Question 47

[Guest]

Dear Dr. Ferdinand,

Do you think problem 47 is still fair game on the MCAT? I never seen anything like that. When I approached problem, I was thinking, "I have no clue".

[admin]

This is a test question. Whether you got it right or wrong the score is not included in your scaled score. The MCAT will not require you to memorize the Nerst equation. If it appears on the MCAT, you will be provided the equation and you will simply be asked to use it. The real MCAT normally contains 1-2 test questions so always remain calm, do your best then move on.

In the next few days, I will place a link on the first page that you see when you log in. That link will go to equation lists that will only contain the ones that you need to memorize.

Keep up the good work!

[emilie.maz5603]

Why is E zero at equilibrium ?


As mentioned above, The Nernst Equation is

E = Eo - (RT/nF)ln(Q).

E can be calculated using published standard state reduction potentials.

Q is the reaction quotient, e.g.: for the reaction:

Essentially, when a Redox reaction within a voltaic cell occurs under STANDARD conditions, Q=1 and therefore Ecell = E°cell as (RT/nF)ln(Q) = (RT/nF)ln(1) = (RT/nF)(0) = 0

and so, Ecell = E°cell.

When a Redox reaction within a voltaic cell occurs under conditions in which Q <1> E°cell.

When Q > 1, there is a greater concentration of products driving the reaction to the left resulting in Ecell < E°cell.

Lastly, when a Redox reaction reaches EQUILIBRIUM, then Q = K. As such the Redox reaction has no tendency to occur in either direction (left or right) and so Ecell = 0.

And the following shows why:

Since, Ecell = E°cell - (RT/nF)ln(Q) which is also equal to,

Ecell = E°cell – 0.0592V/n log (Q) and so since Q = K (at equilibrium)

Ecell = E°cell – 0.0592V/n log (K)

And since by definition, E°cell = 0.0592V/n log (K), then,

Ecell = E°cell – E°cell = 0

So that Ecell = 0

And so the main solution to the question is that Ecell is equal to “0” at equilibrium because at equilibrium Q (the reaction quotient) becomes K and the E°cell (at standard conditions) = 0.0592V/n log K which subtracted by itself (as shown above) will give a Ecell of “0”.

[emilie.maz5603]

What is the relationship between K_sp and Q here?


The reaction quotient (Q) in relation to Ksp is based essentially on precipitation type reactions (ie, solubility). Moreover, the Q for the reaction by which an ionic compound dissolves is the product of the concentrations of the ionic components raised to their stoichiometric coefficients.


Example: CaF2(s) ↔ Ca2+ (aq) + 2F- (aq)

Where Q = [Ca2+] [F-]2


The difference between Q and Ksp is really the fact that Ksp is the VALUE of the product AT EQUILIBRIUM ONLY, whereas Q is the value of the product under any condition. The Q can then be compared to a solution containing any concentrations of the component ions to one that is at EQUILIBRIUM. Thus Q can be compared to Ksp and it can be used to predict whether a precipitation reaction will occur upon the mixing of two solutions containing dissolved ionic compounds.
Conditions in which Q is compared to Ksp are as follows:

If Q < Ksp, the solution is UNSATURATED and more of the solid will dissolve.

If Q = Ksp, the solution is SATURATED, the solution is holding the equilibrium amount of dissolved ions and additional solid will NOT dissolve.

If Q <Ksp> Ksp, solution is SUPERSATURATED and excess solid will precipitate.


Thus, the relation between Q and Ksp is as shown above with a similar rationale as the relationship with K. Thus, the magnitude of Ksp can be compared with the reaction quotient Q in order to determine the relative saturation of a solution. However, to my knowledge, the Ksp and Nernst equation (Ecell) are not exactly related as is the K value alone. My understanding is that the Ecell is related to the K equilibrium constant.