Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on October 29, 2012, 12:26:34 PM
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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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I am totally unsure, but let it be 56.6%.
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It is not 56.6%, so the problem is still open.
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69.518% ? But I havent really verified my calculations
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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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I'll try 88.2%
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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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Could you guys explain how you went about solving this?
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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%.
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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?
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You have three electrode reactions in series, charges passing through each one are identical.
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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).
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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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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).
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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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If the volume doesn't change then there wouldn't be any movement at all.
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If the volume doesn't change then there wouldn't be any movement at all.
Sigh. That was a rhetoric question. Is it possible for the mass to change but for the volume to remain constant?
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Yes, it has to change as the ions are smaller than atoms, but again, it will change at both ends so it is m1+m2 or 2m.
I missed the example, it should be the cathode. So I imagine that there is only a cathode and at the cathode, the following process happens: Hg2++2e- :rarrow: Hg as Hg is bigger than Hg2+ the volume increases, because the solution volume decreases, but the solution won't move. 2 processes cause the move.
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Yes, it has to change as the ions are smaller than atoms
It is completely unrelated, surface moves because mercury gets oxidized and becomes dissolved, so it leaves metallic phase and enters aqueous phase.
You are wrong on so many levels I give up. It doesn't make you right, but it gives me a free evening >:D
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If it only gets oxidized and becomes dissolved then the solution in the middle should just increase in volume.
Okay, I simply can't understand why the move depends on only 1 process, but not of both (oxidation and reduction). If no explanation, then I leave this problem :(.
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I think I got it now.
"...surface moves because mercury gets oxidized and becomes dissolved, so it leaves metallic phase and enters aqueous phase. "
This makes the solution in the middle expand by 3.11cm in one direction, so it can actually move because of one process. The other process causes just that the solution contracts in the other direction by 3.11cm so it has the same volume at the end of these processes.
Borek, I am really sorry for the disturbance, I just couldn't figure it out yesterday :-[. I hope you understand.
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No problem, it happens. Leaving the problem for the night usually helps.
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My method was the same. Except that I thought the equation would be :
Hg22+ + 2e --> 2Hg
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