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Topic: GCSE Level Ionic Half Equations  (Read 3817 times)

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GCSE Level Ionic Half Equations
« on: February 23, 2006, 05:43:34 PM »

Got a GCSE triple science module set coming up, and I'm having a little trouble grasping the science behind electrolysis - specifically copper purification.

I can see the basic process here: copper loses electrons at the positive electrode, and gains them at the negative electrode. However, the associated half equations relating to this don't make sense to me.

According to the BBC:

Negative electrode:
In copper purification:  Cu2+ (small 2+ at top) + 2e-  ->  Cu
Positive electrode:
In copper purification:  Cu - 2e-  ->  Cu2+ (small 2+ at top)

What I can't grasp is how copper losing two electrons can cause it to end up as Cu2+ (and visa versa), presuming the smaller number (eg 2+) represents the amount of electrons copper has at that time?

I've looked around, but haven't been able to find a clear explination as to why it works out as it does.

Any advice would be greatly appreciated.

Kind regards,

Offline mike

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Re:GCSE Level Ionic Half Equations
« Reply #1 on: February 23, 2006, 06:11:17 PM »
The superscript number represents the charge on the ion, therefore a + represents a positively charged ion (cation) and a - represents a negatively charged ion (anion). Now a negative charge is basically an electron. You know that an atom is composed of protons(+) and electrons(-) so when there are more protons than electrons the ion is positive (+) and when there are more electrons than protons it is negative (-).

Now take copper for example, it is the 29th element so it has 29(+) protons and 29(-) electrons in the ground state. If it were to lose two electrons it now has 29 (+) protons but only 27 (-) electrons leaving a net total of 2 positive charges, ie 2+ or Cu2+.

Is this what you mean or have I misunderstood your question?
There is no science without fancy, and no art without facts.

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