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December 10, 2016
Posted by **Ana ** on Thursday, May 8, 2014 at 7:17am.

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

- (23.Chemistry - Science (Dr. Bob222) -
**Ana**, Thursday, May 8, 2014 at 7:21amThe 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. - (23.Chemistry - Science (Dr. Bob222) -
**Devron**, Thursday, May 8, 2014 at 10:32amI'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 - (23.Chemistry - Science (Dr. Bob222) -
**Ana**, Thursday, May 8, 2014 at 1:40pmThank you Devron for explication.

- (23.Chemistry - Science (Dr. Bob222) -
**Ana**, Thursday, May 8, 2014 at 1:51pmThis Q. seemed too easy to me, and I though it must be a trick question or something, nonetheless thank you for explaining.

- (23.Chemistry - Science (Dr. Bob222) -
**DrBob222**, Thursday, May 8, 2014 at 7:24pmDevron 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