I have been trying to figure out this question for a while and have not been able to do so, can anyone help?
Nernst Equation
- Thread starter masala
- Start date
- Joined
- Apr 3, 2013
- Messages
- 848
- Reaction score
- 325
I’m pretty sure to find KEq you take products/reactants( solids and liquids do not get put into that equation). I think they gave you those other equations to throw you off.
- Joined
- Apr 3, 2013
- Messages
- 848
- Reaction score
- 325
I’m wrong and I think I figured it out. Since the reaction is at equilibrium you sub keq in for log Q. Voltage is zero at equilibrium so E is zero.
- Joined
- May 25, 2007
- Messages
- 4,110
- Reaction score
- 954
The standard calculation method from general chemistry is cumbersome and will take several minutes (I will represent it second). But this question can be determined quickly and easily using PoE and intuition.
PoE and Intuition Method:
First, look at the half reaction table. NADH must be reversed, resulting +0.32 V. Pyruvate remains as is, making it -0.19. The overall redox reaction has E = +0.13. A positive E means that the reaction is favorable as written, so Keq is greater than 1. This eliminates choices A and B.
The easiest way to decide between C and D involves looking at the 0.06/n and 0.13 ratio. The value of n is 2 in this case (the number of electrons exchanged in the reaction), so we have 0.03 and 0.13 in the calculation. 0.03 does not divide into 0.13 evenly, so there is a remainder, which means that the first part of the answer cannot be 1.00. This eliminates choice D and leaves choice C.
Traditional Math-Based Method:
If you want a more precise answer, then we can use the Nernst Equation.
Consider that at equilibrium, deltaG = 0, which means that E is also 0. This is also where K = Q, so we can substitute K into the equation for Q and 0 for E.
This means that K = 10^4.33 = 10^0.33 x 10^4.
The log of 2 = 0.30, so 10^0.33 = 2.
This gives us 2 x 10^4.
In all honesty, I think the Math-based solution is far above and beyond the MCAT and would highly recommend the PoE method presented first. I'm not sure why AAMC would put this question on their practice exam, as this calculation seems beyond what they typically do.
PoE and Intuition Method:
First, look at the half reaction table. NADH must be reversed, resulting +0.32 V. Pyruvate remains as is, making it -0.19. The overall redox reaction has E = +0.13. A positive E means that the reaction is favorable as written, so Keq is greater than 1. This eliminates choices A and B.
The easiest way to decide between C and D involves looking at the 0.06/n and 0.13 ratio. The value of n is 2 in this case (the number of electrons exchanged in the reaction), so we have 0.03 and 0.13 in the calculation. 0.03 does not divide into 0.13 evenly, so there is a remainder, which means that the first part of the answer cannot be 1.00. This eliminates choice D and leaves choice C.
Traditional Math-Based Method:
If you want a more precise answer, then we can use the Nernst Equation.
Consider that at equilibrium, deltaG = 0, which means that E is also 0. This is also where K = Q, so we can substitute K into the equation for Q and 0 for E.
0 = 0.13 - (0.06/2)logK = 0.13 - 0.03logK
0.03logK = 0.13
so logK = 0.13/0.03 = 4.33
0.03logK = 0.13
so logK = 0.13/0.03 = 4.33
This means that K = 10^4.33 = 10^0.33 x 10^4.
The log of 2 = 0.30, so 10^0.33 = 2.
This gives us 2 x 10^4.
In all honesty, I think the Math-based solution is far above and beyond the MCAT and would highly recommend the PoE method presented first. I'm not sure why AAMC would put this question on their practice exam, as this calculation seems beyond what they typically do.
- Joined
- Nov 3, 2014
- Messages
- 2,872
- Reaction score
- 2,777
This is a next step practice test; not AAMC.
It does not appear in AAMC full length practice exams like this. There is a Nernst question but more about the relevance to neurons
- Joined
- Dec 5, 2017
- Messages
- 727
- Reaction score
- 477
Like Ad2b said, this is definitely not AAMC style. They ask either about the impact of a change in some experiment or a biological application. You should think about things like what would happen to the voltage reading if there were lactate build up or why is the voltage only changing on the millivolt scale and not the volt scale.
