1. The problem statement, all variables and given/known data Consider a Galvanic cell made of two half cells: Ag+ + e− → Ag(s) +0.799 V Cd2+ + 2e− → Cd(s) − 0.403 V (a) Calculate the open circuit potential, when the concentration of Ag+ 0.08 mol and the concentration of the Cd2+ is 0.8 mol. (b) Calculate the voltage across the cell, if the internal impedance of the cell is 10 Ω and you connect it to am external resistor, which is 80 Ω 2. Relevant equations E = E ° - (0.0591V/n * log(K) 3. The attempt at a solution part a For Ag+2 (n= 1 as 1 electron participate) and E ° = +0.799 V E = E ° - (0.0591V/n * log(1/[Ag+2]) E = +0.799V - (0.0591V/1 * log(1/[0.08]) E = +0.799V - (0.0591V * log12.5) E = +0.799V - (0.065V) E = +0.734V For Cd+2 (n= 2 as 2 electron participate) and E ° = -0.403V E = E ° - (0.0591V/n * log(1/[Cd+2]) E = -0.403V - (0.0591V/2 * log(1/[0.8]) E = -0.403V - (0.0295V * log1.25) E = -0.403V - (0.00285) E = -0.406V Ecell = Oxidation potential + Reduction potential Ecell = 0.734V + 0.406V Ecell = 1.140V part b V = I * R 1.140 = I * (10+80) I = 1.26 * 10^-2 A since current is same in series therefore voltage across the cell is equal to V = I * R V = 1.26 * 10-2 * 10 V= 1.26* 10^-1 V Is this right ?