Question

2Rb+I2---> 2RbI
I need help with finding the element that is oxidized and reduced.

Answer 1

Argh!@! I've typed my answer twice and both times OS refreshed itself on me...

Answer 2

O.I.L R.I.G
oxidization is loss (of electrons)
Reduction is Gain (of electrons)

Answer 3

In the reactants, Rb isn't reacting with anything so its charge is 0. The iodines are covalently sharing electrons, so their charges are 0 as well.

To form the ionic soilid RbI, rubidium gives up 1 electron to iodine since Rb is an alkali metal while I is a halogen. Since Rb loses an electron its charge in RbI is +1, I's charge is -1 since it gains an electron.

When an atom loses electrons, it is oxidized. When an atom gains electrons, it is reduced.

Answer 4

I iodine is a halogen so its atoms have 7 electrons in the outer shell
Rb Rubidium is a alkali metal so it has 1 electron in the outer shell

The molecule is formed as the 1 electron from the metal goes to the halogen
ie Rb has lost its electron, - it has been oxidized
I has gained the electron - it has been reduced