CONSIDER THE FOLLLOWING UNBALANCED REDOX EQUATION:
IO3^-(AQ) + Al(S)-->I2(S) + A;^3+(AQ)
A. WHAT IS THE OXDATION NUMBER OF IODINE IN THE IODATE ION? I GOT IO3^6(AQ)?
B. WHICH SPECIES IS THE REDUCING AGENT I GOT AL^3+(AQ)
C. USING THE OXIDATION AND REDUCTION HALF REACTIONS WRITE THE BALANCED REDOX EQUATION
I is +5 in IO3^-
B is correct.
What's the problem with C?
Al^3+ + 3e ==> Al is one half cell
The other one is
2IO3^- + 10 e + 12H^+ = I2 +6H2O
I assume an acid solution.
A. To determine the oxidation number of iodine in the iodate ion (IO3^-), you need to account for the overall charge of the ion. Since the equation provides IO3^-(AQ), the overall charge is negative 1 (-1).
The equation can be set up as follows:
oxidation number of I + 3 * oxidation number of O = overall charge
Let's assume the oxidation number of iodine is x. The oxidation number of oxygen in most compounds is -2, but since there are 3 oxygen atoms in the ion, the total oxidation number due to oxygen is -6.
So we have:
x + 3(-2) = -1
x - 6 = -1
x = -1 + 6
x = +5
Therefore, the oxidation number of iodine in the iodate ion (IO3^-) is +5.
B. To determine the reducing agent, you need to identify the species that is being reduced. In the unbalanced equation provided, aluminum (Al) is being oxidized from its elemental state (Al(S)) to the +3 oxidation state (+3 charge).
Since the reducing agent is the species that causes the reduction of another species, the reducing agent in this equation is Al(S), aluminum in its elemental form.
C. To balance the redox equation, you need to identify the oxidation and reduction half-reactions and balance the number of atoms and charges on both sides.
Oxidation half-reaction:
Al(S) → Al^3+(AQ)
On the left side, the aluminum is in its elemental state (0 oxidation state), and on the right side, it is in the +3 oxidation state.
Reduction half-reaction:
IO3^-(AQ) → I2(S)
On the left side, iodine is in the +5 oxidation state, and on the right side, it is in the elemental state (0 oxidation state). The 3 oxygens on the left side give us 6 negative charges.
To balance the charge, we need to multiply the reduction half-reaction by 2:
2 IO3^-(AQ) → I2(S) + 6e^-
Now, we have:
Oxidation half-reaction: Al(S) → Al^3+(AQ)
Reduction half-reaction: 2 IO3^-(AQ) → I2(S) + 6e^-
Finally, we balance the number of electrons transferred by multiplying the oxidation half-reaction by 6:
6 Al(S) → 6 Al^3+(AQ)
Now, we have the balanced redox equation:
6 Al(S) + 2 IO3^-(AQ) → 3 I2(S) + 6 Al^3+(AQ)