CHEMISTRY COMMUNITY

So I struggled to do problem 13.1 until I finally got an answer, and I tried self testing myself with problem 13.2. However, I am not sure what to do.

The problem reads: The following redox reaction is used to prepare orthotelluric acid:
Te(s)+ClO3-(aq)+H2O---->H6TeO6(aq)+Cl2(g)
a)Identify the elements undergoing changes in oxidation state and indicate the initial and final oxidation #s for these elements
b)write and balance reduction/oxidation half-reaction
c)Combine half-reactions to produce a balanced redox reaction

I am not sure about the oxidation # for...(reactants)
ClO3- since Oxygen is -2 and Cl is -1 (the molecule is unstable)
H2O is 0
Te is 0

then (products) Cl2(g) is -2 and I am not sure if H6TeO6(aq) is -12?? or -6?

I also tried balancing it, but everything is separated so how can I get two half-reactions for reduction and oxidation?

Please help? Unless, you feel that this problem is not "homework," then I understand.
thanks :D

For part a) you figure out the individual charges for i) Te as Te --> H₆TeO and ii)Cl as ClO₃^- --> Cl₂.

For i) Te = 0 charge since there is no stated charge in the equation. H₆TeO will have to be broken down in order
to find the individual charge for Te.
H₆ = 6(+1)=+6
Te= ?
O₆= 6(-2)=-12
The overall charge for H₆TeO is 0 so the charge of Te must cancel out the charge of -6 brought from the H₆
and O₆; Therefore, Te must be +6. So we can conclude that Te is being oxidized from a 0 charge to a +6 charge.

For ii) ClO₃^- must also be broken down to find the individual charge for Cl. We know that
O₃= 3(-2) = -6
and that the overall charge of ClO₃^- is -1, so Cl must have a +5 charge in order to give the whole
molecule a charge of -1.
Cl₂= 0 because there is no stated charge in the equation; therefore, we can conclude that Cl is being reduced
from +5 to 0.

I am working on part b right now, so I will post the explanation in a couple of minutes.

for part b) we write the oxidation half-reaction (we established from part a that Te is being oxidized so we will write the half reaction for it)

1. Write down the half reaction
Te --> H₆TeO₆

2. balance all elements except for hydrogen and oxygen
Te is already balanced

3. balance oxygen by adding H2O
6H₂O + Te --> H₆TeO₆

4. Balance hydrogen by adding H+
6H₂O + Te --> H₆TeO₆ + 6H⁺

5. Balance charges by adding e-
Left hand side =0 charge. Right hand side =6+ charge (Balance by adding 6e^- to the right hand side)
6H₂O + Te --> H₆TeO₆ + 6H⁺ + 6e^-

Hope this helps. If you follow these same steps you can balance the reduction half-reaction as well.