For parts A & B, see http://www.enotes.com/chemistry/q-and-a/consider-galvanic-cell-based-following-half-333659

Please note that when you post a question, it should only include *one* question. Dividing it into parts A-C does not make it one question :)

For part C, we need to calculate Ecell under non-standard conditions

Ecell = Eocell - 0.0592/n log...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

For parts A & B, see http://www.enotes.com/chemistry/q-and-a/consider-galvanic-cell-based-following-half-333659

Please note that when you post a question, it should only include *one* question. Dividing it into parts A-C does not make it one question :)

For part C, we need to calculate Ecell under non-standard conditions

Ecell = Eocell - 0.0592/n log Q

We know Eocell and have 1.84V (see link above). To find Q, we have to first write the balanced equation

Au3+(aq) + 3Tl(s) --> Au(s) + 3Tl+(aq)

We need the 3s in front of Tl and Tl+ so that the number of electrons in each reaction is the same so that they will cancel out. Now, we can find Q

Q = [Tl+]^3 / [Au3+]

Q = (1x10^-4)^3 / 1x10^-2

Q = 1 x 10^-10

Now we can solve for Ecell

Ecell = 1.84 - (0.0592/3) log (1 x 10^-10)

Ecell = 2.04 V

If we increase the concentration of Au3+, then the value of Q will decrease and the log of the value will also decrease. As a result, the second term will have a greater magnitude so the Ecell will increase because we are substracting a negative term.