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Topic: Electrochemistry Calculation  (Read 1592 times)

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Offline aaazureee

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Electrochemistry Calculation
« on: October 15, 2015, 10:19:31 AM »
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.

Offline aaazureee

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Re: Electrochemistry Calculation
« Reply #1 on: October 16, 2015, 10:13:20 AM »
May someone help me?

Offline Borek

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Re: Electrochemistry Calculation
« Reply #2 on: October 16, 2015, 02:21:38 PM »
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).
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline aaazureee

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Re: Electrochemistry Calculation
« Reply #3 on: October 17, 2015, 02:48:17 AM »
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

Offline Borek

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Re: Electrochemistry Calculation
« Reply #4 on: October 17, 2015, 02:56:54 AM »
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.
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Offline aaazureee

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Re: Electrochemistry Calculation
« Reply #5 on: October 17, 2015, 03:25:20 AM »
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?

Offline aaazureee

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Re: Electrochemistry Calculation
« Reply #6 on: October 17, 2015, 03:32:37 AM »
Oh never mind i found the solution. Thanks for replying!

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