(23.Chemistry  Science (Dr. Bob222)
posted by Ana .
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

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. 
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 
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
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