Balancing Redox Reactions in Basic Solution, Answer Nos. 11-15

11) the two half-reactions, balanced as if in acidic solution:

Pb(OH)₄²¯	--->	PbO₂ + 2H₂O + 2e¯
2e¯ + 2H⁺ + ClO¯	--->	Cl¯ + H₂O

convert to basic solution:

Pb(OH)₄²¯	--->	PbO₂ + 2H₂O + 2e¯ + 2OH¯
2e¯ + 2H₂O + ClO¯	--->	Cl¯ + H₂O

the two H₂O in the second half-reaction comes from 2H⁺ + 2OH¯

note that the electrons are already equal and add (remember to eliminate two waters from each side) to get the final answer:

Pb(OH)₄²¯ + ClO¯

--->

PbO₂ + Cl¯ + 2OH¯ + H₂O

12) the two unbalanced half-reactions:

Tl₂O₃	--->	TlOH
NH₂OH	--->	N₂

balanced, as if in acidic solution and with a factor of two included in the second half-reaction:

4e¯ + 4H⁺ + Tl₂O₃ + H₂O	--->	TlOH
2 [2NH₂OH	--->	N₂ + 2H₂O + 2H⁺ + 2e¯]

There is no need to convert to base. The fact that there are 4H⁺ on each side indicates this. The final answer:

Tl₂O₃ + 4NH₂OH

--->

2TlOH + 2N₂ + 5H₂O

Notice that no hydroxide ions (in solution) appear in the final answer. This is a base-catalyzed reaction. For the reaction to occur, the solution must be basic and hydroxide ion IS consumed as the products are made. Hydroxide ion is regenerated later in the reaction and in the exact same amount as it was consumed, so it cancels out in the final answer.

By the way, the products are not considered to contain hydroxide ions in solution. Both products are insoluble and so would form as solid preciptates.

13) The two unbalanced half-reactions:

Fe(OH)₂	--->	Fe₂O₃
CrO₄²¯	--->	Cr(OH)₄¯

balance, as if in acidic solution:

3 [2Fe(OH)₂	--->	Fe₂O₃ + H₂O + 2H⁺ + 2e¯]
2 [3e¯ + 4H⁺ + CrO₄²¯	--->	Cr(OH)₄¯]

Note that, in the first half-reaction, both the water and the hydrogen ion are on the same side. This happens occasionally; the more usual case is that they are on opposite sides. Note also that I included the factors necessry to make the electrons equal before adding the two half-reactions together.

half-reactions added together, still in acidic solution:

6Fe(OH)₂ + 8H⁺ + 2CrO₄²¯

--->

3Fe₂O₃ + 3H₂O + 6H⁺ + 2Cr(OH)₄¯

add eight hydroxide ions to each side to change from acidic to basic solution. Note that eight waters were also eliminated from each side:

6Fe(OH)₂ + 2CrO₄²¯

--->

3Fe₂O₃ + H₂O + 2Cr(OH)₄¯ + 2OH¯

14)

2e¯ + Cl₂ --->	2Cl¯
H₂O + Cr(OH)₃ --->	CrO₄²¯ + 5H⁺ + 3e¯

make electrons equal:

3 [2e¯ + Cl₂ --->	2Cl¯]
2 [H₂O + Cr(OH)₃ --->	CrO₄²¯ + 5H⁺ + 3e¯]

add the two half-reactions:

2 H₂O + 2 Cr(OH)₃ + 3 Cl₂

--->

2 CrO₄²¯ + 6 Cl¯ + 10 H⁺

add 10 hydroxides to each side, then eliminate two waters for the final answer:

10 OH¯ + 2 Cr(OH)₃ + 3 Cl₂

--->

2 CrO₄²¯ + 6 Cl¯ + 8 H₂O

15)

6e¯ + 6H⁺ + IO₃¯ ---> I¯ + 3H₂O

H₂O + CN¯ ---> CNO¯ + 2H⁺ + 2e¯

multiply second half-reaction by 3 and add together:

6H⁺ + IO₃¯ + 3H₂O + 3CN¯ ---> I¯ + 3H₂O + 3CNO¯ + 6H⁺

note that there is no need to change this reaction from the "fake acid" format. Simply remove the 6H⁺ and the 3H₂O from each side:

IO₃¯ + 3CN¯ ---> I¯ + 3CNO¯

Note that there are no hydroxides appearing in the final answer. This means that the reaction produced as much hydroxide as it used up, leaving no net gain or loss of hydroxide during the reaction. Therefore, the hydroxide was a catalyst for the reaction.