Calculating Ecell and Voltage Across a Galvanic Cell | Nernst Equation Problem

[madman01]

Homework Statement

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

Homework Equations

E = E ° - (0.0591V/n * log(K)

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 = +0.734VFor 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 = 1.140Vpart 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 VIs this right ?

 

[JohnRC]

No, part b) is not right.

Clue: when you calculated the voltage of the open circuit cell voltage, you used standard electrode potentials and the Nernst equation. But none of the result of part a) was incorporated in your calculations for part b) -- which is a correct calculation of *something*.