[peggysue32]

I am doing grade 12 chemistry through correspondence (SCH4U) and I have no idea how to do one question.  My text book tells me one way, and it seems as though my learning guide contradicts it.

I need to balance equations by the HALF-CELL method.  Show half-cell reactions and identify them as oxidation or reduction.

How do I go about doing this?  What information do I need for each element?

One question i have is

    2-          -        +              2+          2-
SO3  +  MnO4  +  H  <==> Mn    +  SO4  +  H2O(l)

Thank you in advance, for any help  
 
 

[Hunt]

You need to determine the oxidation state of each element. Oxygen usually has an ox state of -2 ( except in Hydrogen Peroxide H₂O₂ and OF₂ ) while Hydrogen's ox state is +1 with non-metals. The concept of assigning Oxidation states is to keep track of electrons in Redox reactions.

Next, you can determine which substance is being oxidized ( loses electrons ) and which substance is being reduced ( gains electrons ).

You write down the oxidation half reaction and the reduction half reaction, deduce the net ionic equation, and then balance the Oxygens by adding H₂O and you balance hydrogens by adding:

1-H⁺ if the medium is acidic ( as is your case )
2-OH^- if the medium is basic
3-none if the medium is neutral

The best way to understand this is to solve many problems.

Let's consider the rxn you gave :

SO₃^2-  +  MnO₄^-  + H⁺ <---> Mn²⁺ + SO₄^2- + H₂O

In this case, H⁺ is given as well as H₂O. Sometimes, it might be only mentioned that the medium is acidic, and u would conclude that there's an H⁺.

For SO₃^2-:
Ox state of O = -2
Ox state of S = x

x + (-2)(3) = -2 ( total ox state of a molecule = charge of the molecule )

x = ox state of S = +4

Let's see what happens to Sulphur in the products side.

SO₄^2-:
Ox state of O = -2
Ox state of S = x

x + (-2)(4) = -2 ( total ox state of a molecule = charge of the molecule )

x = ox state of S = +6

Therefore, for S, the ox state has increased from +4 to +6. We deduce that Sulphur was oxidized by losing 2 electrons.

The Oxidation half-reaction which takes place at the anode :

For SO₃^2- <---> SO₄^2- + 2e

Next, we see if Mn is being reduced.

MnO₄^-:
Ox state of O = -2
Ox state of Mn = x

x + (-2)(4) = -1
x = +7

In the products' side, for Mn²⁺ :

Ox state of Mn = +2

Obviously, Mn has been reduced by gaining 5 electrons. ( Ox state reduced from +7 to +2 )

We write the Reduction half-rxn which takes place at the Cathode:

MnO₄^- + 5e <---> Mn²⁺

We have thus established the following so far :

Ox : SO₃^2- <---> SO₄^2- + 2e

Red : MnO₄^- + 5e <---> Mn²⁺

Multiply the 1st by 5 and the 2nd by 2 inorder to cancel the electrons when adding both half-reactions.

5 SO₃^2- + 2 MnO₄^- <---> 5 SO₄^2- + 2 Mn²⁺

Now that Mn and S are balanced, all that's left is to balance the Oxygen & Hydrogen atoms.

On the reactants' side, we have 23 O atoms. On the products' side, we have 20 O atoms. Hence, we add 3 H₂O molecules to the right.

5 SO₃^2- + 2 MnO₄^- <---> 5 SO₄^2- + 2 Mn²⁺ + 3 H₂O

Check for Hydrogen:
On the reactants' side, no H atoms
On the products' side, 6 H atoms.
Hence, we add 6 H⁺

Net ionic Equation : 5 SO₃^2- + 2 MnO₄^- + 6 H⁺ <---> 5 SO₄^2- + 2 Mn²⁺ + 3 H₂O

Notice : the equation is not only balanced from an atomic perspective, but also electrically balanced. The charges cancel on both sides, try it out yourself.

[Borek]

peggysue32: NEVER post the same in the two forums. It is against forum rules. Two people did the same work to answer your question - you have wasted 50% of their effort. Topic locked.

Vant_Hoff: check High school chemistry forum, xiankai answered peggysue question about two hours ago. Huge snack for you