[aaazureee]

We have this cell:
Cu oxidized to Cu2+ E=-0.34V
Fe3+ reduced to Fe2+ E=0.77V
So the cell potential is 0.43V
(a) The initial Fe3+/Fe2+ halfcell was prepared by dissolving 0.15mol each of fecl3 and fecl2 in 1L of water. What is the emf of the cell initially
So i can do this question: E= 0.43 + 0.0592 log 1 = 0.43V
(b) x grams of FeCl2(s) was added to the fe3+/fe2+ half cell and stirred thoroughly so that it dissolved completely. 10.0 ml sample of this solution was pipette and required 16.0ml of 0.02 mol/L K2Cr2O7 for complete reaction. Calculate x and the emf of the cell after x grams was added

So the second part is quite confusing. This is what i've done so far?
Cr2O7 2- + 6Fe 2+ + 14H+ -> 2Cr 3+ + 6Fe 3+ + 7H2O
mol of Fe2+ in 10ml = 6xmol of Cr2O7 2- reacted = 1.92 x 10^-3 mol
so conc. of Fe2+ = 0.192M.
I cant figure out how to do next.

[aaazureee]

May someone help me?

[Borek]

Titration should help to determine concentration of Fe²⁺, concentration of Fe³⁺ has not changed.

Just plug these concentrations into the Nernst equation, as you did before (unless log 1 was something completely random and you have no idea what it means).

[aaazureee]

Thank you for replying
I have calculated the new E which is 0.42V but i still dont know how to find x. The part adding the fecl2 and dissolving is so confusing for me

[Borek]

You know the sum after addition, you know how much was there initially, just subtract one from the other to calculate how much was added.

[aaazureee]

As far as i understand, in 1L of water, the sum of mol after addition is 1.92mol but initially there is 2x0.15 = 0.3mol. How is this possible?

[aaazureee]

Oh never mind i found the solution. Thanks for replying!