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madman01
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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 ?
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