol L-¹, when the cell's AE is zero. Dados: Eº = -0,136 V Sn²(aq) + 2e ?? Sn(s) Pb²(aq) + 2e ?? Pb(s) F = 96500 C mol¹ R= 8,3145 J mol¹ K-¹ E° = -0,126 V Gabarito: a) 0,027 V b) A [Pb²] diminui e a [Sn²2] aumenta, ocasionando uma diminuição do AE. c) 0,022 V d) [Sn²"]no equilibrio = 0,48 mol L´¹: [Pb²*]no equilibrio = 0,22 mol L-¹
a) The given cell is Sn(s) / Sn2+ (0.150 M) // Pb2+ (0.550 M ) / Pb(s) By Nernst equation E cell = Eo cell - [0.0592 / n ] log Q ------------ (1) The oxidation half cell reaction is Sn(s) Sn2+(aq) + 2e- The reduction half cell reaction is Pb2+ (aq) +