## 6th edition 14.97

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

Kimberly 1H
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### 6th edition 14.97

Use the data in Appendix 2B and the fact that, for the half-reaction
F2(g)+2H+(aq)+2e--->2HF(aq), E=3.03 V, to calculate the value of Ka for HF.

The answer key shows that you're supposed to take the square root when you find K of the reaction in order to get Ka, but I'm not sure I understand the reasoning behind it.

Chem_Mod
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### Re: 6th edition 14.97

Because you have 2 moles of HF in your standard reduction potential equation. In Ka expression, it is the dissociation of 1 mole of HF.

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