Balancing Redox Reactions in Acidic Solution, Answer Nos. 6-10

6)

the final answer:

7)

the final answer:

Comment #1: notice that this is no H⁺ in the final answer, but please keep in mind that its presence is necessary for the reaction to proceed. In cases like this, the H⁺ is acting in a catalytic manner; it is used up in one reaction and regenerated in another (in equal amount), consequently it does not appear in the final answer.

Comment #2: this type of a reaction is called a disproportionation. It is often found in redox situations, although not always. An important disproportionation reaction which does not involve redox is 2H₂O ---> H₃O⁺ + OH¯. This reaction is of central importance in aqueous acid-base chemistry.

8)

the final answer:

9) First a bit of discussion before the correct answer. The H₂O on the right side in the problem turns out to be a hint. This is because you need TWO half-reactions. For example, suppose the water wasn't in the equation and you saw this:

O₂ + As ---> HAsO₂

You'd think "Oh, that's easy" and procede to balance it like this:

H⁺ + O₂ + As ---> HAsO₂

Then, you'd balance the charge like this:

e¯ + H⁺ + O₂ + As ---> HAsO₂

And that is wrong because there is an electron in the final answer. You cannot have electrons appear in the final answer of a redox reaction. (You can in a half-reaction, but remember half-reactions do not occur alone, they occur in reduction-oxidation pairs.)

Here are the correct half-reactions:

the final answer:

Notice that the H₂O on the right-hand side of the equation in the list of problems goes away.

By the way, try to balance

O₂ + As ---> HAsO₂

using H₂O on the left rather than H⁺. That way leads to the correct answer without having to use half-reactions. There are some redox reactions where using half-reactions turns out to be "more" work, but there aren't that many.

10) these are the half-reactions:

Only the second half-reaction needs to be multiplied through by a factor:

Adding the two half-reactions, but not eliminating anything:

The final answer: