Redox in basic solution

nautica

Balance in basic (check my work)

NO + MnO4- ------ NO3- + MnO2

4OH- + 4H+ + NO ---- NO3- + 2H2O + 4OH-
4OH- + 4H+ + MnO4- --- MnO2 + 2H2O + 4OH-
Balanced with 5 e- on each side

2H2O + NO + MnO4- ----- NO3- + MnO2 + 8OH-

Thanks
Nautica

chem_tr

You need help, my friend

First, prepare atomic-based electron balance:

[tex]N^{2+} \longrightarrow N^{5+} + 3e^-[/tex]
[tex]Mn^{7+}+5e^- \longrightarrow Mn^{2+}[/tex]

Then balance them.

[tex]5N^{2+}+3Mn^{7+}\longrightarrow 5N^{5+}+3Mn^{2+}[/tex]

You'll see that both electron counts and atom counts are balanced now.

When you write the "real" ions, I mean, NO and MnO₄^-, you'll have to add OH^- and H₂O to the side with less oxygen and less hydrogen, etc.

nautica

Cool, this is what I got

5NO + 3MNO4 = 5NO3 + 3MnO2

It appears that all are balanced. Does this look right.
Thanks
nautica

nautica

Another class mate said that the original reaction was already balanced. He said he worked through it and kept coming up with the original equation.

Could this be right??? Or does mine look fine???

Thanks
Nautica

chem_tr

Well, it seems that atom counts are okay, but you've forgotten ionic counts. This is why redox reactions are hard to study.

You said that this redox should be in basic solution; so just put OH^- ions to the left side. However, there is a serious error; in basic solutions, permanganate is only reduced to manganese dioxide, as you correctly wrote. So the coefficients cancel to reflect this:

[tex]N^{2+}+Mn^{7+}\longrightarrow N^{5+}+Mn^{4+}[/tex]

So if we write its real forms, we'll get this one:

[tex]NO + KMnO_4 \longrightarrow KNO_3 + MnO_2 [/tex]