[aduke]

Would someone please confirm my answers for the following 2 reaction:

1.  Pb + PbO2 + H2SO4 --> PbSO4
2.  Cl2 --> Cl(-) + OCl(-)

My answers:

1.  Pb + PbO2 + 2 H2SO4 --> 2 PbSO4 + 2 H2O
2.  2 OH(-) + Cl2 --> OCl(-) + Cl(-) + H2O

Also, I have no clue what to do with the following equation:

Mn(2+) + NaBiO3 --> Bi(3+) + MnO4(-)

I know the Mn is oxidized and the Bi is reduced, but what happens to the Na in the half-reactions.

Just one more:

CN(-) + MnO4(-) --> CNO(-) + MnO2

I have the CN(-) as being reduced by 2 e(-) and the Mn as being oxidized by 3 e(-).  I have gotten the half-reactions to the balancing of H's and O's step where I have:

2 OH(-) + MnO4(-) --> MnO2 + 3 e(-)
2 e(-) + CN(-) --> CNO(-) + 2 OH(-)

I don't see a way to balance the H's and O's and therefore believe there is a problem earlier in the problem, but what is it.  Thanks in advance for all *delete me*

[sdekivit]

Mn(2+) + NaBiO3 --> Bi(3+) + MnO4(-)

--> make your halfreaction by going out from the following:

Mn(2+) --> MnO4(-) and NaBiO3 --> Bi(3+) + Na(+) (remember that solving a salt will result in the ions of that salt)

CN(-) + MnO4(-) --> CNO(-) + MnO2

the halfreaction from MnO4(-) can be fpound in a table --> basic environment:

CN(-) --> CNO(-)

CN(-) + H2O --> CNO(-) + 2 H(+)

CN(-) + 2 OH(-) --> CNO(-) + H2O + 2 e(-)

[aduke]

For my first problem with NaBiO3, i got this answer:

7 H2O + 2 Mn(2+) + 5 NaBiO3 --> 2 MnO4(-) + 5 Bi(3+) + 5 Na(+) + 14 OH(-)

Can someone confirm that?

On the second problem, CN(-) is being reduced, meaning the 2 electrons in the half-reaction should be a reactant.  What is wrong with this?

2 e(-) + CN(-) --> CNO(-)
2 e(-) + 2 H(+) + CN(-) --> CNO(-) + H2O
2 e(-) + 2 H2O + CN(-) --> CNO(-) + H2O + 2 OH(-)
2 e(-) + H2O + CN(-) --> CNO(-) + 2 OH(-)

Also, where can I find the table showing the MnO4(-) half-reaction.  I was able to balance the MnO4 half reaction for elements, but not charge.  I have (-) on the reactant side and (7-) on the product side:

MnO4(-) --> MnO2 + 3 e(-)
4 H(+) + MnO4(-) --> MnO2 + 3 e(-) + 2 H2O
4 H2O + MnO4(-) --> MnO2 + 3 e(-) + 2 H2O + 4 OH(-)
2 H2O + MnO4(-) --> MnO2 + 3 e(-) + 4 OH(-)

What is wrong with this?  Thanks for all help.

[Donaldson Tan]

2 e(-) + CN(-) --> CNO(-)
2 e(-) + 2 H(+) + CN(-) --> CNO(-) + H2O
2 e(-) + 2 H2O + CN(-) --> CNO(-) + H2O + 2 OH(-)
2 e(-) + H2O + CN(-) --> CNO(-) + 2 OH(-)
MnO4(-) --> MnO2 + 3 e(-)
4 H(+) + MnO4(-) --> MnO2 + 3 e(-) + 2 H2O
4 H2O + MnO4(-) --> MnO2 + 3 e(-) + 2 H2O + 4 OH(-)
2 H2O + MnO4(-) --> MnO2 + 3 e(-) + 4 OH(-)

They are all wrong. You failed to understand the concept behind half-equations.

i show you an example question to illustrate how to do.
Balance MnO₄^- + Fe²⁺ => Mn²⁺ + Fe³⁺

Do not forget that electron carries the charge of -1.

We first identify: Fe²⁺ => Fe³⁺
For Iron(II) to be oxidised to Iron(III), it must loose 1 electron.
Fe²⁺ => Fe³⁺ + e^-

Next, we have to balance MnO₄^-  => Mn²⁺

since we are dealing with an aqeous redox system, it is valid to assume H+ or OH- or  water molecules might participate in the reaction.

we observe that the reduced form of manganese has no oxygen, so H+ must have participated in the half-equation and convert the oxygen to water.
MnO₄^- + 8H⁺ => Mn²⁺ + 4H₂O

the overall charge on the LHS is 7+ and the overall charge on the RHS is 2+. This means the LHS somehow must acquire -5 charge in order for the equation to balance. This means addition of 5 electrons to the LHS. Remember that reduction is also defined as gaining electrons.
MnO₄^- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O

Now, the manganese half-equation is finally balanced, charge-wise and material-wise.

(1) MnO₄^- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O
(2) Fe²⁺ => Fe³⁺ + e^-

We seek to combine both equations. Redox reactions involve the transfer of electrons. This means the number of electron donated by the oxidation half-equation must equal the number of electron accepted by the reduction half-equation.

Equation (1) involves 5 electrons, but equation (2) involves 1 electron. Since the lowest common multiple between 1 and 5 is 5, then we must scale up the equation (2) 5 times.

Finally, we combine Equation (1) + 5 X Equation (2)
ie. LHS of Equation (1) + 5 X LHS of Equation (2) => RHS of Equation (1) + 5 X RHS of Equation (2)
MnO₄^- + 8H⁺ + 5e^-  + 5Fe²⁺  => Mn²⁺ + 4H₂O +  5Fe³⁺ +5 e^-

Next, eliminate all repeated terms on the LHS and RHS, so we yield the final equation:
MnO₄^- + 8H⁺ + 5Fe²⁺  => Mn²⁺ + 4H₂O +  5Fe³⁺

I hope you understood my workings.

[AWK]

For my first problem with NaBiO3, i got this answer:

7 H2O + 2 Mn(2+) + 5 NaBiO3 --> 2 MnO4(-) + 5 Bi(3+) + 5 Na(+) + 14 OH(-)

Can someone confirm that?

This reaction goes only in acidic medium!

[aduke]

Sorry to everyone who has helped me despite my incompetance on this problem.  I got to school today and talked to a classmate in first period who showed me the problem with my work.  I had been working on redox reactions for about 3 hrs Saturday and another 3 hrs on Sunday, so I was getting tired and rushed through this problem.  I misread the book and it is, as AWK pointed out, supposed to be in an acid.  sdekivit showed me exactly how to fix the problem, except for the basic part, and I pretty much was too convinced that I was correct to think it through.  I now feel I have a good fundamental knowledge of redox reactions.  Thanks to all those who replied despite my refusal to listen.  It is great to know that I have somewhere to turn when I have no clue where to start on chemistry problems.  Thanks!