E = Eº + (2.303 RT/nF) ln Q
E = -.300 V + (2.303 x 8.314 J/mol K x 298 K / (2 x 96485 C/mol)) ln (0.39^2 x 8.5x10-2 / (9.0x10-3)^2)
E = -.300 V + (2.303 x 8.314 J/mol K x 298 K / (2 x 96485 C/mol)) ln (0.0014)
E = -.300 V - 0.741 V
E = -1.041 V
Calculate the potential for: Br2(l) + 2Cl-(aq) → Cl2(g) + 2Br-(aq) Use Eº = -.300 V
when the concentrations of the soluble species are as follows:
[Cl-] = 0.39 M, [Br-] = 9.0x10-3 M, PCl2 = 8.5x10-2 atm
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