[Marcos Castillo]

Hello
I've got a paragraph, and a question about it. I write it down and then my attempts to solve them:
"The standard alkaline battery uses Zn and MnO₂. KOH separates them. OH^- ions react with Zn electrode to produce Zn(OH)₂ and free electrons. These electrons move trough a electrical circuit to return to the MnO₂ electrode, where they react, to produce MnO₃ and OH^- ions. Both these quemical reactions end up producing a difference in potential voltage about 1,5 V between both electrodes."
My attempt to translate this into quemichal formulas:
Zn+2OH^- Zn(OH)₂+2e^-
MnO₂+2e^- MnO₃+OH^-
How do I balance second reaction, MnO₂+2e^- MnO₃+OH^-?
Thanks

[chenbeier]

It is not MnO₃, it will be Mn³⁺

The reaction is MnO₂ + H₂O + e^- => MnOOH + OH^-

[Marcos Castillo]

Hello chenbeier

It is not MnO₃, it will be Mn³⁺

The reaction is MnO₂ + H₂O + e^- => MnOOH + OH^-

It is not needed an equilibrium in electrons?. I mean that 2 electrons are released, and only one needed:

Zn+ 2OH^-=>ZnO+H₂O+e^-

MnO₂ + H₂O + e^-=> MnOOH + OH^-

Which is the role played by KOH? Where does it takes part?

[chenbeier]

Of course you need balance the equation, in his way that you can eliminate the electrons.

OH- is needed to complex the zinc.

[Marcos Castillo]

I think it's only needed to multiply per two the reduction reaction of the manganese, and then sum both semireactions:

1-Zn+2OH^-=>Zn(OH)₂+2e^-
  MnO₂+H₂O+e^-=>MnOOH+OH^-

2-Zn+2OH^-=>Zn(OH)₂+2e^-
  2MnO₂+2H₂O+2e^-=>2MnOOH+2OH^-

3-Zn+2OH^-=>Zn(OH)₂+2e^-
  2MnO₂+2H₂O+2e^-=>2MnOOH+2OH^-

4-Zn+2MnO₂+2H₂O=>Zn(OH)₂+2MnOOH
 

[chenbeier]

Yes correct. Zink stays as a complex Zn(OH)₄]^2- normally.

[Marcos Castillo]

Thanks!

[AWK]

Yes correct. Zink stays as a complex [Zn(OH)₄]^2- normally.

A bit incorrect, but you can guess this printing error.