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Quantitive elctrolysis question - help! - HSN forum

# Quantitive elctrolysis question - help!

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### #1bikerdude

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Posted 28 May 2007 - 12:25 PM

hey.

I'll get straight to it!
A Copper comound was know to contain Cu (I) ions OR Cu (II) ions.
It was found that 0.32g of Cu was formed.. electrolysis cell was going for 16 mintutes, current of 1 amp.

determine which copper ion was present, working must be shown.

Cant seem to get it right, so far ive just done the usaul steps, Q=IT then find out how many moles of electrons needed for the Cu (I) ion (which is one?) then gong on to say for this eqn to equal 0.32g, 960c/96500c x 63.5 (one mole of Cu)
I then did the same for the Cu (II) ion in hope of it equaling 0.32 but it didnt work... what am i doing wrong?! Ah!

Martin

### #2BuckminsterFullerene

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Posted 28 May 2007 - 01:34 PM

Firstly, use Q=It and you should get 960 C

Now, set out corresponding ratio's so you get

0.32g --- 63.5g (1 MOLE)
960 C --- x Coulombs (Unknown)

Now you should get that x = 63.5 X 960 all divided by 0.32

After such you should be getting 190500 C, this is the number of coulombs per mole, so you divide this by 1 F (96500) and should get roughly 2, so you you have two moles of electrons and It is therefore Cu2+
Chemical Engineering - Edinburgh

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### #3bikerdude

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Posted 28 May 2007 - 01:44 PM

Thank you Buck!

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