Alkaline Dry Cell. (1 Viewer)

[atBondi]

If my textbook has

Anode
Zn(s) + 2OH- (aq) --> ZnO s) + H2O(l) + 2e-
Cathode
MnO2(s) + 2H2O (l) + 2e- --> Mn(OH) 3(s) + OH-(aq)

For the reactions, what would be the overall reaction?
For overall reactions do i simply just add the 2 equations togethor without the elctrons? But i recall having to look at some " spectator" thing, (which is if it stays the same on both equations, i can just leave it out?)

So if i have a go and try, would the overall be...
Zn(s) + 2OH- (aq) + MnO2(s) + 2H2O (l) -->ZnO s) + H2O(l) + Mn(OH) 3(s) + OH-(aq)


Or do the many OH groups are somewhat spectators and i can leave them out?
Thanks.

 

[Pwnage101]

First off, your second equation should be:

MnO2(s) + 2H2O (l) + e- --> Mn(OH) 3(s) + OH-(aq)

[remember, charges must balance]

Now, yes we 'add' the 2 equations, but we need the same amount of electrons in both equations, so we will add equation 1 with 2 x equation 2 which gives:

Zn(s) + 2OH- (aq) + 2MnO2(s) + 4H2O (l) + 2e- ---> ZnO s) + H2O(l) + 2e- + 2Mn(OH) 3(s) + 2OH-(aq)

[NOtice, we have 2e- and 2OH- on both sides, so they cancel]

This gives:

Zn(s) + 2MnO2(s) + 4H2O (l) ---> ZnO s) + H2O(l) + 2Mn(OH) 3(s)

[1 H2O will also cancel from both sides]

So:

Zn(s) + 2MnO2(s) + 3H2O (l) ---> ZnO s) + 2Mn(OH) 3(s)

 

[atBondi]

Ahh ok makes sense. Thanks a bunch.