Fun......redox problem?

Question

Can you balance these 3 redox equations for me? Since I have no way to type the charges I have put them in parentheses.

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 because I am a beginner chemistry student. I am a step away from having if completely click for me. THANKS!

Actually, chemistry problems are kind of fun once you get a feel for them. I like coming on here and posting a question or problem and then checking my solutions against other peoples' solutions. It gets me feedback I dont get from a professor. Cuz I work alone. Thanks.

Answer 1

Looking at the 1st one, there are 3 oxygen atoms on the left that aren't present on the right. Balance the right hand side by adding 3 water molecules so the reaction becomes:
Sn++ + IO3- = Sn++++ + I- + 3H2O
Ooops. Now there are 6 hydrogen atoms on the right, which aren't on the left. So add 6 protons to the left hand side, so the reaction becomes:
Sn++ + IO3- + 6H+ = Sn++++ + I- + 3H2O
OK so now the atoms balance, but the charges don't. On the left you now have 2 - 1 + 6 = 7 and on the right you have 4 - 1 = 3. So now you need to add 7 - 3 = 4 electrons to the left hand side so the charge on both sides = 3, so the reaction becomes:
Sn++ + IO3- + 6H+ + 4e- = Sn++++ + I- + 3H2O

Same applies for the second one:
NO3- + H2S = NO + S
Add water for O, becomes:
NO3- + H2S = NO + S + 2H2O
Add H+ for H, becomes:
NO3- + H2S + 2H+ = NO + S + 2H2O
Balance the charges with e-, becomes:
NO3- + H2S + 2H+ + e- = NO + S + 2H2O

And the third one:
Mn++ + BiO3- = MnO4- + Bi+++
Add H2O to balance O:
Mn++ + BiO3- + H2O = MnO4- + Bi+++
Add H+ to balance H:
Mn++ + BiO3- + H2O = MnO4- + Bi+++ + 2H+
Add e- to balance charges:
Mn++ + BiO3- + H2O = MnO4- + Bi+++ + 2H+ + 3e-

Answer 2

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 3

Problems will never be fun