For the following reaction in acidic solution.

Cr2O7^2- + I- ---> Cr^3+ + IO3-

give the balanced half reduction reaction. Indicate the oxidizing species.

Didn't bonjo do this for you below?

Actually it is above, not below.

To find balanced half-reaction equations, you need to follow these steps:

Step 1: Identify the Reducing and Oxidizing species.
- The reducing agent is the species that gets oxidized (loses electrons).
- The oxidizing agent is the species that gets reduced (gains electrons).

In the given reaction: Cr2O7^2- + I- ---> Cr^3+ + IO3-
- The Cr2O7^2- is reduced to Cr^3+. Hence, Cr2O7^2- is the oxidizing agent.
- The I- is oxidized to IO3-. Hence, I- is the reducing agent.

Step 2: Write the half-reactions for the oxidation and reduction processes.
- The half-reaction for the oxidation process (loss of electrons) is called the oxidation half-reaction.
- The half-reaction for the reduction process (gain of electrons) is called the reduction half-reaction.

The oxidation half-reaction:
I- ---> IO3-

The reduction half-reaction:
Cr2O7^2- ---> Cr^3+

Step 3: Balance the atoms. Balance the atoms other than Oxygen and Hydrogen in each half-reaction by adding appropriate coefficients.

For the oxidation half-reaction:
I- ---> IO3-
Since there is only one I- on the left side, add a coefficient of 1 in front of IO3- on the right side to balance the iodine atoms.

I- ---> IO3- (balanced)

For the reduction half-reaction:
Cr2O7^2- ---> Cr^3+
Since there are two Cr atoms on the left side, add a coefficient of 2 in front of Cr^3+ on the right side to balance chromium atoms.

Cr2O7^2- ---> 2Cr^3+ (balanced)

Step 4: Balance the charges. Add electrons (e-) to balance the charges on each side of the half-reaction.

The oxidation half-reaction:
I- ---> IO3- + 6e-
Since the oxidation state of iodine changes from -1 to +5, it requires 6 electrons.

The reduction half-reaction:
Cr2O7^2- + 14e- ---> 2Cr^3+
Since the oxidation state of chromium changes from +6 to +3, it requires 14 electrons.

Step 5: Verify the balancing. Ensure that the number of atoms and charge are balanced on each side of the half-reactions.

The oxidation half-reaction:
I- ---> IO3- + 6e- (balanced)

The reduction half-reaction:
Cr2O7^2- + 14e- ---> 2Cr^3+ (balanced)

Step 6: Multiply the half-reactions by appropriate coefficients, if necessary. To balance the number of electrons transferred, multiply each half-reaction by a coefficient, such that the total number of electrons is the same in both the oxidation and reduction half-reactions.

In this case, we can see that the number of electrons is already balanced (6 in both half-reactions), so there is no need to multiply either half-reaction by a coefficient.

Therefore, the balanced half-reaction for the oxidation process is:
I- ---> IO3- + 6e-
(Oxidizing species: Cr2O7^2-)

And the balanced half-reaction for the reduction process is:
Cr2O7^2- + 14e- ---> 2Cr^3+