NOT fun.....redox reaction balancing....bad bad bad stuff!!!?

Question

Thought I would try a different approach to getting at least TWO units of feedback for my question.
Can anyone balance these 3 redox equations for me? Since I have no way to type the charges I have put them in parentheses. Hey if you can balance these I will consider you a genius of the highest level.

Sn (2+) + IO3 (1-) --------> Sn (4+) + I (1-)

NO3 (1-) + H2S -----------> NO + S

Mn (2+) + BiO3 (1-) --------> MnO4 (1-) + Bi (3+)

I need to see if I am on the right track, but I am kind of having problems with balancing the charges. THANKS!

Answer 1

You can see my answers in your first try!

Anyway I put it again here:

First reaction:

Sn(2+) + (IO3)(1-) -----> Sn(4+) + I(1-)

NOTE: Oxidation number of I in IO3 is I(5+)

We could separate in two semirreactions:

Sn(2+)...........------> Sn(4+) + 2e-
I(5+) + 6e-......------> I(1-)

We do have to get the same number of electrons in both sides, so:

3 Sn(2+)........------> 3 Sn(4+) + 6e-
I(5+) + 6e-.....------> I(1-)

We add both semirreactions to cancel electrons:

3 Sn(2+) + I(5+) -------> 3 Sn(4+) + I(1-)

Finally , your balanced equation is:

3 Sn(2+) + IO3(1-) ------> 3 Sn(4+) + I(1-)

That´s it!

Second equation:

NO3(1-) + H2S --------> NO + S

Applying the same method above the result is:

2 NO3(1-) + 3H2S -----> 2NO + 3S


Third equation:

Mn (2+) + BiO3 (1-) --------> MnO4 (1-) + Bi (3+)

The balanced equation will be:

2Mn (2+) + 5BiO3 (1-) --------> 2MnO4 (1-) + 5Bi (3+)

Good luck!

Answer 2

3Sn(2+) + IO3(1-) +6H(+) -> 3Sn(4+) + I(1-) + 3H2O

2NO3(1-) + 3H2S +2H(+) -> 2NO + 3S + 4H2O

2Mn(2+) + 5BiO3(-1) +14H(+) -> 2MnO4(-) + 5Bi(3+) + 7H2O