Cr2O72-(aq) + I-(aq) -->Cr3+(aq) + IO3-(aq)
What is the oxidizing species and the reducing species?
To determine the oxidizing and reducing species in a chemical reaction, you need to consider the changes in oxidation numbers of individual elements.
In the given chemical equation:
Cr2O72-(aq) + I-(aq) --> Cr3+(aq) + IO3-(aq)
Start by assigning oxidation numbers to each element.
The oxidation state of oxygen (O) is -2, except in peroxides where it is -1.
The oxidation state of hydrogen (H) is +1.
The sum of oxidation numbers in a compound must equal the overall charge of the compound.
Now, let's assign the oxidation numbers to the elements in the given equation:
Cr2O72-(aq):
Cr has an oxidation state of +6 since there are two of them and the overall charge is -2, hence each Cr atom has an oxidation state of +6.
Oxygen (O) has an oxidation state of -2. And since there are seven oxygen atoms, the total charge is -14.
I-(aq):
Iodine (I) has an oxidation state of -1.
Cr3+(aq):
Chromium (Cr) has an oxidation state of +3.
IO3-(aq):
Iodine (I) has an oxidation state of +5.
Oxygen (O) has an oxidation state of -2, and since there are three oxygen atoms, the total charge is -6.
Now, determine the elements that undergo a change in oxidation state:
In the reaction, the oxidation state of chromium changes from +6 (in Cr2O72-) to +3 (in Cr3+). The oxidation state of iodine changes from -1 (in I-) to +5 (in IO3-).
The species being reduced is Cr2O72- since the oxidation state of chromium decreases from +6 to +3.
The species being oxidized is I- since the oxidation state of iodine increases from -1 to +5.
Therefore, Cr2O72- is the oxidizing species, and I- is the reducing species.
Oxidation is the loss of electrons and that is the reducing agent.
Reduction is the gain of electrons and that is the oxidizing agent.