Nernst equation question

[Tokspor]

Take the equation E(observed) = E(cell) - 0.0592 log [H+]. If the E(cell) were actually 0.300 V rather than 0.285 V, what would be true of the pH measured by the pH meter, assuming that no correction is made?

A. The pH from the pH meter would be too high, because the E(cell) in the calculation is too high.
B. The pH from the pH meter would be too high, because the E(cell) in the calculation is too low.
C. The pH from the pH meter would be too low, because the E(cell) in the calculation is too high.
D. The pH from the pH meter would be too low, because the E(cell) in the calculation is too low.

I thought it would be B. I simplified the equation to E(observed) = E(cell) + 0.0592 x pH. Now, the E(cell) in the calculation would clearly be lower than the actual, and I thought that was a given from the question stem. What this will do is make the E(observed) look higher than it would be if the E(cell) was actually 0.285 V instead of 0.300 V. Since we don't know that the E(cell) is too high, we would think that therefore it must the pH that is too high which is causing the increasing in E(observed).

But the correct answer is C. According to the text, "if the real value for E(cell) is actually higher than the value being plugged in, then a value that is too large is being subtracted to get E(observed). This means that the [H+] is too high, so the calculated pH is too low."

This doesn't make sense to me, and maybe I'm not getting the logic here, but it would seem to me that 1) we would get a higher E(observed) than expected so 2) we would think that the value for the entire (0.0592 log [H+]) term is less, meaning the pH calculated is more than the actual. Also, answer choice C says that the E(cell) in the calculation is too high. This is clearly not true, it seems to me, because it contradicts the information in the question stem. We're using 0.285 V in the calculation when the actual is 0.300 V, so this is clearly too low.

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[loveoforganic]

I agree with B.

 

[RogueUnicorn]

need rest of passage to know concretely.

 

[ezsanche]

Take the equation E(observed) = E(cell) - 0.0592 log [H+]. If the E(cell) were actually 0.300 V rather than 0.285 V, what would be true of the pH measured by the pH meter, assuming that no correction is made?

A. The pH from the pH meter would be too high, because the E(cell) in the calculation is too high.
B. The pH from the pH meter would be too high, because the E(cell) in the calculation is too low.
C. The pH from the pH meter would be too low, because the E(cell) in the calculation is too high.
D. The pH from the pH meter would be too low, because the E(cell) in the calculation is too low.

I thought it would be B. I simplified the equation to E(observed) = E(cell) + 0.0592 x pH. Now, the E(cell) in the calculation would clearly be lower than the actual, and I thought that was a given from the question stem. What this will do is make the E(observed) look higher than it would be if the E(cell) was actually 0.285 V instead of 0.300 V. Since we don't know that the E(cell) is too high, we would think that therefore it must the pH that is too high which is causing the increasing in E(observed).

But the correct answer is C. According to the text, "if the real value for E(cell) is actually higher than the value being plugged in, then a value that is too large is being subtracted to get E(observed). This means that the [H+] is too high, so the calculated pH is too low."

This doesn't make sense to me, and maybe I'm not getting the logic here, but it would seem to me that 1) we would get a higher E(observed) than expected so 2) we would think that the value for the entire (0.0592 log [H+]) term is less, meaning the pH calculated is more than the actual. Also, answer choice C says that the E(cell) in the calculation is too high. This is clearly not true, it seems to me, because it contradicts the information in the question stem. We're using 0.285 V in the calculation when the actual is 0.300 V, so this is clearly too low.

How can it be B? The question directly says that the E(cell) is INCREASED from 0.285 to 0.3.
0.3 is >>>>>>>>>>>>>>>>>>>>>>>>>>>>> 0.285. Ec
So the answer has to be either A or C.

Now to determine wheter it is A or C you need to determine how the pH will change when E(cell) is increased. We know the equation is

E(observed) = E(cell) + 0.0592 x pH

taking this eqaution and solving for pH
we get...

E(o) - E(cell) / 0.595 =pH

now if we increase the voltage of the cell from 0.285 to 3. we have the follwing two equations

E(o)-.285/0.595 = pH

E(o)-.3/0.595=PH

Which equation will make the pH smaller?

The second equation because the numerator will be smaller when the voltage of the cell is equal to .3.

Hence the pH will decrease

and C is the right answer

 

[Omni]

How can it be B? The question directly says that the E(cell) is INCREASED from 0.285 to 0.3.
0.3 is >>>>>>>>>>>>>>>>>>>>>>>>>>>>> 0.285. Ec
So the answer has to be either A or C.

Now to determine wheter it is A or C you need to determine how the pH will change when E(cell) is increased. We know the equation is

E(observed) = E(cell) + 0.0592 x pH

taking this eqaution and solving for pH
we get...

E(o) - E(cell) / 0.595 =pH

now if we increase the voltage of the cell from 0.285 to 3. we have the follwing two equations

E(o)-.285/0.595 = pH

E(o)-.3/0.595=PH

Which equation will make the pH smaller?

The second equation because the numerator will be smaller when the voltage of the cell is equal to .3.

Hence the pH will decrease

and C is the right answer

ezsanche is right.

(I think this question was a little bit vague, but I'm assuming that Eo is the same for both equations.)

The way I did was more of a qualitative method.
You have this equation:
Eo=Ec-0.0592*log[H] Lets simplify that eqution a bit and just get rid of the constants.
New equation is just this:
Eo=Ec+pH
As Ec increases, Eo increases (and vice versa)
As pH increases, Eo increases (and vice versa)

So now, when we subtract Ec from both sides, we will have somethng like this:
Original Ec--> Larger=pH
Increased Ec--> Smaller=pH

Hence, the increased Ec will produce a smaller pH, the second part of the answer was all ready given in the question.

 

[thebillsfan]

whoa where in the world did the question Eo=Ec-0.0592*log[H] come from? did you guys derive it from the nernst equation, or is it a commonly known derivative? ive never seen that before in my life.

 

[Tokspor]

Take the equation E(observed) = E(cell) - 0.0592 log [H+]. If the E(cell) were actually 0.300 V rather than 0.285 V, what would be true of the pH measured by the pH meter, assuming that no correction is made?

A. The pH from the pH meter would be too high, because the E(cell) in the calculation is too high.
B. The pH from the pH meter would be too high, because the E(cell) in the calculation is too low.
C. The pH from the pH meter would be too low, because the E(cell) in the calculation is too high.
D. The pH from the pH meter would be too low, because the E(cell) in the calculation is too low.

I thought it would be B. I simplified the equation to E(observed) = E(cell) + 0.0592 x pH. Now, the E(cell) in the calculation would clearly be lower than the actual, and I thought that was a given from the question stem. What this will do is make the E(observed) look higher than it would be if the E(cell) was actually 0.285 V instead of 0.300 V. Since we don't know that the E(cell) is too high, we would think that therefore it must the pH that is too high which is causing the increasing in E(observed).

But the correct answer is C. According to the text, "if the real value for E(cell) is actually higher than the value being plugged in, then a value that is too large is being subtracted to get E(observed). This means that the [H+] is too high, so the calculated pH is too low."

This doesn't make sense to me, and maybe I'm not getting the logic here, but it would seem to me that 1) we would get a higher E(observed) than expected so 2) we would think that the value for the entire (0.0592 log [H+]) term is less, meaning the pH calculated is more than the actual. Also, answer choice C says that the E(cell) in the calculation is too high. This is clearly not true, it seems to me, because it contradicts the information in the question stem. We're using 0.285 V in the calculation when the actual is 0.300 V, so this is clearly too low.

I understood the part in bold as "E(cell) remains 0.285 V in the calculations." This means to me that we're still using the calculation (E(o) - 0.285 V)/0.0592 = pH. You would get a larger pH from this than you would if you used E(c) = 0.300 V.

So my problem is just that "assuming no correction is made" indicates to me that E(c) remains 0.285 V (meaning it's too low), which leads to a higher pH.

And thebillsfan, this equation was given in the problem I had, but it's also another form of the Nernst equation commonly written as E = E(standard) - (0.0592/n)*logQ.

 

[thebillsfan]

I understood the part in bold as "E(cell) remains 0.285 V in the calculations." This means to me that we're still using the calculation (E(o) - 0.285 V)/0.0592 = pH. You would get a larger pH from this than you would if you used E(c) = 0.300 V.

So my problem is just that "assuming no correction is made" indicates to me that E(c) remains 0.285 V (meaning it's too low), which leads to a higher pH.

And thebillsfan, this equation was given in the problem I had, but it's also another form of the Nernst equation commonly written as E = E(standard) - (0.0592/n)*logQ.

the .0592 is equivalent to RT/F, correct?

 

[Tokspor]

the .0592 is equivalent to RT/F, correct?

The entire 0.0592*logQ component is equivalent to (RT*lnQ)/F.

 

[ezsanche]

I understood the part in bold as "E(cell) remains 0.285 V in the calculations." This means to me that we're still using the calculation (E(o) - 0.285 V)/0.0592 = pH. You would get a larger pH from this than you would if you used E(c) = 0.300 V.

So my problem is just that "assuming no correction is made" indicates to me that E(c) remains 0.285 V (meaning it's too low), which leads to a higher pH.

And thebillsfan, this equation was given in the problem I had, but it's also another form of the Nernst equation commonly written as E = E(standard) - (0.0592/n)*logQ.

I see now how you would have gotten B with your way of thinking. Now that I think about it the question could have been worded better.