Balancing a chemical equation with ions

[rasen58]

The question was to balance this equation

 

What I did was

Cu2+ + 2I- -> CuI + I2

 

But the answer is

 

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

 

So do 2 moles of ions of an element combine to make 1 mole of that element? Is that what I was missing?

[rktpro]

In your attempt to balance, you see, Iodine atoms are not equal on both sides. Though, I wonder the product is CuI and not Cu2I2

[rasen58]

But even if I did

Cu²⁺ + 3I^- -> CuI + I₂

it would still be wrong, but why is it wrong?

[Sensei]

Check what is said in preparation of CuI

http://en.wikipedia.org/wiki/Copper%28I%29_iodide

[jimmydasaint]

It seems to be a two step process. With the products of the first reaction decomposing to make reduced copper.

http://wiki.answers.com/Q/How_is_cuprous_iodide_prepared?#slide=1

[ashwinkirtane]

Whenever you want to balance a redox reaction, there are essentially few steps involved. If you follow them correctly, you'll certainly arrive at the correct answer.

 

Step 1: Write the two halves of the redox reaction (oxidation half and the reduction half)

Here,

Oxidation Half:

I^- -> I₂

Reduction Half:

Cu²⁺ -> Cu⁺

 

Step 2: Now that you've written the two halves, balance them. If required(not required in this case), add water molecules to Hydrogen deficit side, OH^- if the reaction was carried out in basic medium and H⁺ ions if the reaction was carried out in acidic medium. Since reduction is gain of electrons and oxidation is loss of electrons, show that too. Like this,

 

Oxidation Half:

2I^- -> I₂ + 2e

Reduction Half:

2Cu²⁺ + 2e-> 2Cu⁺

 

^*Make sure that the electrons on both the sides are equal (charge conservation).

 

Step3: Add both the halves:

 

2Cu²⁺ + 2e + 2I^--> 2Cu⁺ + I₂ + 2e

 

Now use some common sense! See, you need CuI on the product side so add two iodide ions on both side and cancel the electrons off course. What do you get?

2Cu²⁺ + 4I^- -> 2CuI + I₂