## Endgame Q.2D

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

lilymayek_1E
Posts: 107
Joined: Sat Aug 17, 2019 12:16 am

### Endgame Q.2D

Can anyone guide me through part D of question 2 on the 'endgame' review? I've correctly calculated part B (which is to find deltaGo, Eo, and K) I'm quite sure, but every time I go through part D I feel like I'm getting a crazy answer.
The galvanic cell is at 25C:
Fe(s) | Fe2+(aq) || Ce4+(aq), Ce3+(aq) | Pt(s)
Ce4+(aq) + e- → Ce3+ (aq) Eo = 1.61V
Fe2+(aq) + 2e- → Fe(s) Eo = -0.44V

Part D is:
If the concentration of the Ce4+ and Ce3+ ions are 0.015M, what would the concentration for the Fe2+ ions be when it powers a 2.20V green LED light?

JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

### Re: Endgame Q.2D

For part D you should use the Nernst equation. E=Eo-(RT/nF) *lnQ
You have found Eo in part b. E is given as 2.20V. And Q=[Fe2+][Ce3+]^2/[Ce4+]^2 (products over reactants)
I believe the answer should be [Fe2+]=8.4*10^-6 M

JohnWalkiewicz2J
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am
Been upvoted: 1 time

### Re: Endgame Q.2D

For part D, start off with the nernst equation : E=Eo-(RT/nF) *lnQ, and since you have every value other than K, rearrange the equation to solve for K.
Since K = ([Ce3+]^2 * [Fe2+]) / ([Ce4+]^2), and you have the concentrations of Ce3+ and Ce4+, you can solve for the missing concentration [Fe2+].
Hope this helps!! : )))))

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