Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Marcos Castillo on April 23, 2020, 12:34:00 PM
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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- :rarrow:Zn(OH)2+2e-
MnO2+2e- :rarrow:MnO3+OH-
How do I balance second reaction, MnO2+2e- :rarrow:MnO3+OH-?
Thanks
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It is not MnO3, it will be Mn3+
The reaction is MnO2 + H2O + e- => MnOOH + OH-
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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?
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Of course you need balance the equation, in his way that you can eliminate the electrons.
OH- is needed to complex the zinc.
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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
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Yes correct. Zink stays as a complex Zn(OH)4]2- normally.
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Thanks!
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Yes correct. Zink stays as a complex [Zn(OH)4]2- normally.
A bit incorrect, but you can guess this printing error.