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2002 - Written Paper - Q14(a) - HSN forum

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2002 - Written Paper - Q14(a)


4 replies to this topic

#1 David3001

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Posted 30 May 2005 - 04:40 PM

I thought I knew how to do this one, the formula I used was:

Weight obtained = (amps x secs x 1 mole)/(96500 x 2)

My result wasn't what was in the answers, so any help would be appreciated, thanks.

#2 David3001

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Posted 30 May 2005 - 04:52 PM

I was also stuck with 14.b)ii.

The formula I was given at the top of the page was

I2 (aq) + 2S2O3 2- (aq) ---> 2I- + S406 2- (aq)

If that's required.

Thanks again.

#3 gary

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Posted 30 May 2005 - 05:02 PM

Q=It
Q=0.97 c

2I-(aq) ---> I2 (aq) + 2e-

2mol Iodine is deposited by 1 faraday

so 2x96500=193000c

0.97/193000 x 126.9

= 5.025x10 -6 moles

#4 Icemaiden

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Posted 30 May 2005 - 07:35 PM

Ahh thankee....I always get stuck on electrolysis.

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#5 David3001

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Posted 30 May 2005 - 08:01 PM

Anyone able to help with the second question? Sorry for asking for all this help lol, but I'm really grateful.





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