chemistry question, Nernst equation? help?

K

Kevin A

New member
A battery formed from the two half reactions below dies (reaches equilibrium). If [Fe^2+] was 0.24 M in the dead battery, what would [Cd^2+] be in the dead battery?

Half reaction......... E°
Fe^2+ → Fe ........-0.44
Cd^2+ → Cd .......-0.40

This is the way to solve it, but i don't know how to continue. please help?

Ecell = E°cell - (0.0591/2) × log (0.24 / x)
0 = 0.04 - (0.592 / 2) × log(0.24 / x)
 
-0.04 = -0.0591/2 × log(0.24 / x)
You can divide -0.592/2 to get it as a whole number

-0.04 = -0.02955× log(0.24 / x)
Divide each side by -0.02955
1.354 = log (0.24 / x)
Use antilog (10^)
22.6x = 0.24
Solve for x algebraically now
x = 1.062×10^-2
Therefore
[Cd^2+] = 1.062×10^-2 mol / L

Hope this helps!
 
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