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Topic: Problem of the week - 29/10/2012  (Read 16576 times)

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Offline Borek

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Problem of the week - 29/10/2012
« on: October 29, 2012, 12:26:34 PM »
Mercury coulometer shown on  the picture below was connected in series with a small electrolysis apparatus filled with a silver nitrate solution. Voltage was applied till the current flowing in the circuit was higher than 30 mA and system was left for about 10 minutes. After that time mass of the electrode in the electrolysis apparatus changed its mass by 19.6 mg, while the mercury solution in the capillary moved by 3.11 cm. What was the current efficiency in the electrolysis apparatus, if the capillary diameter is 0.25 mm?
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Offline Rutherford

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Re: Problem of the week - 29/10/2012
« Reply #1 on: November 05, 2012, 02:43:53 PM »
I am totally unsure, but let it be 56.6%.

Offline Borek

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Re: Problem of the week - 29/10/2012
« Reply #2 on: November 07, 2012, 06:03:08 AM »
It is not 56.6%, so the problem is still open.
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Offline Schrödinger

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Re: Problem of the week - 29/10/2012
« Reply #3 on: November 07, 2012, 09:00:04 AM »
69.518% ? But I havent really verified my calculations
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Offline Borek

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Re: Problem of the week - 29/10/2012
« Reply #4 on: November 07, 2012, 09:10:40 AM »
Still not what I am waiting for.

There is always a possibility number I am expecting is not the correct one, try to convince me to your result  ;)
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Offline DrCMS

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Re: Problem of the week - 29/10/2012
« Reply #5 on: November 07, 2012, 12:02:32 PM »
I'll try 88.2%

Offline Borek

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Re: Problem of the week - 29/10/2012
« Reply #6 on: November 09, 2012, 08:19:50 AM »
I'll try 88.2%

And that's the number I was expecting - so in the worst case we are both wrong for the same reason ;)
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Offline XGen

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Re: Problem of the week - 29/10/2012
« Reply #7 on: November 09, 2012, 09:31:50 AM »
Could you guys explain how you went about solving this?

Offline DrCMS

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Re: Problem of the week - 29/10/2012
« Reply #8 on: November 09, 2012, 10:52:10 AM »
The Mercury coulometer is 100% efficient.

Hg2+ +2e  :rarrow: Hg

The movement can be converted into a volume of Hg (V = πr2l) from which you can calculate the mass of Hg (m = ρV) and then the mole of Hg formed (m/200.59).
The change in mass of the electrode can be converted into moles of silver, (19.6mg/107.8682). 
Dividing one by the other, taking into account the number of electrons in both reactions gives 88.2%.

Offline Rutherford

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Re: Problem of the week - 29/10/2012
« Reply #9 on: November 09, 2012, 11:09:20 AM »
The movement can be converted into a volume of Hg (V = πr2l) from which you can calculate the mass of Hg (m = ρV) and then the mole of Hg formed (m/200.59).
The movement is caused by precipitation and by dissolution, so by both of them. Therefore, shouldn't the obtained mass be divided by 2, before comparing with Ag?

Offline Borek

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Re: Problem of the week - 29/10/2012
« Reply #10 on: November 09, 2012, 12:09:20 PM »
You have three electrode reactions in series, charges passing through each one are identical.
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Offline Rutherford

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Re: Problem of the week - 29/10/2012
« Reply #11 on: November 09, 2012, 12:27:36 PM »
On the coulometer there is an anode and a cathode. At both, there is a process happening. The two processes that are happening make the solution to move, so I need the mass change for only one process to compare with Ag. When I calculate the mass from the volume of the cylinder (πr2l) and the density, I get the sum of the mass that is precipitated at the cathode and the mass of Hg that is released at the anode, but only one of these masses should be used for comparing (these masses are of course equal).

Offline Borek

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Re: Problem of the week - 29/10/2012
« Reply #12 on: November 09, 2012, 12:42:05 PM »
On the coulometer there is an anode and a cathode. At both, there is a process happening. The two processes that are happening make the solution to move, so I need the mass change for only one process to compare with Ag. When I calculate the mass from the volume of the cylinder (πr2l) and the density, I get the sum of the mass that is precipitated at the cathode and the mass of Hg that is released at the anode, but only one of these masses should be used for comparing (these masses are of course equal).

No. You calculated mass of the mercury on one side only. Imagine there is only one electrode (mercury is only dissolved, or only deposited). Would the mass (calculated from the movement of the mercury electrode surface) be different?
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Offline Rutherford

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Re: Problem of the week - 29/10/2012
« Reply #13 on: November 09, 2012, 12:56:29 PM »
If there was only an anode, the solution wouldn't move, because it wouldn't have where to move. If some Hg2+ must be released at the anode (its mass is m1), then some of the Hg2+ must precipitate as Hg at the cathode (mass m2). This makes the movement of the solution, meaning that both processes make it i.e. V*ρ=m1+m2 (m1=m2).

Offline Borek

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Re: Problem of the week - 29/10/2012
« Reply #14 on: November 09, 2012, 12:59:12 PM »
If there was only an anode, the solution wouldn't move

Think it over. You dissolve mercury, its mass gets lower, but the volume doesn't change?
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