## Midterm 2014 Question 8

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

Alex Nguyen 3I
Posts: 100
Joined: Fri Sep 25, 2015 3:00 am

### Midterm 2014 Question 8

I'm confused as to why it is assumed that the potential difference of the cell is going to be negative. We are trying to find the Ka for HF. Is it because we need H and F to be products and HF to be a reactant that we flip the half reactions accordingly?

Chem_Mod
Posts: 19551
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 882 times

### Re: Midterm 2014 Question 8

Can you post the full question? Some of the UAs do not have the course reader.

Alex Nguyen 3I
Posts: 100
Joined: Fri Sep 25, 2015 3:00 am

### Re: Midterm 2014 Question 8

Yeah sure.
$F_{2}(g)+2H^{+}(aq)+2e^{-}\rightarrow 2HF(aq),E^{o}=+3.03V$
$F_{2}(g)+2e^{-}\rightarrow 2F^{-}(aq),E^{o}=+2.87V$
Calculate the value of Ka for HF.
Can some explain to me the idea behind Ka again as well as why they assumed that potential difference would be negative?

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

### Who is online

Users browsing this forum: No registered users and 2 guests