Question

AP Chem Help!
The following galvanic cell at standard conditions has a potential of 0.03 V: Ag+(aq) + Fe2+(aq) - Fe3+(aq) + Ag(s). What would be the effect of increasing the concentration of Ag+?
A. The cell potential will be 0 V.
B. The cell potential will decrease.
C. The cell potential will not change.
D. The cell potential will increase.

Answer 1

well, what is the source of electrons (ultimately the source of current)?

Answer 2

i don't know my worksheet doesn't state that. This is all that is given.

Answer 3

can you figure it out with whats given?

Answer 4

HINT: determine whats being oxidized and whats being reduced

Answer 5

Fe oxidized and Aq reduced?

Answer 6

so when you have a galvanic cell, where are the electrodes flowing from?

Answer 7

electrons*

Answer 8

Fe to Aq

Answer 9

it's Ag. but yeah thats right. So, if you added more Ag, would that affect the potential (i.e. electron flow)?

Answer 10

yes, will it increase then?

Answer 11

nope. because that's not the source of electrons. If you raised \([Fe^{2+}]\), then the potential would go up.

Answer 12

so the cell potential won't change?

Answer 13

nope, why would it? If \(Fe^{2+}\) runs out, and there is \(Ag^+\) left, the battery won't run.

Answer 14

i just checked the back of the book and it says it does increase.
Thanks for you help though.

Answer 15

why would it though? it doesn't make sense

Answer 16

i really don't know, ill ask my teacher about it tomorrow.

Answer 17

if you use the nernst equation: \(E=E^o+\dfrac{RT}{nF}*log\dfrac{[Fe^{3+}]}{[Ag^+][Fe^{2+}]}\)

at 25 celsius it reduces to:

\(E=0.03V +\dfrac{0.0591}{n}*log\dfrac{[Fe^{3+}]}{[Ag^+][Fe^{2+}]}\)

the reaction is a 1 electron transition

\(E=0.03V +\dfrac{0.0591}{1}*log\dfrac{[Fe^{3+}]}{[Ag^+][Fe^{2+}]}\)

at standard conditions the concentrations of all species are equal to 1 M

\(E=0.03V +\dfrac{0.0591}{1}*log\dfrac{[1]}{[1][1]}\)

if we raise \([Ag^+]\), \(E=0.03V +\dfrac{0.0591}{1}*log\dfrac{[1]}{[2][1]}\)

E= -0.1482097574330769 V

Answer 18

we'll what the nernst equation shows is that the potential is actually going more negative.
Which would mean that the potential decreased, so i was wrong.