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Author Topic: How would i go about this question>?  (Read 1080 times)  Share 

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Alexx

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How would i go about this question>?
« on: October 11, 2010, 08:25:39 pm »
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Okay, so we've gone through this in class so many times and each time we're getting a different answer.

Does the 3.50A apply to each of the six cells or the whole battery?

Can someone please walk me through it?

akira88

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Re: How would i go about this question>?
« Reply #1 on: October 11, 2010, 08:37:27 pm »
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"A particular lead-acid battery delivers a current of 3.50A..."
I would assume it is just the battery as it does not say "each cell delivers a current of 3.50A".
Hope that helps :)
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physics

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Re: How would i go about this question>?
« Reply #2 on: October 11, 2010, 08:40:15 pm »
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apparently its the whole thing
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matt123

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Re: How would i go about this question>?
« Reply #3 on: October 11, 2010, 08:47:14 pm »
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"A particular lead-acid battery delivers a current of 3.50A..."
I would assume it is just the battery as it does not say "each cell delivers a current of 3.50A".
Hope that helps :)

yeah agreed.
its the whole thing = 3.50 amp.s

therefore for part I
since n(e) = Q/f

you get 3.5 X 20 X 60 / 96500
which gives you 0.0435 mol of electrons

then since its +2 electrons .. every +2 gives you 1 mol .
hence its 0.0435 / 2 ... = 0.02175 mol of the pb02

hmm part ii)
i would find the mass overall of the battery
then divide it by 6.


some 1 correct me if im wrong

hope that helps

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Mao

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Re: How would i go about this question>?
« Reply #4 on: October 11, 2010, 11:10:48 pm »
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The cells are connected in series. I'm fairly sure the answer to part b i) is 0.0217*6, and the answer to part ii) is 0.0217*mass(PbSO4)

It's been a while since I looked at cells in this much detail, but if I remember correctly (I don't have a schematic handy atm), 3.50A current passes through each of the 6 cells, and in turn produces 12V when one pair of the redox only creates 2V.
« Last Edit: October 12, 2010, 11:32:29 am by Mao »
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matt123

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Re: How would i go about this question>?
« Reply #5 on: October 12, 2010, 07:53:38 am »
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The cells are connected in series. I'm fairly sure the answer to part b i) is 0.0435*6, and the answer to part ii) is 0.0435*mass(PbSO4)

It's been a while since I looked at cells in this much detail, but if I remember correctly (I don't have a schematic handy atm), 3.50A current passes through each of the 6 cells, and in turn produces 12V when one pair of the redox only creates 2V.

just out of curiosity.

since there are 2 electrons
why dont u go 0.435 / 2
giving 0.2 abouts?
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Mao

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Re: How would i go about this question>?
« Reply #6 on: October 12, 2010, 11:32:48 am »
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The cells are connected in series. I'm fairly sure the answer to part b i) is 0.0435*6, and the answer to part ii) is 0.0435*mass(PbSO4)

It's been a while since I looked at cells in this much detail, but if I remember correctly (I don't have a schematic handy atm), 3.50A current passes through each of the 6 cells, and in turn produces 12V when one pair of the redox only creates 2V.

just out of curiosity.

since there are 2 electrons
why dont u go 0.435 / 2
giving 0.2 abouts?

Misread your post, apologies. Previous post fixed.
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crayolé

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Re: How would i go about this question>?
« Reply #7 on: October 13, 2010, 12:47:49 am »
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So the current doesn't get split up between the 6 cells?

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Re: How would i go about this question>?
« Reply #8 on: October 13, 2010, 05:58:41 am »
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So the current doesn't get split up between the 6 cells?

Connected in series --> the same current goes through each cell connected 'head-to-tail', the voltage adds up (each cell is only 2V, a 6-cell battery is 12V).
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