[firefly08642]

The question says:
The standard potentials at 25 degrees Celsius of the half-cells Pt/Fe3+,Fe2+ and Pt/I2,I- are +0.771V and +0.620V respectively. Neglecting activity coefficients, calculate the potential
i) on a platinum wire dipping into a solution containing 1x10^-5 mol Fe(NO3)3 per kg of solvent and 1 x 10^2 mol Fe(NO3)2 per kg of solvent

Please check if I answered this correctly:

Fe/Fe3+//I2/I-

I2   +   Fe2+    ->    Fe3+    +     I-

Q=[Fe3+]/[Fe2+] = (10^-5)/(10^-2)
E=(0.620)-(0.771)=-0.771V
E=E-(RT/nF)ln Q where n = 2, R = 8.31451070, T = 298.15, F=9.648530929
=-0.68V

[firefly08642]

Ummm... No one knows?  

[Donaldson Tan]

nobody knows? you just have to be patient.

anyway, you are wrong.

E=E^o-(RT/nF)ln Q is for finding the electrode potential of one oxidised species to its corresponding reduced potential, from the standard value.

Use the above equation to E_oxd and E_red, then you can find E_cell by using E_cell = E_oxd + E_red

Reminder: be careful with your positive/negative signs

[firefly08642]

Thanks a lot.

I was told today that we haven't covered the necessary material to answer this question, which is why I couldn't do it.  

I'll use your information and a physical book to see if I can answer it before I go the lecture.

[firefly08642]

It should've been: E=(0.620)-(0.771)=-0.151V

I'm still stuck with this question. The lecture didn't really help.  

I'm not sure but I think I found the Ecell as you said above.

But the question is saying about a Pt wire dipping into a solution of 10^-5 mol/L of Fe3+ and 10^-2 Fe2+. What do I do with these concentrations?

Please help, it's due in for Monday. All I can think of is Nerst equation... which you told me is wrong.

[firefly08642]

The question out in full is:
The standard potentials at 25 degrees Celsius of the half-cells Pt/Fe3+,Fe2+ and Pt/I2,I- are +0.771V and +0.620V respectively. Neglecting activity coefficients, calculate the potential
i) on a platinum wire dipping into a solution containing 1x10^-5 mol Fe(NO3)3 per kg of solvent and 1 x 10^2 mol Fe(NO3)2 per kg of solvent
ii)on a platinum wire dipping into a solution containing 1x10^-3mol of free iodine and 1x10^-2 mol of KI per kg of solvent. neglect the possible formation of I3- ion.

What chemical reaction, if any, occurs when the wires are removed and equal volumes of the two solutions i and ii above are mixed? Justify your answer.

I think if I can do i) I should be able to apply the same thing to the rest of the question.

[Albert]

i) You just need to calculate E for this electrode?

E_Fe3+/Fe2+° = +0.771V

Fe³⁺ + e- -> Fe²⁺

E = E° - 0.0592 log (1x10^-2 / 1 x 10^-5)

...someone will be furious with me for this

[firefly08642]

I think Albert is correct - Thank you so much.

I started to work on this question again now, and I was thinking.. why am I doing 0.771-0.620, when the first part of the question has nothing to do with I2 and I-. standard electrode is just 0.771 as albert said. Then because two concentrations are given, I think Nerst equation must be used. I know R, T, n (the equation of Fe3+ + e- -> Fe2+: 1 electron is transferred), F and standard E is 0.771 as above.
Q, the reaction quotient is to do with the concentrations of Fe3+ and Fe2+. The only problem is that I don't know if it is Fe3+/Fe2+ or Fe2+/Fe3+. I looked in my lecture note and it says /[oxd] so I think [Fe2+]/[Fe3+] = [10^-2]/[10^-3] as albert put is right.

Thanks again, I think I can do the rest by myself now.