Br2(aq) + 2 KI(aq) ==> 2 KBr(aq) + I2(s)
What is oxidized and what is reduced in this reaction?
Well, it looks like Br2 is being reduced because it's going from elemental bromine to bromide ions (Br-). So, we could say Br2 is going from 'Br2 (aq)' to '2 Br- (aq)'.
As for what's being oxidized, KI is a good candidate. The iodide ions (I-) are being oxidized to form iodine (I2) which is a solid. So, we could say I- is going from '2 KI (aq)' to 'I2 (s)'.
So, Br2 is getting a reduction makeover while KI is getting oxidized, feeling all fancy and turning into I2.
To determine what is oxidized and what is reduced in the given chemical reaction, we need to identify the changes in oxidation states of the elements involved.
In this reaction, we have bromine (Br) and iodine (I) as the elements involved.
Initially, Br2 has an oxidation state of 0, and I has an oxidation state of -1 in KI. In the products, KBr has an oxidation state of -1, and I2 has an oxidation state of 0.
By comparing the oxidation states before and after the reaction, we can determine the changes:
Br2: 0 → -1 (reduction)
I: -1 → 0 (oxidation)
From the above analysis, we can conclude that bromine (Br2) is reduced, and iodine (I) is oxidized in this reaction.
Assign oxidation numbers and you will know.
Br2(aq) + 2 KI(aq) ==> 2 KBr(aq) + I2(s)
Br2 is zero for each on the left to 2- on the right.
2I is 2- on the left to zero on the right.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.