[lucky_star]

Problem: Balancing the following red-ox reaction:

MnO4- (aq) + As4O6 (s) --> Mn2+ (aq) + AsO43- (g)


Attempt:
1. Assign O.N.

MnO4- has O.N. of
Mn+4(-2)=-1
Mn=+7

As4O6 has O.N. of
4As+6(-2)=0
4As=+12
As= +3

Mn2+ has O.N. of +2

AsO43- has O.N. of
As+4(-2)=-3
As=8-3
As=+5

2. Writing Half equations:
Reduction (O.N Decreases) Mn+7 to Mn2+ therefore:
MnO4- + 5e => Mn2+

Oxidation (O.N. Increases.) As+3 to As +5 therefore:
As4O6 => AsO43- + 2e

There is no import. atoms in either 2 half equations. So we just need to find the L.C.M. (Least Common Multiple)
5e in the first half eq and 2e in the second half eq => L.C.M = 10

Meaning I need to:
Multiply 1st half eq by 2 And Multiply the 2nd half eq by 5.

Red: 2MnO4- => 2Mn+2
Oxid: 5As4O6 => AsO43-

3. Combine these 2 half eq's:
2MnO4- + 5As4O6 => 2Mn+2 + AsO43-

Then balance the Charge (Not the O.N.)
On the left I have 2(-1) + 0= -2 charge because:
MnO4- has a -1 charge, As4O6 has a zero charge

On the right I have +2+(-3)= -1 b/c:
Mn2+ has a +2 charge, AsO43- has a -3 charge

So the left side has 1 less charge than the one on the right!

4. I was told that it is in acidic solution, so we'll use H+ to balance the charge:
:- We'll add 1H+ to the left side, then add 1/2H2O to the right.

H+ + 2MnO4- + As4O6 => Mn+2 + AsO43- + 1/2H2O

Get rid of the fraction by multiply by 2 on both sides:
2H+ + 4MnO4- + 2As4O6 => 2Mn+2 + 2AsO43- + H2O
Count the number of Oxygen:
Left=28 and right=9

Now it does not make sense to me any more
Can any one show me what I have done incorrectly?
Thank you in advance!

[ARGOS++]

Dear Lucky_star;

The As is never balanced during your RedOx reaction!

Good Luck!
                    ARGOS⁺⁺

[UG]

Hi lucky_star,
Just a note on balancing half-equations:
There are a number of steps, for example if I were to do the reduction of dichromate ion to chromium(III) ion in acidic conditions, I would:
-Identify the reactant and the products formed:

Cr₂O₇^2-  Cr³⁺

-Balance the atoms undergoing a change in oxidation number:

+6             +3     
Cr₂O₇^2-  2Cr³⁺

Only chromium is changing O.N so balance that.

-Balance oxygen atoms by adding water on the appropriate side:

Cr₂O₇^2-  2Cr³⁺ + 7H₂O

-Balance hydrogen atoms by adding hydrogen ions to the appropriate side:

Cr₂O₇^2- + 14H⁺ 2Cr³⁺ + 7H₂O

-Balance the charge by adding electrons to the more positive side:

LHS=+14,-2= +12
RHS=2 x (+3)= +6
So add 6 electrons to the LHS:

Cr₂O₇^2- + 14H⁺ + 6e^- 2Cr³⁺ + 7H₂O

Hope this helps. 
UG

[lucky_star]

Thank you ARGOS++  and UG. Ha, I forgot that part!
+Snack for yall!