E = E^0 - (RT/nF) ln([Cd2+])E = -0.402 V - ((8.314 J/(molK) 298 K)/(2 96485 C/mol)) ln(10^-4)E = -0.402 V - (2481.27 / 192970) ln(10^-4)E = -0.402 V - (0.012849) ln(10^-4)E = -0.402 V - (0.012849 -9.2103)E = -0.402 V + 0.118089E = -0.283911 V
Ответ: -0.283911 В.
E = E^0 - (RT/nF) ln([Cd2+])
E = -0.402 V - ((8.314 J/(molK) 298 K)/(2 96485 C/mol)) ln(10^-4)
E = -0.402 V - (2481.27 / 192970) ln(10^-4)
E = -0.402 V - (0.012849) ln(10^-4)
E = -0.402 V - (0.012849 -9.2103)
E = -0.402 V + 0.118089
E = -0.283911 V
Ответ: -0.283911 В.