Question about cathode half-equation of dry cells? (1 Viewer)

[MysteryMoon]

I've been looking through textbooks, school worksheets and tutoring notes, and their cathode half equations are slightly different to one another. They have something like these:

2MnO2(s) + 2NH4+(aq) + 2e- --> Mn2O3(s) + H2O(l) +2NH3(aq)

MnO2(s) + NH4(aq) + H2O(l) + e- --> Mn(OH)3(s) + NH3(aq) <-- there's like a H2O on the other side and an OH on the opposite

Do they both mean the same thing or are they referring to different dry cells?

 

[dan964]

Use the first one. It is slightly clearer. (Although you could remove ammonia on both sides and just write H+ on the LHS, check with teacher though)
The first one Mn goes from oxidation state +4 to +3 which means it is being reduced.
The second one Mn goes from oxidation state +4 to +3 which is reduced.
So it really makes no difference.

 

[dan964]

when you double the second you note that
it is just a different end product, because Mn2O3 reacts with 3H2O to produce 2Mn(OH)3. The question is there enough p.d. for this to occur.

 

[MysteryMoon]

Ok thanks! Makes everything so clear now.