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.

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

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

can you figure it out with whats given?

HINT: determine whats being oxidized and whats being reduced

Fe oxidized and Aq reduced?

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

electrons*

Fe to Aq

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

yes, will it increase then?

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

so the cell potential won't change?

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

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

why would it though? it doesn't make sense

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

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

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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