The correct answer is D.
But my question is how is it possible to have an initial voltage of 1.1V for the cell is with concentration of 1M, if E standard for the cell is 0.48.
I am sure I am missing something here.
An electrochemical battery made from Sn and Cu electrodes under standard conditions shows an
initial voltage of V= 1.1 V. If the concentration of both Cu2+ and Sn2+ solutions is increased to 2 M from
1 M, the observed voltage will: (Hint: use the Nernst equation)
A. increase
B. increase if Sn is oxidized
C. increase if Cu is oxidized
D. remain the same
E. C. decrease
5 answers
I'm not Dr. Bob222, but I play on this site every now in then. I think you are confused because of the wording of the question; you are not calculating Eo based on the information that a chart in your book gives you, but they are just giving you Eo and asking you based on the concepts, what will happen. You are correct: Eo should be equal to 0.48, based on information from your book, but they are giving you a number and asking you how it will change. You can use the Eo value that you calculated or you can use the value that they give you; either value will return the same answer. The Nernst equation is as followed:
E = Eo −(0.0592Vnl)ogQ
As, you can see if you substitute 1M or 2M values for Cu2+ or Sn2+ into Q, you will just get Eo, since log of 1 is equal to 0. And 0 times any number equals 0. So, E will be equal to Eo and the value will not change, or better yet the value will remain the same. D is the best answer choice.
E = Eo
E = Eo −(0.0592Vnl)ogQ
As, you can see if you substitute 1M or 2M values for Cu2+ or Sn2+ into Q, you will just get Eo, since log of 1 is equal to 0. And 0 times any number equals 0. So, E will be equal to Eo and the value will not change, or better yet the value will remain the same. D is the best answer choice.
E = Eo
Thank you Devron for explication.
This Q. seemed too easy to me, and I though it must be a trick question or something, nonetheless thank you for explaining.
Devron is right but I want to add a point or two.
First, Ana is right. It PROBABLY isn't possible unless the reaction is carried out in under other than standard conditions; i.e., basic solution, complexing materials added to the basic mix, etc. Devron is right that none of that matters.
What does matter?
That log term matters. log (1/1) = 0; log (2/2) = 0 and in fact anything they do to the numerator doesn't matter as long as the same is done to the denominator because log x/x = 0
First, Ana is right. It PROBABLY isn't possible unless the reaction is carried out in under other than standard conditions; i.e., basic solution, complexing materials added to the basic mix, etc. Devron is right that none of that matters.
What does matter?
That log term matters. log (1/1) = 0; log (2/2) = 0 and in fact anything they do to the numerator doesn't matter as long as the same is done to the denominator because log x/x = 0
0 answers