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### Topic: Calculating "n" in a battery  (Read 3168 times)

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#### zmasterflex

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##### Calculating "n" in a battery
« on: February 25, 2014, 03:20:29 PM »
Greetings all. How do I know what the value of "n" is in these reactions? (When I say "n" I'm referring to delta G= -nFe)

Zn(s) + 2NH4Cl(s) + 2MnO2 --> Zn(NH3)2Cl2(s) + Mn2O3(s) + H2O(l)

MH(s) + NiO(OH)(s) --> M(s) + 2Ni(OH)2(s)

Does it have something to do with oxidation numbers? Thanks

#### Corribus

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##### Re: Calculating "n" in a battery
« Reply #1 on: February 25, 2014, 03:41:56 PM »
n is the number of electrons transferred from the reductant to the oxidant during the course of the redox reaction.
The easiest way to determine n in most cases is to look at how the oxidation numbers change from one side of the reaction to the other, but keep in mind your stoichiometry always. For instance, in your first equation, what is the charge on zinc as a reactant and as a product? Likewise for "M" in your second reaction. (Your second reaction isn't balanced, by the way, so this will make things difficult for you.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### zmasterflex

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##### Re: Calculating "n" in a battery
« Reply #2 on: February 25, 2014, 05:02:20 PM »
n is the number of electrons transferred from the reductant to the oxidant during the course of the redox reaction.
The easiest way to determine n in most cases is to look at how the oxidation numbers change from one side of the reaction to the other, but keep in mind your stoichiometry always. For instance, in your first equation, what is the charge on zinc as a reactant and as a product? Likewise for "M" in your second reaction. (Your second reaction isn't balanced, by the way, so this will make things difficult for you.)

Thanks for the response. In the first reaction Zn as a reactant has a "zero" for oxidation number. In the product Zn(NH3)2Cl2 we "assume" (or I assume) that (NH3)2 = 0 and Cl2= -2 so Zn = +2 and we have a transfer of 2 electrons. However, how do I correlate that with the other part of the reaction? Regarding the reactant MnO2 the Mn= +4. On the product side Mn2O3, Mn2= +6 so Mn = +3?? That's not consistent. I have a similar problem with the second reaction but whatever answer that I get for the first one will probably answer both (The product in the second reaction isn't 2Ni(OH)2(s) it's Ni(OH)2(s).. typo). Thanks

#### Corribus

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##### Re: Calculating "n" in a battery
« Reply #3 on: February 25, 2014, 05:28:07 PM »
If your equation is balanced properly it shouldn't matter what the other part of the reaction is. Electrons donated by the reductant have to equal the electrons gained by the oxidant. And this is the case in both of your equations. In the Zn equation, you are going from Zn0 to Zn2+, and you are taking TWO Mn's from +4 to +3. This is why I said stoichiometry is important.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman