## 2014 Midterm

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

Blake_Katsev_2E
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### 2014 Midterm

Why in question 8 would the first rxn be oxidized? dont you want a positive cell potential? and also how do you get from K to Ka and what does Ka mean in this context?

Maggie Bui 1H
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### Re: 2014 Midterm

We don't always necessarily want a positive cell potential: if we want a spontaneous reaction, we would need a positive cell potential, but that's not what we're looking for here. The first reaction is oxidized because we need to reverse that reaction to calculate Ka. The reaction that is associated with Ka breaks down an acid into its constituent components, so the products are the individual ions while the reactant is the acid itself. Notice this is the reverse of what we are given, so, we reverse the first reaction. K is Ka in this context. Ka is the acid dissociation constant - the equilibrium constant for the dissociation (breaking down) of an acid.

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### Re: 2014 Midterm

As Maggie stated above, a positive cell potential is not always required (it's only for when you want a spontaneous reaction). The reason why we know we want HF on the reactant side and the H+ and F- ions on the product side is because we are asked for the Ka value, which implies that the acid (in this case, HF) is dissociating into its ions. The way to end up with HF on the reactant side in order for it to dissociate, we need to flip the first equation.

Ariana de Souza 4C
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Joined: Wed Sep 21, 2016 2:56 pm

### Re: 2014 Midterm

at the end of the problem, how come we take the square root of K? it says
Ka= sqrt(K)

Blake_Katsev_2E
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### Re: 2014 Midterm

Thank you that helped out a lot. But still a little confused on the square root part^

Anna_Kim_2E
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Joined: Wed Sep 21, 2016 2:56 pm

### Re: 2014 Midterm

Ka = [H+][F-]/[HF] and K = [H+]^2[F-]^2]/[HF]^2

therefore, when comparing Ka to K, K is exactly Ka square. Therefore you square root the value K in order to achieve Ka.