## Midterm 2012 Q8

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Antonio Melgoza 2K
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Joined: Fri Sep 25, 2015 3:00 am

### Midterm 2012 Q8

Can someone walk me through the math? I'm not understanding how it got -2.33?
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Tara 2K
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

### Re: Midterm 2012 Q8

Starting from the second line after we put in the values for Ecell and E°cell, subtract 0.63V from both sides of the equation. This gives us .03V = -(0.0257V/2) ln ([Zn+^2+]/(.10)). Then divide both sides by (-0.0257V/2), which results in -2.33 = ln([Zn+^2+]/(.10)).

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