2014 midterm #8


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Jineava_To_3N
Posts: 35
Joined: Wed Sep 21, 2016 2:57 pm

2014 midterm #8

Postby Jineava_To_3N » Mon Feb 13, 2017 10:54 am

The question asks to solve for Ka of HF given the standard cell potentials:

F2+2H(plus one charge) +2e->2HF E knot=3.03 V
F2+2e->2F(minus one charge) E knot=2.87 V

I understand how to use the formula to solve for K but I don't understand how the cathode and anode are determined. In the solution, HF is on the reactant side, but why is it not on the product side?

Amy_Bugwadia_3I
Posts: 37
Joined: Wed Sep 21, 2016 2:56 pm
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Re: 2014 midterm #8

Postby Amy_Bugwadia_3I » Mon Feb 13, 2017 11:14 am

Because the question is asking for the value of Ka, we know that the question implies that the acid (in this case, HF) is dissociating into ions. Thus, the reaction equation will be HF --> H+ (aq) + F- (aq). Because the equilibrium constant K is always defined as the concentration of the reactants (raised to their respective stoich coefficients, if applicable) divided by the concentration of products (raised to their respective stoich coefficients, if applicable), we know that Ka equals ( [H+][F-] )/ [HF].

What helps me remember the difference between the cathode and the anode is "Red Cat" - REDuction happens at the CAThode. In order to obtain HF on the reactant side, we need to flip the first equation given in the reaction. There are two methods to calculate the Ecell - the first is to use E(cathode) - E(anode). If you use this method, make sure to plug in the standard reduction potentials given at the beginning of the problem (BEFORE you switch signs). The second method is to flip one equation (thus switching the sign of that E standard value), add the two half reactions together, and then add the new E standard values together. If you use either method, you should end up with -0.16 V. Hope that helps!


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