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Chemistry
OpenStudy (anonymous):

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.

OpenStudy (aaronq):

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

OpenStudy (anonymous):

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

OpenStudy (aaronq):

can you figure it out with whats given?

OpenStudy (aaronq):

HINT: determine whats being oxidized and whats being reduced

OpenStudy (anonymous):

Fe oxidized and Aq reduced?

OpenStudy (aaronq):

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

OpenStudy (aaronq):

electrons*

OpenStudy (anonymous):

Fe to Aq

OpenStudy (aaronq):

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

OpenStudy (anonymous):

yes, will it increase then?

OpenStudy (aaronq):

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

OpenStudy (anonymous):

so the cell potential won't change?

OpenStudy (aaronq):

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

OpenStudy (anonymous):

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

OpenStudy (aaronq):

why would it though? it doesn't make sense

OpenStudy (anonymous):

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

OpenStudy (aaronq):

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

OpenStudy (aaronq):

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.

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