Electrolysis of potassium iodide

Electrolysis of molten potassium iodide

Would give iodine and potassium. At this temperature, (molten potassium iodide), iodine exist as gas. This is the violet gas. The silvery solid is potassium.

In molten potassium iodide, the only cation is K+ and the only anion is I-, iodine vapour and potassium metal were the only products.

I read your http://hk.geocities.com/genevieve_ching/2.doc

i) Since copper deposited on electrode R, the electrode reaction must be
Cu2+ + 2e --> Cu , therefore electron flow must be counter clockwise.

ii) 1) Reaction at electrode S would be 4OH- --> O2 + 2H2O + 4e, gas bubble would be seen to evolved from electrode S.

2) It becomes paler in blue colour.

Metal Q would dissolve to provide e, if its cation is coloured, it would change the colour of the solution.
Reaction at electrode P is 2H+ + 2e ---> H2, gas bubble would be seen to evolved from electrode P.

iii) Set-up X is actually a electro chemical cell.

iv) Since metal would dissolve and not P dissolve. Q is higher than P in the electro chemical series.

+ve electrode and -ve electrode are actually compared by electric potential. the one at higher electric potential is the +ve electrode.

In your set-up. Current flow out from electrode P. This is the +ve electrode while current flow in to electrode Q, Q is the -ve electrode. P is connected to S, therefore S is the +ve electrode and R is the -ve electrode. If you use 3 beaker. Let the left one be the electrode chemical cell, consider the middle one now, the electrode connected to the -ve pole of the electrode chemical cell would be the -ve electrode. then, the next electrode in the middle beaker is the +ve electrode. For the right hand side beaker, the electrode connected to the +ve electrode in the middle beaker is the -ve electrode of the right hand side beaker.