Question

I now have an cell potential problem dealing with the molarity of its constituent aqueous solutions, never dealt with something like this before.

Answer 1

what are you trying to find?

Answer 2

I'm trying to find the Cell Potential of a given electrochemical cell, more information momentarily, OpenStudy isn't playing nice right now >:C

Answer 3

"At 25°C, E° = +1.88 V for a cell based on the reaction
3 AgCl(s) + Al(s) → 3 Ag(s) + Al3+(aq) + 3 Cl-(aq).
Find the cell potential E if [Al3+] = 0.20 M and [Cl-] = 0.010 M."

Answer 4

So I know the molarity of its ion solutions, I know that it's at 25 C, so all of the standard potential stuff is applicable.

Answer 5

@JFraser , could you maybe help me on this? You clearly know what you're doing. I'm not sure what to do with this one at all, really, lol...

Answer 6

you gotta use the Nernst equation

Answer 7

you familiar with it?

Answer 8

@aaronq is correct, use the Nernst equation for cell potential questions when the concentrations of the solutions are not standard (1M)

Answer 9

even though the TEMPERATURE is standard, the CONCENTRATIONS are not. so you have to add a correction for that

Answer 10

Alright, cool, taking a shot at it now, finding that formula as well:

Answer 11

I'm going to set up the equation, but I'm not sure how to do the corrections you're talking about/how that changes the math yet :/ \[E = E^{\circ}-\frac{RT}{nF}\ln(K)\]

Answer 12

(Scrolling issues, will plug in knowns momentarily)

Answer 13

Alright, no double arrow, so the solution isn't in equilibrium and thus E_0 doesn't equal zero; I need to get rid of E_0 somehow, yes?

I can understand the LHS itself just becoming E_0, but that doesn't algebraically make sense to me, with the E_0 term still on the RHS.

Answer 14

you're not getting rid of E^0, that's the standard potential that you'd look up from a table

Answer 15

it's not a chemical reaction, it's an algebraic equality, so \(E\) and \(E^0\) are NOT the same thing

Answer 16

\[E=\frac{(8.314 \ \rm J \cdot mol^{-1}\cdot K^{-1})(25+273.15 \rm \ K)}{( 3 e^{-} \ \rm mol)(96,500 \ s\cdot A \cdot mol^{-1})}\ln(K)\]

Answer 17

almost

Answer 18

you need to use the \(E^0\) value at the start, it's not equal to zero

Answer 19

\(E\) is the voltage you're solving for, \(E^0\) is the standard voltage of 1.88V the problem gives you

Answer 20

\(K\) is the value of the equilibrium constant, which you can find because they give you the concentrations of the ions

Answer 21

I'm going to leave units out for the remainder due to how messy it's getting;

I'm trying to remember how to determine K, isn't it something like the molarity of gases of the products to the power of their coefficients, divided by the molarity of gases of their reactants to the power of their coefficients?

Answer 22

Yeah

Answer 23

OH, that's where E^{0} comes in, didn't even catch that, thank you.

Answer 24

the value of \(K\) will be\[K = [Al^{+3}]*[Cl^{-1}]^3\]

Answer 25

since you're given the concentrations, this is easy\[K = [0.20M]*[0.010M]^3 = 2*10^{-7}\]

Answer 26

so plug into the Nernst equation:\[E = E^0 - \frac{RT}{nF}*\ln(K)\]

Answer 27

Alright, cool; I think I'm going to move on rather than numerically calculating it, just to cover as many remaining things conceptually as possible. I'm going to have to trust myself that I can do the Algebra at the right time in two hours, but if I don't cover most things conceptually, it might all be for nothing.