Focus 6 Exercises (53, 57, 73) [ENDORSED]

[Vivian Leung 1C]

Hi,

I just needed some help on these Focus 6 Exercise problems (see attachments)

Question 53: Could someone explain the reasoning for the different parts of this question?

Question 57:

Why is the anode an HF reaction?
How do you tell that the cathode is 2F2 +2 e- --> 2F-?
Why is Ka= root of K and not 1/K? (since I thought that K of the reverse rxn= 1/K of the forward reaction? )

[Chem_Mod]

6.53

The anode side is the product side (reaction starts with lower concentration product), 0.0010 M CrCl3(aq).

Diluting [P] therefore makes Q << 1, which makes ln Q even more negative, and therefore E more positive.

Work through the concentration cell example I did in class. Lower concentration product is on anode side.

6.57

Cathode: The only way to show reduction is F₂ + 2 e- --> 2F-

Ka = root of K because you have the reaction 2HF ---> 2H+ + 2F-

But asked for: HF ---> H+ + F-

Write out the K expressions for these two reactions to see Ka = root of K.