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Copper (II) and Iodide

Just looking at converting two half equations into an ionic one.

I'm looking at a reaction where aqueous Cu 2+ ions are mixed with aqueous iodide ions. So, iodide ions will be oxidised to iodine ( 2I - ARROW I2 + 2e-)

and copper 2+ ion will be reduced to copper 1+ ions

( Cu2+ ARROW Cu+ + e-)

So to combine these two half equations to get the ionic equation, I need to multiply the copper half equation by two to balance the electrons.

Doing so and adding the two half equations I get:

2Cu2+ + 2I- ARROW 2Cu+ + I2

but apparently the answer is

2Cu2+ + 4I- ARROW 2CuI + I2

I'm not sure why this is the case, why is there a 2:1 ratio between iodide and cooper (II) ions? Also where does the Copper Iodide on the LHS come in?

Thank you!

Reply 1

EierVonSatan

You've added the two half equations together just fine. Does the question state that CuI is insoluble in water or that the reaction forms a precipitate? This is your cue to have CuI_(s) on the left side of your ionic equation - since you've in effect added 2I- to the left you need to do the same to the right for it to balance :yep: