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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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