## Redox Reaction: prepare orthotelluric acid

FrancoNancy_Sec1L
Posts: 26
Joined: Fri Sep 20, 2013 3:00 am

### Redox Reaction: prepare orthotelluric acid

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?

thanks :D

martha-1I
Posts: 76
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Redox Reaction: prepare orthotelluric acid

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.

martha-1I
Posts: 76
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Redox Reaction: prepare orthotelluric acid

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

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.

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