Question

What is the standard reduction potential value for Mn(OH)2--->Mn?

Answer 1

Look up in a table? Or use Nernst equation to calculate it?

Answer 2

I calculated it from a Frost Diagram. I got -1.6. A table I found on the internet gives it as +1.56. I am confused. O.o

Answer 3

Do sound a bit weird.

Answer 4

The top is the one I found. The bottom one is the frost diagram I used to calculate it.

Answer 5

Maybe it's just a typo. O.o

Answer 6

haha. Now I see it.. what you did it correct.

Answer 7

Ok. Thank you. I just have one other question? If you got time?

Answer 8

Shoot.

Answer 9

To calculate the reduction potential of the couple MnO2/Mn, I would use the formula:
E= n1E1+n2E2/ n1+n2
Right? So I can substitute the values as \[{ (0.34\times5)+(-1,6\times2) }{ 5+2 }\]

Answer 10

Oops. That's divided by (5+2).

Answer 11

0.34 is the reduction potential of MnO4---->Mn(OH)2
-.16 is the reduction potential of Mn(OH)2---->Mn

Answer 12

That's -1.6.

Answer 13

You don't need help... you simply rock at it.

Answer 14

Looks right to me and nice formulated as well.

Answer 15

Actually, the question says: Reduction potential of MnO2/Mn in acidic solution equals 0.02 V. Find the reduction potential in the basic solution.
When I use the same formula for acidic solution, I don't get 0.2V. :(

Answer 16

*I don't get 0.02V.

Answer 17

Well as you say we should simply use the following theorem:
For two redox-processes with belong standard reduction potential:
\[\large A + z _{AB} \to B ~ ~ ~ ~ ~ ~ ~ E _{AB}^{\Theta}\]
\[\large B+ z _{BC} \to C ~ ~ ~ ~ ~ ~ E _{BC}^{\Theta}\]
it then follows:
\[\large z _{AB} E _{AB}^{\Theta} + z _{BC} E _{BC}^{\Theta} = \left( z _{AB}+z _{BC} \right)E _{AC}^{\Theta}\]
manganese(VI) oxide or is it Manganese(IV) oxide? you wrote different things in your outline

Answer 18

not redox but reduction processes* my bad.

Answer 19

It's Manganese (IV) oxide. \[MnO _{2}\]

Answer 20

The rule is that two find the reduction potential of nonadjacent species, you find it by combination of two other couples. I think I found my mistake.

Answer 21

I should be calculating the reduction potential from \[MnO _{2} \rightarrow Mn(OH)_{2}\]
and then \[Mn(OH)_{2} \rightarrow Mn\]

Answer 22

for comparison, that is.

Answer 23

I was going from \[MnO _{4}^{-} \rightarrow MnO _{2}\] earlier.

Answer 24

Sorry. *\[ MnO _{4}^{-}\rightarrow Mn(OH)_{2}\]

Answer 25

I get your drift :)

Answer 26

So would that be correct?

Answer 27

It follows the form of the theorem I set up so I suppose it should.

Answer 28

Okay. Thank you so much for all help! ^_^

Answer 29

No problem. However I got to say this was a pleasure... when people find out their problem them self it is just awesome looking at!

Answer 30

:)